.-q.F"-
---
. r+j:-ri:Y!4S;F*;;+-.
;""8F;:;-
::
C{ONTEI!}ISI
'
CIIAPIER ONI
SYSTEUS OF LTNEAR EgUArIOIvS
Pagles
Articles
Llst of books mltten bY the author
Professor Md. Abdur Rahaman'
1. College Linear Algebra
2. College Modern Algebra
3. College Higher Algebra
CHAPTER, fiTO
DEf,ERMINANTS
of Mathematics)
@asiI agJura & Fundamentals
one)
Methods
[Volume
frathematical
4. b-oU.g"
(Spec."ial Functions & Vector Analysis)
two)
5. boU"g" Mathematical Methods [Volume
Problems)
Value
(Integral Transforms & Boundary
6. +rqq"EEE< aq"rfalg
",,
g*W'm* lffitH HHH##P*'*")
Algebra
8- Ideal Solution of Cotlege Linear
rTfl{rq
;
rrrrsi Ers< Aq{Flrc< c"t{
I
1.1 Introduction to systems of linear equations
4
equations
linear
1.2 Degenerate and non-degenerate
1.3 SoliUon of a non_homogeneous system of linear equtions204
L.4 Solution of a system of homogeneous linear equations
ETERCISES .I
t
ib. E-;"F* qilfie e +imi-fiq< qm{'Tflttq
2.1 Introduction
2.2 Deflnition of determinant
2.3 Sarms diagrams for determinants
2.3. f Sarms diagram for determinant of order 2
2.g.2 Sarms diagram for determinant of order 3
2.4 Minors and cofactors
determinants
2.5 Fundamental properties of (kplace's
expansion)
determinant
2.6 Expansion of
equations
linear
to
2.7' Application of determinant
(Crammer's rule)
Multiplication of two determinants of finite order
2.8
2.8.r Multiplication of two determinants of the same order
2.8.2 Multiplication of two determinants of different orders
2.8.3 MultiplicationtheQrgm.:' .
EtrEEQIqpp - (Al
'
2.9 AdJointdeterminant
2.lO Inverse or reciprocal determinant
2.ll Symmetric and skew-symmetrlc determinants
2.1 1. I Froperties of symmetric determinant
2.11.2 Ortho-symmetric determinant
2. 1 1,3 Properties of skew-symmetric determinant
2.12 Skew-determinant
2.13 Differentiation of a deterlnina$t " . ,'
EXERCIIIETT - 2 EI
4L
4l
42
42
42
43
43
44
45
46
47
48
49
i
,d
i
1
7t
79
79
82
83
84
85
85
'"4
..:.1
,til
't,
'1
'i't
t,
9l
J
CIIAPTER THREE
MATRIX AI.GEBRA
**;;
Artlcles :
:
Introduction
3. f
.o
r!)
3.2 Definition of matrix
5:5 aJaiiiot "rra-"".l"t multiplication of matrices 95
5:; M;irit multiPlication
39
,
3.5 TransPose of a matrix
97
matrix
a
of
(or-conjugate)
3:6 Compiex con3rrgate of a coriplex matrix
98
g'.; c""jig"t" tti.rEfo*i
'matrices wlth-examples
,98
5,8 Sp.tii types of
1O3
3.9 Theorems o" ii"""pose.matrix'
104
a
matrix
of
conjugate
**pi""
3.10 Theorems o,
a
of
3.I1 Theorems on the 6on3ugat-e tlanspose
106
comPlex matrix
matrices
skew-symmetric
and
!O7
3.12 Theorems on syrm.ttic
3.13 Theorems o"'li"i*iuuo and skew-Hermitian matrices lto
113
3.I4 Theorems on idempotent matrices
ll4
3.15 Singular urrJ ttott-"ingular matrices
115
3.16 Inverse matrix
1'6
3.17 Adjoint or adjugate matrix
of a square matrix ::'5. iA Pr6cess of maiirg the inverse
r/
3.19 Theorems on inverse matrix
I19
3.2O Theorems on orthogonal matrices
l2O
g.2l Theorems on Unitaiy matrices
122
3.22 Theorems on i,,"ol'iorY matrix
122
matrices
applying
by
ilrg sor.:uon or ii.r.* .qr.jtio.r"
r42
ETE,RCISES - 3
I'
4.L
4.2
4.3
4.4
4.5
4.6
4.7
CHAPTER FOITR
RANT OF A MATNtrT:
153
Necessary definitions applied itt tTF
154
Trace of i matrx and properties of trace
r58
Deflnitton of rank of a matrix
162
form
normal
the
to
n"J"-"Uo" of a matrix
167
law
Svlvester's
t67
C6"aiu"" f"r consistency of a System of linear equations
r82
Sweep out method and its applications
urDRcrsDs - 4
184
CHAPTER FIVE
VECTORS IN IRN AND CN
5.1
5.2
5,3
5.4
5.5
&6
5.7
5.8
5.9.
5.10
5.11
5.L2
5.13
5.14
.:jP'
Pages
195
195
Artlcles
Introduction to vectors
E;iid;r, n-space IRn
Vector addition andscalar multiplication.in
in IR' under
;;";;r;;oerties of the vectors
multiplication
scalar
,'
;;;;';deition and
lro veetors in IR;;;; scatar produet ofIRI
and horm in
Oi"t^t
- - -*n
""
of dot product in tRS;src properties
CauctrY- S-ctr*ar1 ineq-UafitY
Minkowski' s -inequatilY'
Curves in q{^
iit in IRn
--B
"" vectors and f, ,f, li' notationt T*'*
Spatial
iro"" Product of two vectors in IR"
Qpmplex numbers
Vectors in C"
200
20t
202
203
203
204
2A4
205
210 :;
i
':,.
,,
Binary operau-ffifriforry1ositior)- on a 9et ''
O"ntiti.i, of group with o<ainplEs' ,
Definition of ring with examPles'
Definition of ffeld with examples
.
6.4
6.5 Field properties 9f real nqmbers.
6.6 Definltion of a vector Spaie (or iineat' siiace)
6.7 ExamPles of vector sPace
6.8 SubsPaces of a veitor sPace'
of subsPaces
6.9
- -WarrrPles
t
-Odd w.", ""*a'"ett"" 't ofr
^
/
r-^
^f
^..I^^*aaac
-.1-
1s6
196
198
198
199
EIGRCISES-5"''
rmcIgR SPACT,S
6.1
6.2
6.3
If
213
?13
2t+
2t5
215
2r6
217
224
225
233
238
244
,t4s
.--.!l
i6*
l
CTIAPTER ONE
SYSTEMS OF LTNEAR EguATroNS
l.llntrodtrctiontosystemsoflirrearoqrratlons
as Ilneer Alglebra
Matrlx theory l{i rnore generally knovm
systems of linear
which is orl$i,atecl i, .he st.udy of
(or variatrles) and in the
equatlons in se'reral rrnhnowns
for their solutions' An
attempt to find general methocls
(or unkno$rns) is llnear if
equation in two or'rnore variables
is' if it
of s;c'r:ond degree or greater' that
it contains no terrns
the
<lf the variablcs or roclts of
contalns no protlucts or po-lvers
to the ii.sr power- and do
variables. Ail variables occur oniy
logarithmic' or
not appear as arglrrnents for tri$onometric'
exponent"ial f'unctions'
be
can
A straight line in the cartesion ly-plane
of the form
represented al$ebratcally by an equation
constants (or nunrbcrs) and
ax+ by = c where a' b and c are real
kind is called a
x and y are variahles' An equation of this y' Sirnilariy'
'the variables x autl
linear equation in
in three variables
ax+by +c:t,+d=oisalinear equatlon
in three dimensional
x, y and z which represents a plane
if lt is of the
In general, an equation is called llnear
$pace.
inrrna-rx, +a2x2+..' "' *o., lh=b(l)whereavaz"' "' '&''
which are to be determined'
lfi:=o'then{lJiscalleclahornogeneotls'linearequatlon
llnear
and if b * o then {1) is calied a non-homogeneout
equation.
ExarnPles of linear equations'
(i) Y*mx = o which is a homogienefius linear equatiort
through the
representing the sllaight line passing
origin.
linear Algebra-l
"
:
COLLEGE LINEAR AT,GEBRA
SYSTEMS OF LINEAII
.i
(ll) 2x + 3y = 5 which is a non-homogeneous linear
Ca6€ I
GLee tr
eiluation'iepre.denting a strai$ht line not passing
through,the origin.
case m
(llU ,i,+ 2y + 5z = 2O which is a non-homogeneous linear
r eeuation representing a Plane.
(iv) \ - k - ax,-2xu= 3 (Non-homogeneous)
(v) xt + k+ ...'.. + r; = I (Non-homogeneous)'
EOUATIONS
3
No solution if the lines are parallel
l'rccisely one solution if they intcrsect
Inlinitely nlanv s;olt'ttions if they coincide'
These cases are illrrstrat.ed bY the following exarnples
:
Exampl* l. The linctri si:i1eln i I I ; 3) nt" no solution'
are
since the lit:es representccl by tbese two linear equations
Examples of non-Ilnear equations
(i) 2fi +3Y=1
parallel.
(ii) x- xy = 2
(iiil * nY'+4x+4=4
(M a-# +2hry + bY2 =o
M 6l + 13.rY+ 6f -5x-5Y+ I =o
v
(o,2)
Equatlon
(t) represents a Parabola'
(it) represents a hYPerbola,
(tii) represents a circle,
(iv) represents a pair of straight lines passing through
the orlgin and
(v) represents a pair of straight lines not passing
;
tlrrough the origin.
I.et'IR be the set of real numbers. 'I'hen a sotutlon of the
linear equation alxi + a2x2 + ... ... + g',-rh = b is any n-tuple
(o.d2,
on) of elements of IR such that the equation is
saUsliea when we substitute x, = ar, h, = 02, ,,. "' ' .rn = s,r' The
set of all such solutions of this linear equation is called the
solution set.
Now we conslder the following two iinear equations :
4r x+-brY = crl /
arx+bry=czl
If we interpret x, y as coordinates in the 'qr-plane, then
cmir of the al:ove two linear equations represents a straight
lgme and {o,g) is a solution of the above two equations if and
orrly if flne point P with coordinates c[,P lies on both lines.
Hence there are three possible cases :
i.
Example 2. The linear system |i'r==tr) has only one
solution, since the lines represented by these two equations
intersect aL(2, l).
';Li.{{
-I
COLLEGE LINEAR ALGEBRA
SYSTEMS OF LINEAR
i
3r rxr + anh, + "' "'
Exarnple 3.'Ihe linear svstem ;:l;r'=u) n"" infinitely
rn.rny sloltrtiotrs, sincc the lirres regrresented by these
ecluatiorrs coincide.
^r,"i
+
EOUATIONS
dt,''\' = P'''l
* oi+ * "' "' +'l:..o'..] o'l
,r,
J
ar,iXI + Ar..,2xz+ "' "' + z1-"''r;=btn
#;t; thecoeflicients au' i = l'2' "" n
j = 1,2, ...,
are real
the free terms bt' i = l'2' "':m
If the b1 are all
the set of real nurnbers'
m'
f'o*
taken
numbers
sJrstem' If at
(1) is called a homogeneous
systenr
the
zero, then
a non'
then the system (1) is called
least one bl is not zerc!'
crn is
of numbers crl' 0b' ""
o
homogeneor" "y"tlm' ""ot"""e
given by (1) if
of linear equations
system
the
of
called solutlon
equation in
cln ls a solution of every
=
(t1,
;
"'
x1 =
h=%, '
the system (1)'
(1) is a vector x.whose components
A solution vector of
it has at
of (1)' If (1) is homogeneous'
of the variables and
(o,3)
(3,o)
*-*-*4 x
'..
-3
Y-\
'x-.
:;. .\
-6
{
I
L.2 $egenerate and non-degienerate Hnear eqrrations'
'fhe ijeneral linear eqrration arxl + a2x2 + "' "' + q/i = b is
E
l
l
constitute a solution
ff
&
$
also callt:d non-degenerate llnear eguation'
ir
Alirrearequationissaidtclbedegenerateifit}rastlrefr:rrn
the
oxl +oJd2 + ' ' '" +'o'rcr' = b' Tltat is' if every coefficient of
variable is equal ks zero.'Ihe solution of such a degenerate
linear equation is as "follows :
(i) If the constant l: * o, then t'he abol'e equation has no
solution.
(ii) If the constant Ll = o, then every vect.ttr u = (crt ' %' .'"" c'/J
#
leastthetrivial(orzero)solutlonXl=o,.xz=a,...',i'=o.
of the homogeneous
lf x =(or, cz""'or,) is a solution
or
then it is called a rron-zero
and at least one s' + o'
\
system
is a solui-ion of the allove equaUon"
1.3 $tolution of a non-hcmogeneous system of linear
eguatiouo: .
A system of linear equations tor a set cf :l sis:riltatreous
linear equations) ln n variables [or urlknr:wns) xt'xt' "" "" 4'
is a s.e[ of equations of the form
I
t
J
I
N
Il
I
homogeneous system'
non-triviat solutlon of tl:e
is called conslstent if it has
A system of linear equations
A
lnconslstent if it has no soluUon'
at least one solution and
determlnate if it has a uniq'ue
consistent system is called
ir it r'""-ilo[ thau one solution"
solution .,a itaJuEli-tf,
has an
of linear equations always
An indeterminate system
i"n"n" number of solutions'
are called equlvalent if
Two systems of linear equations
the second
system is a solution of
every solution of the llrst
and converselY (vice versa)'
':l
7
SYSTEMS OF LINBAR EgUATIONS
COI,LBGE'I,IN EAR ALGEBRA
3t tXt + atzxz + ... ." * atrr 4,
+ "'+
a'21r+L
System of linear equations
^''jr\r+
= bl
I
dz'4' =o''l
tr*r
e)
I
d' n.z\z+ a'-ir*t \r*t + "' * 2'-r,4. - b'*J
denotes the first unknown with
and so {2 * xr '
a'2Jz* in an equauon excJpt ttre first' Here D I
whJre"ar, *b.ri
Inconsistent
*4
t,
r;l
,l
Unique
solution
A slraighllbr',vard method of solution, known as Gausslan
elimlnalion involves a successir,'e "Whitling away" of the
var.iatlles irr order to isolate their values. This method is
birsed on l itc following three eiementary operations which
alter the Ibrm of the equations' but not the solutions :
(2) of equations excluding
I
i!
;\
il
.
Proocsg I
(i) Interchan$e equations so that the first unknown x, has
a non-zero co-efllcient in the first equation i' e, ar, * 0'
(ii) For each Dl, apply the operation
Ir -+ -a,r L, + ar r\
That is, replacc the linear equation Lr by the equation
,gbtained by multiplying the first equation L, by - air ' and the
ith equaticin L, by a,, ancl then adding' We lhen obtain the
following system which is cquivalent to the system (i)'
system (1)'
equations and fewer unknowns than the original
subsystern
Repeating the above process rvith each new smhller
we obtain by induction ttrat' the system (1) is either
inconslstent or is reducible to an equivalent system of the
following form w'hich is known as echelon forrn
=
&rrxr * anh +..' "' + ar,
{,
Pll
u'or\r* u''Jr*' \r*t+ "'+ dt, x,- o''!
(i) Interchange a pair of equations.
(ii) Multiplying an equationthrough bya non-zero number'
(iii) Adding a multiple of one equation to another equation
er, equivalently, if we consider a system of m llnear
equattons tn the n unknowns xr.&, ...., .r1 given by (l)' we can
reduce lt to a slmpler system as follows :
_ It is to be noted that the system
the first equation, form a subsystem which has fewer
'^;
x
.i!
:
B)
I
or\, * d r r*r{,.'r' * ... + a'* )i, - J
zero
where ,. rr,...1, t.d the leading co-e{ficients are not
b',
i. e, arr *O,drrr*O, ..-,4:, *0'
Deflnltlon : In reduced echelon form the unknowns x'
(i * 1' J2'
which do not appear at the beginning of any equation
note that
... j.) are known as free variables' We also
(i) if an equation 0x, t 0.rA * "' + O{, = b' b* O occurs' tlren
the system is lnconslstent and has no solution'
(ii) if an equation Ox, + O& * "' + O-r1= O occurs' then the
equation can be cleleted without affecting the solution'
Process 2 : Consider the following system of m linear
in n
equations (or set of m simultaneous linear equations)
unknowns \, x2,...,x"
t'
dttxt * a:I2)c)'+'..". 'ct,r{,
a2txl + azt )Y 'i-"' rr i{2, {,
d-rx, *.\rt-f:
i!
*l',
9;rz
,l
czz
= bZ
-i' ... + ilrr,{.,= brr.,
i
(1,
cqz
I
czz
times the second equation {iom lhe third equation
times the second equation form the fourth equation
1
We reduce the s.r'stem (1) to a simpler slrsterl as follows
:
Flrst step Dilminations of xl from the second' third' ""
mth equations. We may assume that the order (rute) of the
equations and tlle order (rule) of the unknowns in each
equation such that art * O' The variable xl can then be
eliminated from the seconcl, third,...,mth equatlons by
subtracting
9,t*""
8rr
SYSTEMS OF LINEAR EOUATIONS
il
COLL}] G E LINEAR AI,G EI]RA
1
{
.J
I
t
,l
thc second equatlon form the mth cquation'
ii'. ,rr,n"r steps are now obvious' In the third step we
eliminate xr, in the fourth step we eliminate xn etc'
This process will terminate only when no equations are
left or when the co-efficients of all the unknowns in the
the form'
remaining equations are zeto' We have a system of
S!-'2- times
arrxrl atzk +...+arr,)i' =.b., I
"r.r.?
the lirst equation from the second equation
*
"'*3o .i'l
c- + +.'. + k*); =b. I
the first equation from the third equation
$-t,i*""
8rt
_a
o= br+t
B)
I
I
tirst eqrration from the mth equation
1*
ott ti*"" the
Thts glves a ncw systenr of equittiotls of thc form
allxl + anXZ + ..'* flI,,'t'r, =.1).t I
cz*z + '..+('a,-\,=D z
I
{2)
Jr* x2 + ... * crrrxn = b* n,l
Any solutiolr of lhe system (l) is a solution of the system
(2) and converselY.
SecondstepEliminationof.rifromthethird,fourth'......'
czs ""
mth equations in the system (2)' If the co-efficients err'
c*r,, h the
system (21are not all zero' we may assume that the
that
order (rule) of the equations and the unknowns such
c22* O. Then we nray eliminate x, from the third' fourth' ""
mttr, equations of the system (2) by subtracting
b,r, )
o=-1
where r s m. We see that there are threc possible cases
(i) No sotution if r < m and one of the nunll;ct:-ifi, ' t ' " i:* is
not zero.
(ii) Precisely one sotrution if r = n nnd b,*t..... i:- if 1:r'esent
are zero.
This solution is obtained by solving tlie nth ctiualion of
so
the system (3) for.r1, then the (n-1) th equation frrr "r;',-t and
on up to the line.
(iii) Infinitely rnany solutions if r < n and b, * r. ..., bm if
present, are zero, Then any of these solutions is obtained by
ihoosing values at pleasure for the unknowns xr+l' "" 4r
solving the rth equation for x, then (r-l) th equation for xr-,'
and so on uP to the line.
12
COLLEGE LINEAR AT,GEBRA
'l'hus we obtain the equivalent system (with the same
rrolutlous as the sYstem (l).
xr+2xr-xs=2)
-r2 -)ca=l I Ql
b-xe=l)
*'n'Z-i
'\.v <'\_/
=?) .u
Thlssystem(3)isinechelonformandhastwoequations
in three unknowns and so it has 3-2 = 1 free variable which is
.q and hence it has an infinite mimber of solutions'
Letx3 = a {where ais arbitraryrealnttmber); then k=|+a
and x, - -a,I'hus the general solutlonis x, = -a,x2 = I +a'
)h = a, whcre a is anY real number, Now a particular solution
can be cll,rli,rinr:tl l.)"'/ .tlj\Iiilg any value for a, Irt a =1, then xt= - 1'
\-2..t:,, '. i t'1. in other worcls, the S-tuple (-1'2' 1) is a
particr.rli-tt' soltttion of the given system.
E)rample 3. solve the following system of linear equations:
3xr-xr+x" --21
x, +1xr+2xr=6 I
2xr+3x, +)b =O )
Solutlon : Reduce the system to echelon form by the
elemeptary operations. Interchange first and second
equations. Then we obtain the equivalent system'
xr+1xr+2x" = 61
3xr-xr+x* --21
2xr+1xr+h= OJ
We multiply first equation by 3 and 2 and then subtract
from the second and third equations respectively. Then we get
the equivalent sYstem.
xr+1xr+2x, = 61
-l6n-5x. =-2O I
-7xr-Sxr=-12)
EOUATIONS
We multiply second equation by - A*d
13
then ad<l with the
third equation, Then we obtain'Equivalent system.
Since the second and the third equations of the new
systey (2) are i49-tt-tlggLwe can oilggera a-ny_93_9 o[_them.
-Hence we can simPlY wrtte.
SYSTEMS OF LINEAR
xr+5.xr+2x"= 6 I
-l6h-Src" =-20 |
13t
13
- 16'rt =- 4
J
This system is in echelon form and has three equations in
three unknowns. So the system has a unique solution. From
the third equation we have \ = 4' Putting xe = 4 in the second
equation, we get xz = O,Again putting xz = O and 4 = 4 tn the
first equaUorl, we $et x, = -2.
Thus x, ='-2, h = O, lt = 4 or, in other r"'ords' the 3-tuple
(-2, O, a) is l44Nque solutlon of the giver] systeln'
Eramff 4.JJProve that the ftrllow'ing system of linear
equations iMnsistent :
?;i3;:?*,;;'I
6y@
3\-xz+2d=z
J
Proof : Reduce the system to echelon form by elernentary
operations. We multiply tirst equation by 5 and 3 and then
subtract from the second and third equations respectively.
Then we obtain the equivalent system.
xr+2xr-3x" =- 1l
7I
n-7xz+ll:;.
/ )m + tl.rc, = 1O )
seconcl equation from the third equation.
V/'6" srbtract
t'fien we get the equivalent system.
xr+2xr-3x. --11
-7xr+ ll.rco = 7 |
O+O = Il )
or, equivalentll,,
-- 11
"r;?; i& =71
:0 = 3J
l4
SYSTEMS OF LTNEAR
COLLEGE LINEAR ALGEtsRA
't-h"" th: Si"."
"y"t.- @chelon
x|+2tri+8x"-xn =$
O = 3 (which is not
.l
3x, +4xr+2\+3xn=-21
xr+2xr+8:6-xn -8 f tU
7xr+9xr+.rt+Bxn =g )
Solutlon : Reduce the system to echelon form by the
elementary operations. Interchange the first and third linear
equations of the system (1). Then we have the equivalent
system
xr+ 2x, + 8x, -xn
=
g I
3x, +4xr+2x"+Sxo=-21
zxi + sxl + s.rl + xnI Q)
=s
7xr+9xr+\+8xo =g
I
-2*r -22x" + 6xn = -26 Lo,
-x2- llxr+3xn -- 13[ t
-1xr- 55.16 + lS.xn = -56
J
Divide the second linear equation of the system (3) by -2.
Then we have the equivalent system
xr+2xr+8x"-xn =$ l
x2 + Llx"- Sxn = 13
-xr-11.r6+3xn =- 13f (4)
-?tr- 55.t6 + lSxo = - 56
kt us respresent the four linear equations of the system
I
J
(41 by Y
r, U r, U and U n respecUvely.
"
Apply the operatlons
U. -+ Li + L',
"dU: +Vn * W,
Us: -k - I l.r. + 3x, = -- 13
L'z:4,+11.16-3x, - 13
U"+Ur:0+O+O
)
=Q
O=O
Ln : -5xr-5516 + 15xn =-56
i. e.
I.et us represent the four hnear equations of the system (2)
by Lr, Ie, k and Lo respecUvely. Reduce the system to echelon
\Ur:1xr+55x, -75xn =65
fonn by the elementary operations. Eliminate x, from the
.Uq + 51, : 0+O+O =9.
second' thtrd and fourth linear equations by the operations
l, -+Lt - SLt, h -+f: -2L, andLn -+Ln -ZLr.
Iz :Bx,+ 4x, +2xr+Bxn =_)
- 3L, : -Jx, - 6x, - 24\ + 3xn = _24
Iz-SL, : 1\-22xr+ 6xo =-26
l*:2xr+Sxr+5\*&
2L,
: -2x, - 4x, - L6x" + 2xn =- 16
l**2Lt: -x2-Llx"+Sxn
=-15
Ln :7x, + 9x, + x" + 8xn
16,
l'hus we obtain the equivalent system
true) Flence th.e
stem is incons_is!9n_(. !:_e,t!e system
has no solution.,I
Etample 5. solve the following system of linear equations:
Zxr+3xr+Sxu+xn -S
EgUAnONS
=Q
- 7L, : -7x, - l4x, - S6x, + Zxn = - 56
L4-ZLr, -5h--55.r6 +l5,xn =-56
i.e. O =9.
Thus we obtain tJre equivalent system
,. ., i
,
:
xrt 2tc, + 8.rc, -xn - 8 I
xr+llr,r-3?==lr,
,
lU,
Q= 9 )
Divide the fourth equatiort of the systenr (5) by 9 and
interchange it with the third equation we get the new sygtem.
xr+2xr+8.r6-xn -Bl
h, + ttx, - 3x*_=
V_
i
1t | (6)
I
I
O='O l
t0
COLLEGE LINBAR AI,GEBRA
MS OF LINEAR EgUATIONS
17
cuntafirs an €Ouatiorr of the fo'rn O = 1 which ic ne,t-tnde;-tletTcg
the given system is inconsist€rrt i:e th€ system has Io
" (i) =
true. Mot'eover if x= ct is a solution, i. e' ao = b, then * = 3'
Theorem : Given the following system in echelon form :
al rxt + apxz + arslql + ... + ah.r; = bl
%r 4r+ a4z * 1\2 *t + "' +arr'ri = b,
Thus the equation has a unique solution as desired'
New let us assume that r > 1 and the tJreorem is true for a
system of (r-1) equations.
+ "' +ar,'r; = b,
a2J2\r*
a4, t.
nH. *r {. *r +"'+ qrr4=b.
where I < jz <....j. ,rd all * O, aar*O,...,\. +O
Then the solution of the given system is as follows :
There are hvo cases
(i) if r = n, i. e. if there are as many equations as unknowns
ttren the system has a unique solution.
(ii) if r < n i. e . if ttrere are f,ewer equations than unknowns
then we can arbitrarily assign values to n-r free variables and
obtain a solution of the system.
Proof : The proof is done by induction orl the number of
equaUons in the system. If r =l then we get the single linear
equatlon alxl + %12 +.., + q,.ri., = b. where a, * 0.
The free variables are jrz, ..., xr,. I-et the free variables be
assigned arbltrary values, say x2 = (b ri = q3, ... , 4, = 0!r.
Putttng these values into t-he above equation and solving
for x, u,e &t
I
,, =i |E--uror.-ilr,rs " .,.-q,qJ
These values give a solution of lJ:e e<1uation. Since putting
these values we get.
", [+ {b -
az {x2
It is to be noted that x= | ,"
"
%, *112+r
.
arj,x1,, +41, *rt. *l *." +a-x" =br'
as a system in the unknowns {r' ..', {,'
Now the system is in echelon fonn. By induction we can
arbitr.arily assign values to the (n - jz+ 1) - (r-I) free variables
irr the reduced system to obtain a solution
(saY {z= %2,..., r(-, = cr,,)
As in case r = l, these values and arbitrary values for the
Yreld
additional y- 2free variables (say x2 = d2' ..,, \z-r=
'0:r-r)'
a solution of the equation with
t
xr=;} (br - arzGz -... -ar,or,)
(Note that there are (n-j, + 1)-(r-l) + Uz-21 = n-r free
variables). Furthermore, these values for x1 , x2, """,'rqnalso
satisfy the other equations, slnce in these equations' the
are zero.
co-efficients of x1 , ...,
\-r
Now if r = n then j2 = 2. Thus by induction we obtain a
unique solution of the subsystem and then a unique solution
of the entire system. Hence the theorem ls proved.
xample 6. Express the following system of linear
- ... -- ,,,,r,,.)f +'azdz"i- ... * a,.ft, ',, b
or, b = b which is a true statement.
(t)
Furthermore, if r - n = l, then we have ax= b where a* O.
v"
b is
solution. because
18
CoLLEGE LINEARALGDBM
sYs'r'EMS OF LINEAR EOU.{TIONS
Solutlpn :. I*I qq {Eitresent the four liqear equations of the
;',iven system (I) by Lt,Lz, L3 ,and L4 respectively. Reduce the
systern to echelon form by the elementary operations.
Eliminate xl from the second, third and fourth linear
Dlvldlng the third equation of the system (3) by 2 we get
-.ys +& -2x7,= - l. which is identical with the fourth equation.
So wt ('an clisrcgard one of them.
tlen<'e thc system (3) reduces to
Xt-X2 *Xs-xq+4
=l I
xz*g+2xa+24=g
I
-xs*x4-2xs --lJ
equations by the operations
La- La- 2Lr,Ia + Ia - 3L, and La -+ La- I-1 respectively.
-
la:2x1- x2*3x3
-ZLtt-2x1 + 2x2-2x3 +2xa-2.x5 ,1=2
L2-2L11 X2+ xs42xa+2x5
=Q
L3:3x1 -2x2+24+xa+xg
=I.
*,3{. *q*n-3"s :9_
:4I;.,,-3",*3?
.
t,g-3l,ti xz-xs+ 4xa'2x5
=-l
Multiplying the third equation of the system (4) by
-Yt-Xz *X3-X4 i')6 =l I
x2 * i3 + 2xa + 2.'--5 =Q L
:-2
x2 -.tb + 4xa -2.r5
* -*s'{ =-1J
,.,, ..
1
kt us represent the four linear equations of the systern (2)
byL{,lat, L3', artd La'respectively. Eliminate x2from the
*
'l
third ariil the fourth linear equations by the opierations
L'z
:'
I
Ttrus the sy.-stern (2) reriuces to,
4+ l'+a-2b+2a,+2b=O
ar, x2= - (1 + 3a)
+l+3a+1+a-2b-a+b=l
or'Xt=-l-3a+b'
l{ence the general solution is xr= - I - 3a * b,
2= - (1 + 3a),
)6= 1 + a - 2b, h = a,4= b where a and b q5e any real n r^Arorr/:
x1
:,-:i,.:;;^3riJ=il t'r'
*iz"-:,-;::?l
three equations in the five unknowns: hence the srystem has
an irrfinite number of solutions and 5 - 3 = 2 free Variables.
Since the three equations begin with the three unkno\MrlS x1,
x2 and *3 repectively, the other two unknowns xa and x5 ore
free variables which may have any real values desired. To
find the general solution
us say x+ = a, 'and xs = b where a
'let
and b are any real ndmbers. Putting these values in the third
equation of the system (5) we get xs = 1 + a - 2b. Puttiqg the
values of x3, xaand )6 in the second equation we get
Again, putting the values ot'x2, xs, x4and x5 in the first
ccluation of the systemJS)we get
Y't1 L'2:-2xg+2x4 -4xs =-)
L'q- L'2, -x3 +xa -2x5 =- I
:.
','
Xr-xz*xs-&+)6 =1 I
x2*xs+2xa'r2k=O I tO
x.g -xq+2'x5 -l
)
Now t& bystem is in echelon foim and there are on\r
L+-Lt:x2+O+3X1 +O
=- l.
.l'hus we obtain ihe following equivalent svstem ((Witli the'same
solutions rs the system ( I ) ).
14' -+Lg'- l-z' ?fftd L'4 '+ L '4-
6)
we get
-n:--
;
19.
(er
S}nsflTMS OF LINEAR EgUAfiONS
cgllruoni of the glven system
COIIECE LINEARN.GEBRA
2A
8y.tcm.
hooogleoeors tlaoar cqtilrfoos'
1.4 Solution of a systcrn of
be homo$gneons if
of linear equations is said to
system
, .$
b- of the non-hornogeneous
all the constant terms br'b2""
system has tlre form'
system are zrro; ttrat ls' the
=:l
a11X1 + 412X2 +"'+arn)h
ar21x1 +a22X2
'r;',
+"'*3rrr4r
='|
*t*, x2 + "'+ a,nrrxn= o J
conslstent
system of linear'equations is
Every homegeneous
the system' This
o is always a solution of
Sincexl =O, k=O, "' ' ';h =
If the other solutions edst'
solution is called the trivial, solutlon.
solutloqs' Thus the above
they are called the noa-trlvlal
be reduced to an equlvalent
io*o*un.ous system can always
forrr:
homogeneous system in ebhelon
+ +a rn+r
a11x1 + a1f02 + arsxs "'
I I
l*"':t* )h =O
s
J
solutlon.
ls less than the number
(ii) if r <n i' e' the number of equaUons
t
I
I
Thus we obtain the equivalent system
xt+ h. +2 xs =-Ol
5x2+5x^ =O t
xz+xs =OJ
tlre equivalent qrstem
Hencewehavethefollowtr4ltwoposslbilities:
equatlons ls equal to the
(i) if r = o I' e' the number of
the system has only ttre zero
number of unknowns then
i
lo z - 2x1 +3x2+ xs = O
ZLrs2rl+2.x2+4xs =O
=O
la+zl,teS:q + S{3
E
Dffilngthesecondequationoft}resysterr(3}B5.weget
;;rir:.':n"'n.:.^+"'+ a..2.,4' -1Ol
air\r*rdr* 1x].*
Al
equauons of the system
I.ct us represent ttre three linear
the system to an
(2) ty Lt.Leand L3respectivellr' Reduce
operations' Eliminate *1
echc)on form by the elementary
operation' la'-+la + 2L1
from thc secrnd equation by the
,..(I)
...rry
I
(1). Then we get the equivalent
xt +h + 2x3 =-O|
=91
-2xt-+bcz+xs
x2 + x3 =QJ
,
2I
of
has non-zero solutlon'
unknowns, then tfre system
solution of the following system
-'Excrrplc 7' find the non-trivial
of homogeneous llnear equations:
=0..l
x1 + x2 +2xs
I
(r)
=o
h, +xs
-2x1 +3x2 + xs =oJ
echelon form by the
Solution: Reduce the system to
second and third
elementary operations' Ipterchange
xt_ xz+2 x. = o 1
tii:;3i *
are ldentical we can
Sirlce t]le seeond and third eqrraUons
have the equivalent system
dtsregard one of them' Hence we
is in echeron rorm'
*'*?o:;:=3
)rn'"n
two equations in three
In this echelon form there are only
an infinite number of nonunknowns, henee the system has
2 I free variable wl:tch is
zero solutons. The systenr has 3 - =
is x1 = - a' x2= - o" xs-- d
x3. [,et xs= a.Thus the general solution
or (- a. - a, a), where a is any real number'
l' then x1 = - l' x2- - 1'
F'or particular solution' Iet a =
system'
xl. l, or (- l, - I. l) is a particular solution ofthe
22
COLLEGE LINEARAI,GEBRA
SYSTEMS OF I,INEAR
Example 8. Find the solution of the- following system ol
hornogeneous linear equations:
xt- x2 - x3 - x4
x1 +3x2-xs+)Q
=O-l
-Ol
3x1 '7x2-xs-6x4
=Q |
2x1 +2x2-2x3
=Q I
6x1-2x2-4 xs - 5x+ =0 )
(1)
Solutlon: Let us represent the five linear equations of the
system (1) by L1 ,L2,L3,La, and L5 respectively. Reduce the system to
echelon form by the elementary operations. Elimlnate x1 from the
second,, third; foirrth and fifth equations by the operations
L2 -+ L2 - Lr , I+ -+ 14 -3L1 , La -+ La - 2L, and'
la"- Ls - 6L1 resPectivelY.
-o
la -Iit: 4x2+O +2xa
Ia: 3x1 - 7x2- xe-6& =Q
- 3L1 : - 3x1 + 3x2+ 3x3+ 3:q : L
I,3-3Lr: -4x2+24*Sxa
la: 2x1+2x2-24
QQUATIONS
kt us replesent the forrr linear equations of the sy5tem (3)
lty l,'1,L'z,L'3, and La' respectively' Eliminate x2 from the
third arrd lburth cquations by the operations L i3 - L '3 * L;'2,
'and L'4 4 I-'q - l,'2
L'3 * L'21 2x3- xo =g
L'+ - L'2 :2xs - &=O
Tttus tle systeln (3) r-educes to
xt-xz-xs - & -01
4x2
- +24 =0 I .
..1
2lo-i =ol
(4)
2r, -xq =O )
In the system (4) the third and fourth equations are
identical we can disregard one of thern. Thus we obtain the
equivalent systemi
xr-xz-xs - &=Ol
45 +2\ =0 l
24 -x4 =o ]
-O
-O
-2L,y: -2x1 +2xy+24 +24 =O
L4-2Lr : 4x2+O +2xa=Q
I4 : 6x1- 2x2 - 4xs- 5xa = Q
-6L1 : -6x1+6x2+6xs+6& =O
where a is any real number. Then h=t,
k-6l-r:
'Itrus the general solution is x1 = a, x2 =-|, x3= 3, &= a.
4x2+2xs+&
=O
(
& =01
=ol
=o>
-4x2 +2xg- 3xa
4xz + 2xa =Ol
4x2 +2xg + x4 =o
a)
In the system (2) the second and fourth equations are
identical we can disregard one of them. So, the system (2)
4.u.2 +2 x4 =O I
Axz + 2-r3 + xa =O
.1
ana xr = a.
\
For particular sclution, let a = 2. Then xt= 2, h= - l, xs = 1,
Xq= 2 or, (2 - 1, 1, 2) is a particular solution of the given
system.
Example 9. Find the solution space of tlrc following
homogeneous system of linear equations:
x1 +2x2 +3.rq +15 =q
xt- x2-x3 - X4 =q
-4x2+24-3xa =0 I
aa
x2= -|
or, [a'-;,i,"t
_,l
reduces to
(5
In this echelon fomt there are only three equations in four
unknowns, hence the system has an infinite nurnber of
solutions anird 4 - $ = I free variable which is x4. L,et xa= 4,
Thus we obtaln the equivalen:;f^fl
4 xz +2&
23
(3)
2x1 +3x2 +3xa +:6 =O i
\ +x2 +xS+2X+ +.rB =O I
3x1 +5x2 +6xa +24 =O
2x1 +3x2 +2x3 +5x+ +24=O
'
I
El. P. 1982.
24
SYSTEMS OF LINEAR
COLLEGE LINEARAI-GEBRA
Solution : Reduce the system to echelon form by the
l' 3
elementary operations' We rnultiply lst equation by 2'
and 2 and then subtract from 2nd, 3rd, 4th and 5th equations
respectively..Then we have the equivalent system
^ lT-.,7-;*;;l
+ xg -x4 =O
-x2
Hence the solution space of the
Exanple 10 Determine the values of tr so that'
-xz -Sr<c - x5 =o I
=O f
O+O+O
=o I
2xg+24 + x5 =o )
x1 +2x2 + 3.ta + -16, =-q
xS+Dc4 +x5
'
-x2 - 3x+ -x5 =-Ol
xs+2xa +)6 =t)1
2xs+2xa +x5 =Q )
from 4th
We nrultipfy 3jd equailon by 2 and then subtract
equauon. then we have the equivatent system
x1 +2x2 + 3x4 +)6 =O1
)
- l'l
2x +3y +)'z =3 I
x+)'Y +32 =2 ]
x+Y-z
" eguation. we get
h+ 2a- 3a + ?a=O, i'e X1 = - a'
tO. U. IL
lSA
Solutlon : Reduce the system bo echelon form
by
eglation by 2 and 1
elementarlr operaUons' We multiply flrst'
ard tfle ttrird equations
arrd then substract from the second
respectively' Then we obtairn the equivaleet Erst€m'
x+Y-z
=l I
y+()"+28 = 1l
()t-t1Y +42 =rJ
(X *U sd then substract from
We multiply second equation by
the thir<t equaUon. Then rve obtain the equirraleot
x +Y-z =l
-x2 - Sxa -)6 =Ot
xt+2xa +x5 =OI
-2xa'- xr =0 'l
f'his system is in echelon form ha.Yrng four equations in
variable
five urrknowhs. So the system has 5 - Q = I free
which is x5 and it has non-zero solution'
2a 7rr
l-et x5= 2a where a is any real number' Putting x5=
a' Puttin$
tlre 4tlr equation we get - 2x+- 2a = O' that is' x1= we get xz= 'a and
& = - a and x5 = 2a inthe 2ncl & 3rd equations
is= O. trinally putting xz= d, x4= - a and xs= 2a in the lst
the
x' y and z has (i) a
following linear system in three variables
(iii) no solution:
unklue soloution (ii) rnore than one solution
:x :*, i:: )
x1 +2X2 +SXa + 15 =\
25
given syst€m is
W = {F a, a, o, - a,2alla€ IRI'
f
We subtract 2nd equation from 3rd' 4th and 5th equations'
'Ll:cn we irave the equivalent system
E9U'aTIONS
{4-a:lxl
system'
)
y+()'+2)z =l
i
+2)V =2-)\")
x+Y-z
=l
y+Q'+2lz =l
l
(6-l -12) =2-),)
=l I
x+Y-z
=l f
y+(r'+2lz
(3 + l.) (2-Nz=2- A )
1
Or,
Clr.
solution if
This system is in echelon form' It has a unlque
i' e' if
tlre coefficient of z in the thrid equation is llon-zero
is
)"*2andr! + -3' In case)' =2 third equation is O = Orvhich
true and the system has more than one solution'
26
SYSTEMS OF LINBAR
COLLEGE IINEARALGEBRA
In case i = - 3. the thirri equation is O = 5 which is not true'
andrlrence the systenr has no solution.
nte 11. For what values of .]. arid p the following
Gfstem of linear'equations has (i) no solution'
(ii) more than one solutiotr (iii) a unique solution :
rt[*r
x+Y+z =6 I
x+Zy +32 = lO I
x+2y+)z =$ )
lD. U. H 198q D. U. H (Stat) 19821
Solutlon: Reduce the given system to echelon forrn by the
elernentary operatiorrs. We subtract lst equation frorn 2nd
& the system will be in echelon form having two e{uations in
three unknows. so it'has 3 - 2 = I free variable which is z and
hence for I = 3 and F = 10, the given systerh will.have more
than. one soltltion.
(iii) For a unique solution the coefficient of z in the Srd
equation must be non-zero i.e )" *3. and p may have any
value. Thus for l,*3 and p arhitrary the given system have a
unique solutlon.
Exanple 12. For what values of l. , the following linear
equations have a solution and solve them completely iri each
case:
.
I
Y+ 2z=4 i
I
x+2Y + 4" =\^l
x+4Y+LOz =)'2)
x+Y+ z
+2Y
*n=l:! II
We subtract 2nd equation from 3rd equation. Then we
x+ 4y + \Oz =)'2 )
Reduce the given system to echelon form by the
elementary operations. We subtract'lst equation from 2nd &
Srd equations respectively. Then we have the equivalent
have the equivalent system
y+ 2z =l
x+y+z =l
Solution: The giiien system of linear equations is
y+Q,,-llz =P-6 )
x+Y+z =6
27
(ii) lf l =3 and $ = IO,. then the 3rd equation of (*') vanishes
and 3rd equations. Then we have the'equivalent system.
x+y+z =6
EOUATIONS
I
I
().-3)z = pt-10 )
system.
(*)
Now from the last sYstem (*') we have the following three
eases;
(i) If 1= 3 and p + l0 then the 3rd equation of (*) is of the
formo = a wherea= p- 10 + Owhichimpliesthatoisetlual
to a non-zero real number which is not true. Thus we conclude
that for l" =;3 and p + 10, the given system has no solution.
x+!.+z = I
y +'32 =I-I II
3y+92 =)'2 -1,j
We multiply 2n<i equation by 3 and then strbtract from the
3rd equation. Then we have the equivalent system
. x+!*z=l
I
x+Y+z =L
I
y+32 =,t-l
f =Y*9, =tr-I
I (-)
?&-3X+2)
o+o =L2'37.+21 o =
2g
coLLtsGE LINEARALGEBRA
EgUATIONS
E:XERCISEA - 1
SYSTEMS OF LINEAR
lf )\2 - 3)," + 2 * o, the given system will be inconsistent ancl
'if ),2 - Nt+ 2 = o, the'above systern will be in echel'on form
having two eguatlons in three unknotrrns. So the system has
3 - 2 = I free varlable whtch is z' So the system has non - Trlro
soluflonfcx)'2-3I+2=O
that ls, (L l) ( X-21=O
=+I=l or'I=2
Tfrus the $ven systerr is consistent fur 2' = I and ]" = 2'
Cas I uihrn).= 1
The qrstem ({ will be
x+Y+z= lI
y + Zz=OI
\rrhH ls in echctron form where z is afree vartable.
Iretz = a xrtr€tt a is arbitmry real number.
.'.y=-3a.x = l+%
Henc-t(lthegiyensystemoflinearequatlorrshisinfinite
number of solutions for I =1 In particular, }et a =' 1, tlren x = 3'
. y = - 3, z = 1. ls a parffcular solution of t]re $iven qlstem'
@
E Whl'=2
The system ({') wiU be
x+Y+'=ll
y+Sz=ll
which.is in echelon form where z is a free variable'
LeLz=bwhere b is anyreal number. "' y = 1 - 3b, x=2b'
Hence the system hers infinite number of solutions for I = 2.
In particular, let b = - l, then x = 4,Y = - 2, z = - | is a
particular solution of the $iven s5r5[s1n.
29
l. Which of the followtng systems of linear equations are
inconsistent :
xl+)r2=l)
(i) ';::ffi 1)
x1+2x2=31
(ii)
x1- x2 =) )
x1 +3x2 +Zxs = 7 )
xr* xz+ xs=1.|
2x1 + x2+3xs = 8 [
3x1 + 4x2 +6xs = 16 [
(iii) Zxr+2toz+2xs=1Y
Sxi + 3xz +3x, =21
6x1 + 9xz + llx" =29 1
'Aaswers:
(i) Inconsistent' (il) Inconsistenl' (iii) Ineonsistent.
(iv) Inconsistent
are
2. Which of the following systems of lI:ear equations
conslstent? Find all sohJtions of the consistent system'
,, ir i'*:W"=. i l @
I - r1l,'S:I ]
AnswErs : (i) xr = L, x2 =2, xs =4
{it) x, =- 3r -L+. =-2a-92x3 =a
where a is anY real number'
3. Express the following systems of linear equations ln
echelon form and solve them :
x1' xr+2x.+ 4 =-Ol
3.xr +2x4 = 2
... -*i +
tr,2xi+ x2 - xo=lf
L
2x., + 2x2 + x" + 3& =L4 )
xr+2xr+ xo =-11
6x, + x2+ )ca =-4 I
(ii)2xr-3*r- *, = O f
- xr 7 x2-2xs= 7 t
xrh,
= l' )
Aasrrers, (!-*, =1,&=2, 4=-l'*=)t
@ "' =- l' x2=-2' xs:7
4. Solve eactr of the following linear systems of equations :
2x1-3x2--21
2x1+ x2 = 1l
3x1+2x2= Il
3x1+2x2- xs =-15
d,uf,,,,, 5x1 + 3x2 +.P4 =
4x1-Bx2=12l
WW
Llx1 +7x2
(r) x, =
4x+ 3y -22= 14 )
=-3O
11
I
(i) ,T,-.n?,; ix.-;fi . if r,,r -i": ;,:..ru"'..r; :3 I
5x1 +x2-Sxs-24=l)
X] *x2*x3 *4=l)
.{nsrrrers :
28 8
-69 x,r =-SEi,rQ
=B
EE,
94
685
30 79
(iil x1 =679, h=62g, 4 =W, 4 =W.
64
\
8. [i) Express the following system of linear equations,in
cchelon form and solve it :
x+ 1r
--3 I
2.x-2y -z =-B I
4x *z=-I+l
Answers .: ti) xr = ?,- x2 = &, xs = - 2
,,[DU,t&,,rl.FQ
x-3Y-z =-5 )
'!.
where a is. any'real number
(ii) Reduce the following systern of linear-equatious to an
I
(ii) x1 =f g + !i. x2 =-ia + 3, xu = - 3a. .rq - a
where a is any.real number.
6.:find the solution sets of the following systems of linear
equatfons .:
,,
x1 i2x2'+ xs+ X4
X4
X1- X2*X3- o['
-61
(1, x1 +8x2+ xs+54
'2x1
I
7. Which of the following systems of linear equations are
consistent? Find ail solutions of the consistent s5lstem :
3x1 + 2x2-x3 +4)E =61
x1 -5x2 +4x3 +4=4
\ - h*xs*x4
=
x1+2x2-XZ-X4
= ?l
(rr, 24- 2x2 * xs - x4. =
-3x2 + x3 - xa =- 9,J
'
=i (8 - 2b), 4 = B, )e =b
(i) xr =85, xz=
=31
'/r,,n\3:::i?)ix:
x1 -3x3 -24=91
V
l
(2 +b- 3a), ri
q
(i) Inconsistent
.
Cu).xi = 3 + 2a, x2 = 4, where a is any real number.
fiii) xt * - 4. xz = 2, xs =7.
. M x= 2-a,y =2 +2a,2=a
.
where a is any real nurnber.
I 5. Solrra the following systems of ]inear equaUons :
o .''1
ll
j
[ii) x1 =r1. xz=-ll, xs=-1T,4=-11,
:
I
+x3 + 2* =3)
where a anrl b are arbitraql real numbers.
p.u.s. 1980, Bl
@ atrs#ers:
I
Anrwers:
1
3x1 + x2*3x-r.= tl
x1 + 2.x2 + 3xs''+ 44 = q
1l.r', +llx, +4xr"
=l
(ll) i!x'1 +4x2
+& =21
lx,
9f
l,Ii) 3x1 -6x2-2x1 +4x, =4
x+W-32= 6l
2x- y+42= 2 I
Q
3l
SYS'TEMS OIT LINEAR' EQU/\TIONS
COLLEGE LINEAR AI,GEBRA
30
+7:x2+24s+4xa
= z5l
= 2Ci)
'.
xt-x2+2x3=51
2x1 +x2*xs=2 |
tD.U.P.lgqu ';
2x1 - x2 : xs = 4 |
tD. U. s. 1981, 198u1
x1'+ 34 +2i" = 1)
Ahswers : (ij x = !nr"- Le, y =ff_. z= a
:
:
.i
where a is any real number.
(ii) xr - 2, &, =- 1. xs = l.
'
1 . *:: ';:1: i .i
_ ."
..",1,
33l;
SYSTEMS OF
COLLEGE LINEARAI,GEBRA
92
. 9. (il Reduce the system of linear equations
(ttl
5x1+2x2-7x3=11
7xt - *, * i* '- 9l
2xi+1xr- xs=5J
P' u' s' 1s2l
into echelor-r form and hence solve it'
of
(ii) Use echelon method to solve the following system
linear equations :
x1+x2+3
=Ol
x1'-3x2-xg
=O
An$rers : (i)xr
=m,*='ffi,*"=#'
I
4xi'xs+r4 - =9
+5
2xt - 2xz - xc + 8 =^o
lD. u. s. r9s4l
x;-ff ^, * '=ff ^',xr. .= awhere- a !s anY real number'
solutions
Vd.T'ina the non-zero
equatlons :
of l-inear homogeneous
x1 +3x2+2x3
=ql
(i) ri_:__{;"i".i":[
x1 +
l7t
2x1+x2*xs-5x4=9
(iii) xl,+f;*-"*=31
of the following systems
xr -Sxz-2x3 =91
],,,, *,:**rr,
(iv)
: BI
tD.u.&,lgeffI''
2xt - xs +Jq =91
xi + 2i - 3x3 +-3xa =P
' -xr - x2 +2xs-2x+ =O
.x2
'
I
J
(ii) xr =Lf ! , * =7,
Aagwere: (t) ,t =-* a' x2==|' x3=a
X3 = a where a is any real number'
10. (il Solve the following linear equations
' xr + x2 + 2xs *]'n = 1]
3x1 + 2x2 - xs + ll& =-?l
4xi + 3x2 + x" + !4 =lll
where a is anY real number'
(ii) Xl = - a, x2 *- &1 X3 =2
wht" * i" an albitrary real number'
:
.
j :
2xi+x2-3x3+24=l)
(ii) Find aII rational solutions of the following system
linear equations :
xt+h -x3 -_& = -Il
34+4x2-xs-2xa
= 5)
\+2x2+x3
Aosscrs : (i) xr = 5a-b - 4' xe = -7a+ 9' x3 =
where a and b are arbitrary real numbers'
-'--'
*t a'" a and b 4re arbitrarlr real numbers'
(iv) xt '= a - b' xz = a-b' )6 = a' x4 =b
*t"*. *a b are arbitraryreal numb€rs'
2'{+ I
ol
In particular, leta =2,-b-= I thenXt =
'"
1' & = 1r *s =
W# il':T:iiffiii"-lili:"*" ",
=
rhe ro,,owing
a' & ?b
{ii) x, = &'+2b-7,'xz=-Za-b+6'x3 =a'&--b
where a and b are arbitrary real numbers'
'Qd art rmine whether each system has a non-zerc
solution. If exists, tind the non-zero solutions'
x1 +x2+xr]=gl
+2x2+ X3 =91
l',
=ol
(i) xt.-xz*"i=of (ii) 7xi-xz+2L4
xi +12x2 -374=91
xt + h-*;
=Ol
AnswEfe: (i) xr =O, h. ='O' 16 =0'
t\
Aursr: 0t *, ='?;E =?' )b=a
where a lc arrirea!number'
Llnear AlScbra-8
-r
t
JI
5
tt
34
I':r:i ir
i i:
S'STEMS oF LINEAR EQUATIoNS :r | :i
tor,iuBb ur.rBnn ar,oBsRA
Annrorr:
(ii) xr =-ia,x2=-2'xs=ra'xa=a
where a is anY real number'
(iii) *, =ai, *r=?i, xs = a'-& =a
where a is anY real number'
of linear
14. Solve the following homogeneous systems
eQuatlons :
(t)
r''
3x1 + 6x2+8;; *xa-+S'{ =o J
x1 +2x2 -2x3 +2!n * F =Ol
(iv)
" x1 +2x2--; . #- -ziu
=9[
.l6=u)
2xt+4x2-7x3+ &+
Arsrart': $l xt=7, *r=T'x5=z
wherc a ls arrY rcal number'
x3 2'/Y. =9'
Partlctrlar solution i Xl = A' xb = 3' =
(ll) Solve the following homogeneous system:"of,''llnear
cqrtallrlns :
3xi + tt'2+2xs+ x+ =9[
2xv-4x2+6x3+xa =^uf
=91
ixi+zi +24 =o
)
Z*i - a*i *
Aosycf, ! X1 - - a, )h, = x3 = &, )Q =O
e*"
#t
i'*i* 10i2+ xs+6xa+5'c5 =o J
x1 + 2x7 + 2x3 -x4 + 3)cg' = q
(iii) x1 + 2x2 + 3x3 * x4 + -16 =.o !
=a
J(1 -3x2++*"- -'q =9.)
xt-Zth+2x3=O1
2xi+ x2-2xs=Ol
3xr + +x2-oxr=o1
3ix1 +4x2' +4 +2xa +34 =9
(ii) ui;**'*,**#,:?Jff
a2a
*, =!,4
General solutlon : x1 = o, x2=;,
35
u;.
tD' U' H' 19861
o'u' H' 1ee8l
where a is anY real number'
(it) xl, - 3a- 5b' 4 =2a+ 3b' JF3 = 4 h= O' :6=b
=
*t"r" a and b are arbitrary real numbers'
c"
'tiii) xr =-2a+ 5b- 7c,tt'z=o"x3= -2b+2c'xa =fo'x5 =
where a' b and c tire arbitrary real numbers' l
(M & =-%.-4b,ry=u'xs.i, b:3 =b';6=O
wherea and b are arbitrary;real numbers'
solution of
a general soluticn and a ilticylar
lfilFind system
of homogbneous linear equations :
the following
:. i :qa'lrli;
- 3x2' .t ' ' +**- =9)
",
xq+3xs- &=01
2x',+
Sxi -lZi,*;6ft3'+:7-f,ard=OJ " :,,ti'.,.
rrr
where a is an arbttrary real ntrnber'
of linear equauons :
0ll) Solve the following system
xr * x2 +2X3 + .lq= .5^l
2x1+3x2+ Xs- &=lPl
2I'2;Xe:&=,;)
...,r;-/"!A;
2'
Arlgwer ? x1 = !' h '= 3' x^'= l' 4 = -
equations
t{gdReduce the following system: irof linear
' 'l"r;
''
:
'r''
"
''
an echelon form alr0 solve it :
x+2Y-32= 4 )
x+3y* z=lli
2x*'5Y-47=13 | '
(u) Solve the folioruing system
x+ 2y - P,= 6- ]
2x- y+42\;21
E).U,s.tg8ol
'
2x+61*4':22)
to
,i',. ,"U'""T,-"ffit"ft,'
. i, [D.,Ui;$, 198O
.
uslng echelon form or otherwise'
i
,
1
(tmlrovemcnt)
36
COLLEGE LINEAR +I'GEBRA
SYSTEMS OF LINEAR
bllowlng linear equations by using echelon
form :
ESUATloNs
37
lg.Sotvethefollowingsystemofhneareqtrations:
x+2Y +22 = 2
3x-2Y- z =
5
ID.u.s. 198ll
2x-5v +32 =-4
x+iY+62 = O
-;;;iv * +" = - 10 J
A5rer i x=2,y=-3,2=1.
llasircrB:,
(i) x = L,Y =3,2= L
(li) x'r= ? -
^,
2O. Solve the following systems
y = 2a + 2, z = awlrere a is any rea-l number'
1,
17. Find the general solution and also
solution of the following system of llnear
a particular
equatlons :
2xt- x2- x3'+ 3xa=1'l
4x1-2x2- xs+ &=: I
6x1 - 3x2 - xs - Jq =-v2i - x2 +24 -12& =lO,
p. u. 3. 1e8o I
I
Aaswers : General solutton ,
t
Partlcular solutlon ixl = 2'xz=2'xs=-2'4=-L'
equatlons :
,:*1{. fiotyethe following system of 'llnear
+2x+ + 3'rs = '5' )
X1 t x2
+1& +1xs =- I L
x3
+4x224
+5'rq +2x5 =-?^f
xi +3&
+
Sxi * 7.r2 -3x, 9xa + -r2x'' = -i6J
.:":t-;.r,;\-
7'tq = -
-..o- rr*-:*i,
x1 +x2+ xs =
'
il
Xt-Xc+2x3=-31
(iii) Bn
- i, + 1xs = -.21
zxi-n- x3= 4 )
Answcra : (i) Inconsistent
(ii) Inconsistent (fli) Inconsistent'
21. Find all solutions of the following
Aogtper 3 x1 = t, &, = 2a' xs = d'
where a and b are any two real.npmbes'
i
:
-
system of linear
ID.U.IL r9@l
I
Wr] 3o * u
., ':
,0 = 2,.":x2 =- 3, xs = l' &.= o'4=2'
..:]i
i,
fi #,i'#,=* #i:Y:# f )'""'1esu
#,:
=
2i2 - 4x, - 3x+ -9x5' =i}
S"r * !r;'{.2:#.:Er
xt =ia+b+2
Lxa =b
,*, *q.v, - 4x3 + 2x4
of linear equations :
equations :
(l
,
.' .m-:
=-l l'
'x"+iv+32
5 I
=
/Zx-v
-z'
';".'3u;; =-2
I
d/
\tt^{Y ax+5v+42=3 I
n
i"'. I
x4 = -3b' x5 =5
numbers'
where a and b are arbitrary real
:
22. Solve the follorvin$ linear equations
x1 +2x2 - 3x3 + 1*, =^1]
z,xi + 5i2 - 54 + 6&
x, +4xc- xs
-* I
{ D. u. H" 1s871
=O
2xy + 3x2 - 7xs +lO4= |
+2b' xs=a' & =b
Answer ! x1 =- I + 5a -Bb' n= I -a
I
1
numbers'
solutions of the following homogelleous
where a and b are arbitrary real
23. Find all
syslt'nts of linear equations :
x+2Y-32=Ol
Itl
2x+5v+22=O
I
\"
3x- Y-+z=o )
tD.u.H. re86l
SYSTEMS OF LINEAR EOUATIONS
COLLtrGE I,[NEAI?. ALGT'BRA
38
x1 +4x2 +Sx:i +3& =9'l
2xi +Si + 5x3 + xa =O L
(ii) 3x, + 2$ +5x3 - x+ =9 [
4xi+x2+5x3-3-ta =OJ
ID.U.P, 1e851
lD. u. P. 19881
- xi+2xs-3xa+rb=UL
{iv)-xr xi * xz -2xs - rq =9
xs +.q+.rc5=0
where a and b are arbitrary red numbers'
In particular xl = - 85' x2= 50' xs=- 19' & = L')6 =o
into
27. Reduce the following system of linear equations
I
I
solution'
.&rswers : (i) x= Y = z =O' No non-zero
(ii) x1 = b -a, x2= - (a + b)' x3 = aarrd'q = trwhere a and b are arbitrary real numbers'
(iii) xr = 2,a'x2=-?'x3 =a41%
. where a is arbitrary real number'
(M xr = - o.-b, xz = &' x3 = - b' xr = O' 'rj; =tl
where a and b are arbitrary real numtrers'
'" 24."Reduce the following system of linear equations to
echelon form and solve it :
xr * x2* x3+24+?;g =Ll
-3xs =,P !
- 2xr - 2x2 + x3 - 3xa +cxs=t-Yl
+&
x1*x2+2xs- &+9x5=I8J
,,eryy"'
i x1 = g
-''
"#,:';,::# ;,1;f..",*.
25. Solve the following system of linear
x1 +2x2 + 3x3 + 4x1 = 5l
equatiorrs :
2xi + x2+Axs*-x, =?'!
3xr +4x2+ x3 +5x1 =91
i*i +3i2 +5x3+2xa=31
15
,
-tl
io.
2x2+ x3+3xa- -4
=-ll
x2-O4 +74+-r3* = II
5xi + I Lxl +7x^ +124 -l0xs = 4)
15b'
Ansrrers ! xr = - M-21a- 24b'& = 39 + lla+
xs = - 15 - 4a- 5b' r+ = 3' xs =b'
xr*
oxi - zxi'- 4xs - 5xo= I
2x1 +2x2 - xs
- +x5 =9]
3x1 +2x2
solution
26. Find a general solution and also a particular
of the followtng system of linear equations :
3xz -x3 + lxa - 24 = 4, l
2x1 +
2x1 +
x1- x2- X3- &=9
xi +Sx2- xs*-&=9
(iiil 'lxi -Zxi - x3 - 6xr =!
=O
2x1 +2i2-24
39
83
Ar"y:"i lr= )1,x2=-11' xs= zl' xr= 7t'
echelon form and solve it :
x1 * 2x2 +' 3x3 + 4x4 + 55 = t,l
z"i + 3i2 + 1i3 + \xa + 9o" =:t[
3xi +5x2+ 6x3+ 7+*-*=?l
icxi+Si+ 9x3+l0xa+ S'rs= 3)
7b'
Answer i xl = *7 * 4b' h' ='7 + a + lob' x3 = - 2 -2a-axi +7x2 + t0x3 +I-3xa + lPxs = I I
numbers'
equations into
linear
system of
]$-Reduce the following
echelon form and solve it :
4x1 + 2xz + 5x3 + 7x4 +
-h =?)
X7 * X2* x:t+ 4 *!'q =ll
2n +3x2 + 4xs +5& +6:tb = I I
xA = a.xs = b where a and b are arbitrar)' real
3x1+9xz+7x3 + & +8x5=?l
5x1 + h, + xs +6xa + .16 =UJ
Answer: xl = l, xz =O, '6 = 1,;8 =- l'xs =O'
2g.Determinethevaluesof.},suchthatthefollowing
system in unknowns x Y and zhas
(iii) more than one
{i) a unique solution' (ii) no solution'
solution :
lx+ Y* z=L I
x+)'Y+ z=l l
x+ Y +)'z=l
B.U.H 1S5l
I
Arcw?rs : (i) tr* I and L*-2,(u) i'=-2' {iii} }' = 1'
t ir.rnan ercbgRA
4o
,y'\
Determin'e the values of ]...such that the following
systEm in unknowns x y and z has (i) a unique solution, (ii) no
solution (iti) morb than one solution :
,( [9
^ x
-32=-3 )
)zx+xy- z---2I
1 x+2y +Az= I
AttJw.*: (i) tr+2and l"+-5'(li) l'=-5,(iit) l'=2'
31. Deterinihe the values oI )' such that the following
J
I
system of linear equations in unknowns x, y and z has {i) a
.r.riqr" soluilon (ii) no solution (iii) more than one solution :
-
x+ Y+)'z=L )
x+l"Y * z= ?tl
Ix+y + z=)tz )
Answers: (i) )' * -2, )"* f (fi) ?t=-2 (iii) lt = I
32. Find out the conditions on o, p and y so that the
followtng systems of
a solution :
'non-homogeneous linear equations has
x+ 2Y- 3z= cr I
(i) 3x- Y+ 2z= Pi
2x- IOY + 16z=21 )
x+W- 9'= g I
(ii) 2x+6Y-tt'=-9 I
2x-4Y + l4z=21 )
(ii) scr =29 +Y.
Answer:(i)2cr=P-Y
33. Find out the condltlons on a, b and c so that the
of non-homogeneous linear equations is
consistent and also solve the system for a = l, b = l "'d " ='- 2'
2x+Y+ z=al
x-2Y + -z =b I
x+ Y''22=c')
Answer : Condition for consistent is a + b + c = O and
.-the,Emerul solution is x= cr- l,Y = d' - l'z=a
where a iS arbitrary real number.
A particular solution,is x= l, Y = l, z = 2'
:..
, .
:,,,:
XifiBEAICPnnA
&1 Iatrpfrcfu
J. J. $y{,tprter was t}rc ffrst man sho intnoduced the uprd
'metrf in 1850 and later on in 1858.fr&ur Crytq, de\rclopad
the theory of matrices in a systemaflc
-way. Matrjx ls A
powerful tool of modsn, mathematics wh{h ie origtuated in
the study of llnear equa$ons, and it haq wlde appliqa,sons itr
every branch of sclence and especlally tn Physic+ Che-rnistry,
Mathematics, Statistics, Economics, and Engineertngp.
3.2 Dcfri6oa orf nair|r
A metrir is a rectangular array of nunrbers (real or
complex) enclosed,by a palr of brackets (or double vertical
rolls) and t}re numbers in the array are called the enfies or
the clemcntc of the matrix. that i1 a rectangular artay of (real
or complex) numbers of ttre form
is callqd a natfk. The number.s d1r,ilre, ... ..., amn are
called the entrlec or the etreroeuts of the matrix. The above
matrix has m rows and n bolumns and ls called an (m xn|
mat1k {read "m by n matrix"). The matrix of m rows and n
colunrns is said to be of order "m by n" or ra x n. The above
matrix is also denoted by Iary],
..']
.,';rl'3:
Tlre m horizontal n-tuples (arr, are, ..., ar.,), tazt, d.zz, ...,
az.,), ... ..., (ornr , &m2,..., 2*r) are the m rows of the matrix and
lhe n vertical m-tuples
::
MATRIX ALGEBRA
.COH,.EGE LINEAR ALGEBRA
""n "of"t
Addltion of matrices is deflned onl5l for the matrices
having same number of rows and the same number of
I
L
;-,1 L .,", I
L a,,,, r
are its n columns. The element aU, called the ij-entry or
ii+omponeat, appears in the i th row and j th column.
A matrix consisting ol'a slngle row is called a row matrix
(or now vcctor) and a matrix consistir:g of a single colunrn is
called a column matrir (or cotumn vertor).
Matrices are generally denoted by capital letters A, B, X, Y
etc. Square brackets [ ], or, curwed bra'ckets ( ), or. TWo pairs
of parallel lines I I I I ar. used for the mathematical notations
of matrices. In this book we will use the notation [ ].
Emmples of mstriees
ymprc ,.o = [; -3
columns. kt A and B be the two matrices having m rows and n
columns.
[-
fl
:l
Then the sum of A and B is
+b11 ap+bp
I
,rra
J
31n *b1n I
a21+b21 a22+b22
z2n+b2n
a+g=lI ...
I
I
b-1 a.,,2 *b,,1 ?inn * b-r J
The multiplication sf matrix hy numbers (scalars) is
[-ar,1 +
defined as follows : The product of aq (m,r n) matdx A by a
number k is denoted by kA or Ak and is the ( m x n) matrix
obtained by multiplying every element of A by k, that is,
r2
0 i-1
-i 1 4llsamatrixof'order3x3
I r+i -s sJ
[-
karr ka,
I
...
lrarr.l
u""
ta = l ka'' kazz
D"ramfle2. B=l
L k.-, ka-z
over tJ:e complex lield C.
The rows of B are (2, O, i), (-i, 1, 4) 4nd (1 + i, - 5, 3) and,.its:
zir
-i l, I r I ana [+ l.
[ t*i/(.-5)
I
\3)
the5r are identical i.e ,if and only if they contain the same
az,-
J
[-a1 1
*. (;) LS) ,"o ft)
TWo matrices A = [ a1 I and B = [ bu I are equal ifand only if
oh I
I
-?l i1 a matrix of order 2 ;i'il;
/ 2 . , Or
a,r dn.
a2r dzz
, L;, ; ::: ;;"
br., I
I brr brz
n=rhr= | ::, i.: .:: ?"
b-"
L.:b*r bse
i. e. A = 13,,1 = |
tlle real ffeld IR and also over the complex flcld C.
The rows of A are (l; O, - 5) and (2, -3, z) and its columns
columns are
nunfpUolUon of matrlces :
3.3 Addttton
irii'l'f arz I
[ri" I
lorrll^rrl
l^r,
1,1.1,,...,1,1
I-
95
I
number of rows and the same grmber of colurnns and a1, = b,
for all values of i and j.
I
I
ka*,, J
We also define -A = {- 1) Aand A- B =A + (- B).
If the matrices A, B,'C:are conformable for addition and if
k is any scalar, then we can state that
(i) A+B= B+A(Commutatlvelaw)
(ii) (A+B) +C =A+ (B +C)(Associativelaw)
(iii) A+O=O+A=A
(iv) k(A+B)=kA+kB={A+B)k. :
.t: ..
a'-'
".-
where O is the zero matrix of the same;order.l r ri$';rill:.,,.ir'. ,; 'r,i
,d
97
MATRIX AI,GEBRA
g6,
coLIEqE UNEAS, AITEBRA
] *"= [-3 'nJr,rs,^*",;Ill,uu, n'J;{ }=[-i 3]
For examph, ,r^= [
I
-?
I= [8 To ]
^=l?iL ?.u?
-?--'ul =
ena e-"= [:?-u
[-l i ]
g.4 llatrtx h$!fifontton
TWo rratrlces A and B are conformable for multiplication
if the number of columns iu A is equal to the numder of rows
in B.
R,
'
Ib' I
I.etA =,tir az .... a.l and
"
=L
jl
rn
The AB = erbr *,adb2r ... + a,,hJ =L
.,Jr
I
P=r*o, -.,
Alaln, lOt tle rn x p matSlx 6 = [agl and the p xn mptrix
h l. thenAB i*&e m x n *"T C = [ q"l wtrere
p=I
cs=anbr1 *"oba +... +fu h
ti= | .l',
?3ib5
sums and products, we havg the foIlowing properties :
, , (it (AB)c = A(BC) (Awcl,ntive law)
A(B +
(ii0 (A+BlC=
"' :f"
(iv) k(AB) = (kA)E = A[kBlwhere k is any scalar.
lgSl (Eistribudve lars)
RemarLs : In the matrix product AB, the matrix A is called
lhe pre-rntltlpllcr (or Pre-factor) ancl B is called the
postmuttlpltcc {or p6t factor) i
"
=[
;
-1 I
-1
l*0"=[-i 4lol
I
[] I;5:l-,1-?;'i:3 L 'ti:';:Lf,1,?'.
-::l
= t??
I
= t;13:'f -'r-*'Zl3
-B
uo =[ -l il
-il
t;
L3 0l
(-l)'o l^. I * (-1) (-11 I
r 1' + Ftl'z t'
4 q'].
o
=L
=l i-;r[; i i-it-&)++ ?)?* o' (-r) I
thenAB=
(-3) +
1
b-: rio- z s Fs) +'o o
r-1 -3 6l
6 -I4 l.'.AB+BA=l 6s -e
151
L
'
u,l
I
3;5 TxansPose of a matrir
real field IR ' then the
If A is an m x n matrix over the
the matrix O OI yTt"U lo-:o*t
n x m matrix obtained from
of
as rows is called the transpose
as columns and its columns
,s' that is' if A = I aij I is an
A and is denoted by the symbol
an n x m matrix'
m x n matrix then 'S' = [aJi Iis
',r
' If the matrices A, B, C are conformable for the in&cited
(,)
For examples, let
-a
Forexamples,
leto=[] 3 ? -[l'
3.6 Compler co4iugate
T 1 21
ur".,N=l
i6.1
| -z
g
I
(or conjugatc) of a matrix
z' x.r iy is the
The conJugate of' a complex nurnber =
A is an m x n matt"ix over tire
cornplex number z = x- iy' If
A is
Lhe conJugate o[ a rnatdx
r:ornplex field, then we say ttral.
l'hc <:nrtjui':l1cs
i!tc nrll'[rix A whos* fi:lttTteljJ-si n.,..11 11.1.11,;pf:Ctivciy
t'
is'il'A=[n'i ]'tlren A =[Ziii l'
r-,il.lre:elernentsofA.'l'ha1
1.irtr".ri- lUgebra-7
r1
98
Z.
COLLEGE LINEAR ALGEBRA
ifl
l+ i I
I
,
For examples . if A='l -i
Ls 1+2i
rl
-t
A=li
t-
-i
- ,t 5
2
2
L-2i
2+3i
9g
MATRIX ALGEB.RA
igslan8lular matrlx : The number of rows and coluntns of a
-#
-,--_.
I
-5 J
1-i r
2-sL
_t I
.
-5 l
,rrrl =.. :*
=aral,
er of rows and qoltlmns qf ths
then the matrix js known as the
reetangulas.maEix:
/
A matrix A-lq_-c,allecl real provirled it satislles the relation
A=A
Definition Imaginily matrix
A matrix A is called imaginary Provided it satisfies the
relationA=*A
@.:IEEIe*r,I
1)re conjugate of the transpose of a given complex matrix
-,.furexamples,
-i 21 3I
-l
2 o s lrrxl_
?6lyli
I -L
are rectangular matrices'
Dj1g9".1 m"lg : Arytelatrix whose element: :iL: O
wry"s$511*
Forexampre",Ii 3l*.LB
3 ?]
are diaeonar matrices'
A is said to be the conjugate transpqse of A and is generally
denoted hy the syrnbol A*. That is,
if A = [aq ] is a comPle-x matrix, then
* * 1(e= IQil.
i, 'rl;, I
o=l--l
.o,.*rffi
I s t+2i -5 I
1'
i 1-r, l
4,,r.=[-1
L r-i 2-3i -5
-l
/*raif B = lt'*'n" ;f;, ?;z:
/*"r,8*=['-* ,:i1
[2+5i
-l
3-2ii
S/Gpecial types of matrices v{th*exarnples
Squite rnatrix : A gglru< with t hc _same- nr:mbet :1l9r""3"d
colunrrrs js t'alled a square matrix.
fr
Lt"*il--i"
;ilr.".[Tl
3] ",otg
[[]
arezeromatrices
l"r
Upper and lower triangular matrices :
an
A square matrix whose elements aU = O for D j is called
upper triangular matrix and a square matrix whose elemcnts
trr J
a,j = O for i <
J
;,=.ffi
matrices of order 3 and 4 respectively'
zero matrix or null matrix-: A qattl']5lnJvhich--srcry
,ero is called a rllrllJnatrix 61 a zero rn?trix'
' s lor. square tnatrices.
j is called a lowgr triangutar matrix'
i
roo
MATRXATGEBRA
COLLEGE LINEAR AI,GEBRA
Forexamp'.",
[Lii] *,.[B i
2 3l
In this case dlagonal elements of the matrices will be real
i 5l
3-2
o 7)
numbers.
I
Skew-Hcrmltlan matrlx : If A = [ a,J I is a square matrix
overthecomplex{Ield, andA*=( 6t;=-A i'e 3u=- 6ir for
ij = l. 2, ..., nthen A is called a skcw-Hemltlan matrix'
are upper triangular matrices.
r 5 o o.l I I 3 3 3].*lowertriangurar
matrices.
-i;l
"'dlL-r2ol*rol-Li
sztj
For exampres, A =
Symmetric matrlx : A matrix equat to its transpose i'e q
square matrix such that a1; = oii for I s i' j " t' is said to be
symmetric. In short we can say a square matrlx A will be
L-s i
are sJrnmetric matrices.
LA and l.B are also symmetric if l, is a scalar'
Skbw-symnetrlc matrk : A matrix equal to the negative
of its transpose i,e a square matrix such that 2U = - a;i and in
-l
- L-e-folland Lz-s
oi
|
are skew-sYmmetric matrices.
Hermitian matrix : If A = [ ai.1 I is a square matrix over the
complexfield and A* = (Ar)=A i'ea1= aji fori' j= l' 2""' u;
rr. i
I
;,1 -.1
l- t *zi
.J
J
I --i
I
:j
: _-;
li
jrli'r: aii.'r'S;
lareorthogonalmatrices.
_*
I
t
L-s s
l
0j
;;
r
I zz -3J
.aI .+i | ,r.,rl
I
t_-t
.
' 12 2-3i
i'rrt. ex;irnples,, A,= |I 2-rlji 5
':
,/
4t
lq
9 9J
L9 -L
r | 2 2-1
l'55 3l
2l
B=lI zt -A
t-21
g=lIo O 3
then A is called a Hermitian matrix.
ol
r l 8 41
ls
o -0z
I a, 4 l
For examPles, A=l O -O -O l*d
./.,
which therefore, du = O is said to be skew-sSrmmetric'
Fore:<amples, A=l-h
rt "",
.]
In this case diagonal elements of the matrix will be either
zero or wholly comPlex number.
said to be
Tdhogoaal matrk , A t@is
orthogonal tf AAT =AT A = I.
ttrat is, if.S = A-l (the inverse of the rnatrix A)
i:landB=L_A ?
ro Ohgl
f
L-1-8, _r:i
r i l-1 5l
2i i Iare skew-HermiUan matrices.
n= l-r-i
l- 12 -3,r)I
t-ahgl
exampres,"=L!
21 2-31 3l
f
L-A; -i;l
symmetric if .Ar =.{.
}OI
it
v
{
\
idempotent ry!!"_if A14
'
-41
3
I
-s
-3
Forexamples,
I
lanal-t
' [-t 3 5] L I -2 -34ll
are idemPotent matrices.
COLI-EGE LINEAR ALGEBRA
LA2
Nilpotent matrix : A square matrix A is called a nilpotent
matrlx of ordern if A'r = O and An-l + O where n is a positive
integer and O is the null matrix
3.9 Ttreorems on transpose matrix
Theorem I If A and B are comparable matrices and Al and
Br are the transpose matrices of A and B respectively.
3t
For exampt.", e =
:
r03
MATRIX ALGEBRA
[3 ] I ,'n ,=[-i 3 sl
then (i) (Ar)r = A (ii) (A + B)r =;S + Br.
(iii) (AB)r = ISAr (iv) (crA)r = cr AJ. where n is a scalar'
are nilpotent matrices of order 2.
Periodlc matrii : A square matrix A is called periodlc if
Am+l - A where m is a positive integer.
If m is the least positive integer for which Am+l = A, then A
Proof : (i) rrt A = [ au l, l=rl;1'...,'7
then bY definition ,{' = [a; lr = [Ei I
Now (AI )r = [qi] r = [ a,:l = A .'. (,$)r =,{.
is said to be of 1rcrlod -j
r ,I -2 -6-t
(ii) t€tA = [ air ] and B = [ bu ] whereS
ill f'...: l'
then C =A+B is deflned and [ -qL I = [aiJ I + [ b1 I.
Now by the deflnltion ol'the transpose of C
we have (A -t B)1 = Cl' = [glr = [qrl = [a1i ] + tQi I
period 2.
-]horotutory matrlx : A square matrix A is called an
qff
iivolutory matrix f A2 =1.
For exampl., o = [-3 -? I ," an involutory matrix.
={
qf+{\
lr =.S + sT
.'. (A + B)r = Ar + Bf.
(iii) LetA= [au I i= 1,2,..., m; j = l, 2, ...,n
Unltary matrlx : tet A be a complex square matrix, then A
is called a unltary matrlx ,l** = A*A = I or equivalenfly
A* = A-t.where A*=.:fi'r= 61).
Then AI = [aiJ ]T = [ ajil is arr n x m Inatrix
gr = [h. l, = IbrE I is a p x n matrix.
Normal matrlx : [rt A be a complex square matrix, then A
is called a normal matrix if A*A = AA* where A* is the
conjugate transpose of A.
no, example,
[ = [ '.ui ,,.L,] is a normal matrix.
,'t"
:
ThusABisamxpmatrixsothat(AB)r is a p x m matrix.
Also BrAT is a p x m matrix. Therefore, (AB)r and BrAr ltave
$
same dimensions.
NowAB = [ ct ]where (i, k) th element of AB is
cu.
3 ; where i = l,2, ...' m
=i , ?U DJr,
k = l, 2, ..., p.
J=l
104
since i 1 +22 = {Tz2where z1 and
"lherefore, trre' (t;iith''i:lement of (ABlr
,.:i=I
=bul=[rFo,,]r"t"u1<i sm'I sn'(2
sism, 1 sJ <n'
Therefore, ATB = [ fu;b;] r"tru I
A + B and AJ-B are
Since the corresponding entries of
are matrlces of order
equal and also both f + E and A;E
<J
n
T
L.
j=l
Jl
= (k,i)th element of BrAr
Hence (A6;T =g1O'
I
.'.
$ = [a1 lr = lajrl
Now (ctA)r = [cxaijlr = [craii l= c taril = aAI
(ae)r
crAr'
=
"'
amatrix
3.1O Theoremson complex conJugate of
matrix' then
x
complex
n
m
is
any
Theorem 1. IfA = 1 a1; i
A=A.
Proof : By definition of complex conjugate of a complex
ao isttre
matrix, wehaveA= taulforlll I <i<m' I SjSnand
in the i th row and
complex conjugate of a1; i'e iu is ttre entry
< (
and
j thcolumnof A. Again.E= Iiul for { I < i< m' I j n
(the entry in the i
E. i" th" complex conjugate of l.i i'e 5'o = ai1
# -," ""a i th colum, of tnt mitrx A)' Therefore' the entries
tJre corresponding entries of A'
of the complex conjugate of A =
Hence A = A'
Also A and .E are matrices of the same order'
T.heorem 2. lf A and B are two complex
conformable to addition, then ffi=[
matrices
+ B'
of order
Proof : Let A = [ai1 I and B = [br] be any two matrices
< i< m, 1 Sj S n'
m xn. Thenwe tar.e + E! = [ au + biil for alt 1
Also by rlefinitioll, we have A = [tul and 3 = [biil
.A
z2 are arry two comPlex
numbers.
+ b1]
Again 6- +B = complo( conjugate of lary
of c,i where -Q; = ar1 + b,
= complex conjugate
nn
=f,' alib;x
- =Ij=I ajrt b$r
0v) LetA=[aij
105
MATRX AI,GEBRA
COLLEGE LINEAR AI,GEBRA
*e = 6,il * trul = t a'F5rr,
=, =
*,
rJ,=,
=
r,
A;B ='6' +g'
m xn, from (1) and (2)' we conclude that
B [bplbe
Theorcm g. I-etA= [aljlbe anymxnmatrixand =
the complex fleld i'e A and B are
any n xp matrix over
;E = A e
conformable to the product AB' then
to the product AB'
Proof : Since A and B are conformable
I < i J m'
Ib1r.l=[cL lwherecft = atlQ1 for all
"oeg=[ai1lx
lsk <pand 1Sj<n.
Rtso A=1ir;forall 1<i3m'lsj sn
and 6=l h* J forall 13jSn, I 5 kcSP'
NowABis defrnedandwehave AB= tqt x [bEl = F" t ttt
I<i<n'
wtrere a,*=- aq+.;x firrd-+<i5-m' 1<k<<pand
Again ffi = comPlex conJugate of AB
or. AB = [cr lwhene cil = qJ h.
a,r brk
1i, u-,r
-J^ t
-- f[ -u
"Jl( I -= r-u
= [d*l (2 since
"{'" = ZrZ,
for any two complex numberd z1 artdza '
1<ism, t=Oi:pand 1 (jsn'
Sinced,k=irEuforall
A B'
Combining (1) & (2) we conclude that AB =
matrix' then
Theorem 4. L;et6 = [a11] be any m xn conrplex
),A = tr A. I ueing any complex number'
r07
MATRIX ALGEBRA
COLLEGE LINEAR ALGEBRA
t06
Proof : By definition of the conjugate of a matrix, we have
A =t;ul foralll<i<m,
1<jsnand 2ii isthecomplex
(ii) BY de{inition' *: h^'" ".
(A+B)*= 1A+n)'=14+B)'
=(A)r * tB' since(c+P)t=cI+Dir'
A* = ([)''
= A* + B*, slnce
conjugate of atj.
trtforall I <i<m, I sj<n'
^;;l=t ^ "11l
slnceilz'2 =;tr2where \ andz2 are any two complex
atso 1.e- = [
numbers.
Again 1 A, = h1l*t.t
Hence 1.A = )' A .
3.11 Theorers on the conJugate transpose ofa cotnPlex
matrix.
, Theorem 1. kt A*and B* L,e the conjugate transpose of A
and B respectivelY, then
(i) (A):A
(ii) (A + B)* = A* + B*, A and B being conforrnable for
!**
addition
(iii) (AB)* = B* An, A and B being conformable for
multiPlication.
k being a complex number'
Prqof : (il l.et A*= B, then B = ( Ar) = (A)'
and Br = (t At')'= A
Again (E)t = 1fr = F= e
Therefore, B* = A, since B* = ( E)t = (B')
Hence (C)* = A, since B = A*'
= (AB)t
= F)t(A)r since
bl= Ifu forall I si<m' I5j Sn'
=t l4lforall'l<ism, 1<jsn (21
Thus from (1) & (2)' we conclude that the corresponding
entries of l'A and 1 e are equal and they are of same order'
(iv) (kA)* = [ A*'
(iii) (A8)* = (ABf
(CD)r =ilCI'
= B*A*-
=fA*
(iv) GA*= o,Af = G[)t =[
matrix A' (IA)*= IA*
Corolla y : For any complex
(R)"
where l, ls a sealar'
9.12 Theorems on sYmmetrlc
and shew-sYmmetrlc
matrices.
matrix can be uniquelY
Theorem 1' Every square
matrix and a skew-
a symmetric
expressed as the ""* of
symmetric matrix'
and 'S be its transpose'
Proof : t et A be a square matrix'
Then we have
+c (sav)
(1)
i 6 +.tr) and c =| te * ffi
Now Br =*jo+ F)r =| o+
=ir* *\= n
(s)r)
j
and cr =i,ro- ,{)r = r+ -
(2)
o= i B * f) *; te-'s) =B
A1)
where u =
=jr+-ot
= -Lro- ,$) = - c'
r09
MIITRIX AI,GEBRA
r08
COLLEGE LINEAR AI,GEBRA
Thus B is a symmetric matrix and C is a skew-symmetric
matrix. Hence from (1) we conclude that a square matrix can
be expressed as the sum of a symmetric matrix and a skewsymmetric matrix.
To prove the unlqueness of the representation of (l), let if
possibleA=P+O (3)
where P is a sJrmmetric matrix and O ls a skew-symmetric
matrix so that PT =PandOr=-O
(4)
Ttren,S = (P+QIr =Fr+ S=P-O
Adding (3) and (4) we get
A+Ar =2p ...f=2(f +A)
Again subtracting (4) from (3), we get A -,$ =2O
I
...8=;(A-s)
This establishes the uniqueness of (l).
Hence the theorem is Proved.
Thcorcm 2. lf A is a square matr'rx, then A + Ar is
symmetric and A - af is skew-s5nnmetric'
Ploof : Eirst Portion
If A is a square matrix, then
6 +.$)T =,$ + (S)T =Ar +A =A+'qr
Hence by definition A +,S is symmetric'
Second Portion
(A - Ar)r = d - (,s)t =.S - (.{r)r =Ar - a = - (A -'S)
Hence by delinition A - Af is skew-symmetric'
Theorem 3. If A is a skew-symmetric matrix then
AAr =,SA and A2 is symmetric.
Proof : Flrst Portion
Since A is skew-s5rmmetric, Ar = -A
...A,4r=A{-A)=-A2 (i)
andAlA=(-A)A=-A2 (ii)
From (i) and (ii), we get AAr = Ar A.
Second Pordon
(a.6t;r = (N')r Ar = AS Since (,s)r= A'
and (NA)r = Ar (N)r = 'SA
matrices. Therefore,
... Iu\T and .qT A are both symmetric
-A2isasymmetricmatrix.Againsince_lisascalar,
A2 is a sYmmetric matrix'
Note : If A is a symmetric matrix' than
scalar'
s5rmmetric matrix'where k is any
kA is also a
matrices
Theorem 4. If A and B are both skew-symmetric
then AB is s5rmmetric'
of same order such that AB = BA'
matrices' then
Proof :.If A and B are both skew-symmetric
AT = -Aand BT =-B
i'e A=-N andB=-Bf
Also given AB = BA = (-Br)
(-AI) = BrAr = (AB)r
(AB;r'=66'
"'
Thus AB is a sYmmetric matrix'
Theorems.IfAandBaren-squaresymmetricmatrices,
and B commute
then AB is symmetric if and on{ if A
(i.e AB = BA).
AI =Aand Bir =B'
Proof : Since'A and B are spnmetric'
Now (ABlr = (BA)r =
NyriP
.'. (AB)r =AB.
./
Thus AB is a sYrlu{retric matrix'
ConverselY , t/=
(rrB)'i - IfAr= BA
S;inc'd llr'= IJ a-nc1 Af = i\'
I-lenee A ancl lJ cutnrtrr-tte
Retnrtrk:.[.lrllrttittrixl}1.A1]issYrrriltr:tri<li'ii"Sk(.r1.\jl]yrtlrTlctri(:irr'cortlirl$il:iAisjsyt::.trtt,:[i:ic1i1.5[,;111.,,-l:i.ytnrrretrit''
tll
MATRIX ALGEBRA
COLLEGE LINEAR ALGEBRA
tIo
3.13 Theorems on Hermitian and skeu'-Hermitlan
matrices
can be
Theorem l. In a complex field every square matrix
matrix and a
expressed uniquely as the sum of a Herrnitian
skew-Hermitian matrix'
A* be the
Proof : t,et A be a square matrix of order n and
writ€
conjugate transpose of A' Then we can
(r)
*;=j;.;.
L*o-A*)=P+Q(sau
where, =L,W+ A*) and g =
Now P* =
'
= BA +AB =AB
Hence nb + BA is a Hermitian
Again (AB - BA)*=
+ BA'
matrix'
= tsA-AB
=-(A8-BA)
:
=){a"*A) = P
Hence AB - BA is a
skew-Hermitian matrix'
Theorem 3' lf A and B
are Hermitian' then AB
is
. Hermitian if and only if A and I commute'
g* =i,(A - A*)* =f, ra* - (A*)*)
=jta*-nt
1*
=
Hermitian matrices' 'we
B* are the complex'coniugate
o.=-o1]ro';{; *i"t" ". and
,r.rr"ro""" of A and t"ll.t"tT-"1'r;
(AB)'' + (E}.|)
Now (AB + BA)" =
= B*A* + A*B*
Proof : Since A and B are
B',I_ f.?-
j e -a*t
i ,o + A*1*= i {t + (e*)*)
*.T:l"es then
B
and
A
rf
2.
Theorem
T:.T'*"t11
t'la AB - BA is skew-Hermitian' have
AB + BA is Hermitia"
-)o- A*) = - o.
rnatrix'
Thus P is a l{ermitian and Q is a skew-Hermitiari
expressed
be
can
A
Hence from (1) we see that a square matrix
a skew-Hermitian
as tlte sum of a Hermitian matrix P and
*"1T
represe.ntalion^ (1. l."i' o
$rorr" the uniqueness of.
A R + S (2) where
possible, A be also e"ft"""iUfe in th.e.fbrm =
such that
R is Hermitian and Sls skew-Hermitian
R*=RandS*=-S'
Norv A* = (R + S)* = R* + S*= R- S
A*) = P
(3). we Set R =
(3)
)W.
Subtracting (3) Iiorn (2). lve $et S = )W-A*) = 8'
(1)'
which establishes the uniqueness of
Hence the ttreorem is Proved'
Hermitian'
Proof : Since A and B are
(1)
A*= A and B*= B
AB' since AB = BA'
No*'p1gi* = B*A* = BA =
He:mitian'
Thus'by definition' AB is
BA
= 9u)" = 9*6*=
Ag;ain
"o""'""''';
Since A and B ale Hermitian
\
\
1
can be uniquerv
matrix
H::l#t?:'ffi$"square
g
*t'ol p and are Hermitian'
'
expressed as P + i8
of order n and A* be its
matrix
square
the
be
Proof : kt A
conjugate transPose'
Then we can write
{.,
1
*i
;=i;.^.' ;;;; -a*)=.1>(a+e*) le-a*t
=
*
-
*
*o*t and Q = | (A - a-)
I) + iQ (savt (r)"wrrere P = |fo
Now P* =lrrn* A*)* =
| {e* * (A*)*}
ll3
MAf,RD( AI,GEBRA
r"rii'r'i
COLLEGE LINEAR ALGEBRA
tL2
O,ta Scorrammi'lqofg't'otrnrtrlcca
rnatrlces ttrpn AB is
f : f iJ e *" idempotent
tlicarcm
ra"mit rrt 663 = BA'
=;d+A)=in*A*) =P
n. = L\ 1o - o*t)*=# 16* - 1a*)*l
=-fita* -A
=fio-A*)=o
(1)'
Therefore, P and Q are Hermltian matrices. Thus from
and Q
we see that A can be expressed in the form P + rO where P
are Hermitian matrices.
(1) let if
To prove the uniqueness of the representation
A = R + i S (2) where R and S are Hermitian' that is'
possible
R* = R and S*= S.
NowA* = (R+ i S)* = R* -iS* = R-iS
(z) + (3) gives A + A*= 2R
are idernpotent matrlces
Proo(: Slrnce A; B
Also gPen 63 = BA'
A2 =Aand BF = B,
'.
(BA) B = AjaE) B
rro* gos;i = AB BB) = A
!
(B'B] = tP * = ltB'
= [A'A]
matrix'
Hencc AB is an ldemPotent
then A + B
are idempotent matrices'
g
a;tB
If
:
trkor@
only if AB = BA = O'
*iU il Ue*eot"nt tf and
areidunpotent matrlces
P.imf I Sirtee A and B
"/,}=AardFF=B'
(A + B)2= 6 + B) 6 + Bl
:X:#J#ii,;il=o
= A + E slnce
I
R=; (A + A*) = P
(2)- (3) gives A - A* = 2i S
...s=*1e-A*)=o
whichestablishestheuniquenessof(t).Hencethe
theorem is Proved.
Remarks : (f) If A is any n-sguare complex matrix then
A + A* is Hermil"ian and A - Al is skew-Hermitian'
(2) If Ais any n-square complex matrix then AA* and A*A
are both Herrnilian.
(3) If A is l-lermitian matrix, ttaen i A is a skew-Hermitian
matrix.
(4) If A is a skew-I{ermitian matrix' t-hcn i A is a
Herrnitian malI I-\.
(5)Aisl-lcr.rrtil'iariiIirrtr]oniyif.Ai.stlerrnitiirrr'
t6) .A is sltrw-i'lct'tltiiialr if :r'nii ';tlh' ii A lSr sl'ir'"iv
Ilerrtril i;tn.
:
Nor O* = ;; = 0' ttren
(3)
.'.
'
A? ='&'E?'=B'
HenceA+BisanidemPotentmatrix''
matrix' then:
Agairr if [A + B) ts an
idempotgnt
"'
'; i:';': 'r]
[A+BP-A+E
I
or, A2 +AB+BA+BP'=A+8"
or, A+AB + EiA+ B =A+ B
or. AB+.84r-0" - :'' ''
or, AB=-AA (11 .
:'
:'
'
'r ' :"-'r:r':' lii
r:'
"''
' ^"0'b-.44-'*aOu'6FqA'=-(ABIA
'
!'
:.-.t.t;FAts 7,EA?'t-"Be
-'^.'
+ BA:
o
r
we get aa + AB; - BA
Addkls f rf
' I .,
i. e. 2AB*o, .:.aP,t.o+-BA.'.i_
-^*
if and only lf AB = o = EiA'
Hence A + B'is'ide.inpotent
therr 13 = I *A is
3. If A is an iclempotent-rnatrix,,
Jtzl'
.:'
Theorem
idemPotentandAB.=O=.E$'' -''
:'
l:\1
coLlFrQPrr,rN.p4rrncEBRA
= lA = A
:'Al$o:since. A is idempotent, A.2 A
=
Again since I and A are square matrices, so I A
- is also a
squzrre matrix.
Now ( l- A)? = (I. A1(I-A) = I _ tA_AI +A2
Asain
,'. AEI = O
... EIA = O.
n such
*,rampre
= B, then A and B are idempotent.
Proof c ABA = (AB) A = AA = A2 (l) since AB A
=
AlSo AEA = A (BAl = AEl.= A (2) since BA B & AEt A.
=
=
, F.tom (I) & (2), we get A2 = fi ... A is idempotent.
Similarly, we have ,
. :
Therefore, A and g are tnvertble and
other. That is, .[' = B ard d =[:
-]l
are inverses
',o
-il"u'=f-'3
i i]
r'-o'ndcr.,.,o=[i
^-Loi? aJ- L 5-2 2J
12 O -lll- 3 -l ll
.
p""*=l3 I gJ[,9 _; -Z)
D = O, the matrix A {qcalled,the sl4gulif nirtrlr,and tf
D * O,
the.matrtx A ts call,ed the loa-flng& rnetrt.
As for e<amples
2 3']
8 eJ
.=li 5 6l
are singular matrices,
sinceDl = h[=O,D2= lB l=OanODs= lC l=O.
[-3
-?r**'nl=[l
&16. Silngular end non&3iular ntrdGor ' I,et D be the determrnant of the sqtlare matr!( A, then
if
i;]*
*
ll
" ta -"1
?I='
=[3-3
tll ]l= [-3;3 3-.11 =[i ?l =i
uo= [-3
E
of each
= B tAB) = EIA= B (S) strrqeAB =A & EI{ = B
AIso BAE| = [BA)B = BB = 82 {4) slnce EIA * B
Ftom (3) and (4) we get B?
=B.
... B is idempotent. ,,
l
^=I; 11,"=l_i
'
*."*= [3 ll I-3
BAB
.
"
' ''
,'
I
3.16 Inversematrlx i
there exists a
if
lnvertible
A square matrix A is said io be
I where I is the unit
unique matrix B such that AB = BA =
lnverse of A & is generally
matrix. We call such a matrix B the
inverse of
;i|"'u o, ot Here we have to note that if B ts the
A. then A is the inverse of B'
'u
=
1. I€t =
BA = (I -A) A = [A - A2 =A-A= O
Theorem 4.If A and B are squ.rre matrices of order
" Ll
]ri;;' = tit=-6+oandD2= Bl=18*o'
by definition I-A is an ldempotent matrix.
Again AE! = A (I-A) = AI - A2 =A-A O
=
that.AE! = A arld BA
12r3l134-11
=
_; ?]' =Ll _? ?l
are non-singular matrlces'
=I-A_A+A=tr_A.
'So
r15
MATRIX ALGEBRA
hoof : We know.t*ratan
:
r 6+ O- 5 -2+A+2 2+O,-2-l
=l rs-rs* o -5+6+o
L o-15+15 o+6-6 3:3131
rl o o1
=l o I ol=I
.13 Lo oI rJ1r2 O -ll
-l
=L-!u -8 -z
3l
llr I
i
COLLEGE LINEARAITEBRA
116
F 6:5+O O-l+l -3+O+ 3l flOOl
-l-so+3o+o o+6-5 l5+o-15 l=lo I o l=I
L ro-ro+o o-2+2 -b+o+ oJ Loo tJ
I-ar r
I-etthematrix
other. That is, .[l = B and d = A.
ah I
f-arr arz
azz 42n I
, I azr
... ...
.-.
I
t
a." J
then D= la l=
r
1...
"'
..f
l;t * ,"' t """
azn
Lt'' z1e "' Bnn
a t =DAdjA=-m
L€t D be the deterrninBnt of the matrlxA,
a,
ral" I '
l"' ]atz
lazr 4zz . ..i azn
i
a22
- la"
A=1...
of the matrix A' Evaluate the
I-et D be the determinant
fuas no
roatrix Atg atngElsf, and:it
determinant p; if D = 0' thi
ls noa-eingufar and A-l erdsts'
inverse, if D * O the matrix'A
of ttre matrix A then
Find the adjolnt matrix AdJ A
r
AdiA'
3.f76dJoht or adJugPtc natrlr
:..
L"", q,
lnverrce of a squarc matdx
aln
iltz
3.18 Irnoccc! offrdng t.hc
Therefor.e. A and B are invertible and are inverses of each
Let A=l
I ...
rt7
MATRIX N.GEBRA
Art Azt
Ant
An
ET El
EI
I ,l ,
,'. rfl =
I
Irt4{ $1 ,,?,$ .::, nandJ = l:2,3: ...,rt '
.:
EI E]
matrlx
' , . t. .:- i..,
Aarrdb{Bnenallydenotedby,AS,A.,
:
A"r -l
Arz
Ar'
[-Arr
l- l'Arr Azt
I Az, Azz Az,,' l' I Arz l,1zz Anz I
.:
:" : I :' '1 "'' "'
I "'
tA"r. loz ,-. Aoo J ', Lt$,+,, &,
, i i ' "''; '.
AS fOf e1alnDleg ,; :i r ;,i I t'. ';';;.-,i
7.;r
r
firi.gf
i !''
Let R=l I 3 -l luren
1j ''i i
I
An,, J
3.19 Theorems on lnverse
of the n-squarex$atr,( A'
Theo,r.n I : If A-l is the inverse
then A^FI = A-rA = I' where
order.
I ts the unit matfix of the ""*
,, .
,,,r'1.
ff'*1; 'D is
then,A;-r =
Eoof : lrt A = Iarl i'J = t,2,...,n
(D * O) and Ag are the
determinant of the matrix A
the
co-factors of D'
"
I .'
rl
AA-r = Iaylxo I&l = D I"'i t&l
Lz I oJ
ate
?22
zln
Azr
azn
!:'
L"r, a,,2
ann
[-ar r
1I
L-rl 4 lJ L-s fl : lJ r, ,'
1i
Anz
Art Azn
Forrr .the matrix [&11. Ttren the,trylspose of the matrix
I&l k called the adJolnt or the tgjugat! nat;k of the matrix
iai
"''
r l _2 _E.rr r l . & ':_i, i ,,,''
AdiA=l
" 3 -o sl=['.2,'-i +l;
EI
EI EI
Azz
I
:
' r it
=DlI "'...
]H:
Azrt
Hl
;;. l
ll8
COLLEGE LINEAR AI]GEBRA
oI o
O
ol
Ot
oo
ll
r19
MAi!'EIX AICEBRA
Since lA l+ O, we can let b11 =
l=r
(i, J = t, 2' ..., n), where &J are the co-factors of ag in
I
A'
Irt B be the matrix [bryI. Then
Similarly, we can show that A-I A = I
ThusAA_r =A_l A=I.
Theorem 2. If A and B arenon-slngular matrices,
(AB)-t =B:r A-1. Also O-,f1 =Aand (A-r)r = (.,Sft.
hoof : Since, (AB) 1B-t A-r) =A(BB-t) A-r = AIA-r- AA-r =I
and (B-rA-r) (AB) - Br (A-I A) B = Et-rIB = FrB = L
Thus E)-t A-l is the Arverse of AB i.e (AB)-r = B-r A-l
If A and B are two matrices such that AB = I, then A = B-l
*=[i=, ",- ",]=
[a, ""ry]=[+r]=,
'
h* *]=[L, ffi*
simirarry, ,o =
]=,
[i= t
Hence B is an tnverse of A and every non-singular
A has an inverse.
matrix
i
Finally, we have to show that the inverse of a nonand
singular matrix is unlque. Suppose that A is non-singular
have
we
then
A,
of
A-l-is its inverse. If EIis another inverses
on th9 left by A-t''
AEI = I Multiply each side of this equation
then we O-t 6[i) = A-l I =,fl.
or, 1e-rd B =A-l
and B =A-1.
Therefore, A = B-l = (A-l)-r ... A = 01;l)-1.
Ar (A-r)r = 0rrA)r = (Ilr = I shours that pt-t;r = (AI)-r.
Corotlary : If A1 P*r, ....A., are non-slngular matrlces of the
s.rme order, their product is non-stngular and
(ArAz .... &) -t = &-l .... Ar-l 6r-t.
Theorem 3. A matrix has an inverse if and only if it is
non-slngular. A non-slngular matrlx has only one lnverse.
Proof : Suppose that A is of the type (m, n) and that m <n. If
A has an lnverse A-r the products AA-l and A-t A are both
defined and hence A-r must be of the form (n, m). It follows
that A-l A = Ir,. IfA has rank r, tlten we have r < m < n s r, since
tl:e rank of I, is n and the rank of the product of two matrlces
ls less than or equal to the rank of either factor. Hence m = n
= r dDd A is non-singular. If n <m the result follows sirpilarly,
using AA-l = Im.
Now suppose that 6 = [arJl is non-singular
Arr
EI
or, IB = A-l
or, B =A-l
Hence A has onlY
fi
$
,i
,I
I
{
i
,
l:
one inverse A l ,
number of
[The rank of a matrix A is the maximum
linearly independent rour (or column) vectors of A'l
patrlccs'
8.2O Ttcerclrc on otthogoral
each of
Theorem l. If A and B are orthogonal matrices'
order n then the matrices AB and BA are also orthqgonql.
Proof : Since A and B are n-roived orthogonal *"ll"-ti-'
The matrix product AB is also a square matrtx o{ o,,i,,dSr
aud{AE)r(AB) = (Br S) GB)
=ntr (l\relB
Bir I,rB
.
n
r2O
Now (AB) (ABlt= (Rg)'(n*a?)r
Itrus AB is an orthogonal matrix of order n.
Similarly, BA {BAir = (BA) (ArBr) = B (A.{r)Br
= AIA*
Hence BA is an orthogonal matrix of order n.
-AAt=l
Theorcm 2. $ Ais an orthogonal matrtx, then ^[l is uho
orthogonal.
Proof : If A ls orthogonal. we have
fu{r - Ar A = I where I is the identity matrlx.
or, (AAr) l= (l{ta) '= ['- t
I
(At)
SimilarlY, we have
.^s)*!q,):::l'ffi
or,
d 5 .a' larf'=1
-l
-r
-l
-r
or, ( allr
A'=
A'(
C')t
=I,sirrce (.{Tf'=6-'f
=BtB=I
Hence AB is a unitarY matrlx-
Theorcm 2. IfA ts an unitary matrtx' tfren .il ls also unitary '
Hence A-I is orthogonal by deftnition. That is, inverse of
an orthogonal matrix is also orthogonal.
Theorem 3. Transpose of an orthogonal matrix is also
Proof : By dellnttion if A is an unitary matrix, then we
have AA* = A*A = I.
orthogonal.
Proof : Here we have to show that if A is an orthogonal
matrix, then Ar is also orthogonal. By deflnition if A ts an
grthogonal matrix, then we have AAr =Ar A= I.
-t
{d
.1s
unitary.
Proof : By delinition if A is an unitary matrix, then
AA* .=, el* =.t where I is the unlt'matrlx.i :' " : ': '
Hence AT ls orthogonal by deffnition.
i' Redart : For any square matrix A, we have if ArS = I, therr
-.i
AA
and
*=A*A=I
BB* =BnB=I
or, (A*)r At =.$ {A*)'=I
or, 1ar1t6r =4t {Ar)t = I Since (A*)r= (4,}*
(U
121
:
(aa+;t=1A*6;r=F=I
Theorem f . If A and B are unitar5r matrices then AB is also
a unitary matrrx.
Proof:IfAand B are unitary matrices, then by definition
we have
1ee+;-t=(A*A)-l= i'=I
'
or, (e*l-'A-r =A-r n*il= I
*
or, ( C') .Cl = A-' (C')*= I Since g1*f' = (A-ll*
.'. C' i" an unitary matrix by delinition
Theoreio 3. Tlanspose of an unitary matrix is also
or, (a6rF=(AfA)r=F=I
or. (Arfr .S =,S (AT = I
$lgt Theorcms oa rmttarymatrlccs
'
= A (BB*) Ar
= BITTBT = BE|I = Ir,.
Arrr=
: I.
t2l
MATRIX AI,GEBRA
COLLEGE LINEARAI-GEBRA
1
Hence by delinition.S is an unitary matrix'
nemart : For any square matrix A"if AAf= I, then A*A= I'
CoLLEGE LINEAR AI.GEBRA
3.22 Tlreorems on lnvolutory matrL.
Theorem 1. A matrix A is iavolutory if and only if
(I -A) (I +A) = Q.
Proof : Let (I -A) (I + A) = O + I -A2 =O
=+ A is involutory
Conversely, L€t A be involutory. Then by definition,
we have A2 =l
Theorem 2. lf A is an involutory matrix,
II
Then i 0 + A) urd; (t - A) are idempotent.
G ,, + a)l'z =|a, + 2A + N)
Again
I
{} u - o,}'=i
Suppose that
R=l
L;;, ;*
or, X = A-lL
Nl
X1
X2
ll
That is,
=i0-A).Thusi 0-A) is idempotent.
xn
Hence the theorem is proved.
3.23 Solutton of ilndar equations by applldng matrices
(i) When the linear system has n linear equations in n
unknowns
A1 txt
l,€t
* O.px2 +
rxi + azzh +
an rxr + d1pX2 +
+ Om)Qr =lr I
* d21,X1, =tz L
. * arrrr.lc.,
[l,l
I L*"J
l
or, IX-A-r L tX=X
=;(l -2A + I) since A2 = I & P = I
2.2
[",]
1* l, *= l:: lr"al= l'i
::: ;;"
Li"l
multiply both sides of (3) by A-1, then we have
A-IAX = A-lL
fsince A-IA = I
jII 0 +A). Thus2 tt + A) is idempotent.
(P - 2A +
It]
hence ttre system has no solution. If D * O, A is non-slngular
So A-l exists and hence the system has a solution' Now
=fi (l + 2,4. + I), since ,{2 = I & P = I
=
L;;, ;*:: ;"" 1 L*
(2)
L,et D. be the determinant of the matrix A. Evaluate the
determtnant, if D = O, A is slagular, so A-l does not odst and
Hence the theorem is proved.
I
^r,
Then equation no (2) reduces to A)( = L (3)
.'. (I -A) (I + A) =l - A2 = I - I = O.
Pnoor,
matrix-form as
I-arr a.tz aln I fxl
dzz a2n I I x2
|
a,rz ar' I
[",,
azr d2z
=+A2 =1.
'
123
MATRIX ALGEBRA
L22
Ar r Azr
4rt
lAl
EI 8I...
Anz
Azz
Arz -[al
E] "' EI
At., Azn
lr
IIll
12
t}J2
(say)
Aert
EI EI.. EI
1,,
ffin
Then xl = (rll , xz =trr2...., 4n = mn is a solution of the given
system of n linear equations.
{t is to be noted that the solution of the system of
(1)
=rl
The above linear equations can be written in
equations can also be found by reducing the augmented
matrix of the given system to reduced row echelon form'
MATRD( AI,GEBRA
l2s
Multtplylng both stdes of thts equation by 6-t on the
left
COLLEGE UNEAR AI,GEBRA
I24
(tt) Itrhcn the cyster has m linear equatlons in' n
rrrtnotmga[dm<D"
constder the followlng system of m linear equatlons ln n
sothatA
Iil=,
L*J
unknowns xt, h, ...,4t
*
312X2 + ... + arrr{n =
O11X1
+a2fc2+...*a26X,.=t2
A21X1
u,esetAr
ll l
l
tft
^[i]=*'o
a-rxr i'^r*a"*... i"-r'q" = l" J
inwtrich m<l and fu afrd [ ' i =1, 2, "',m,
j = 1,2,.'., r\ are constants (scalars)'
The gtven system (l) can be written in matrix-form as
a'tz ar" I lxrl [], I
[a,, azz
d2n
1",
I
lI'
t2l
I
l= [ ?
;;" J L,"J Lu,J
L;;,
matrix of the system is
augmented
Now the
;
::
4ln : lrl
,-l?
"^
..-I
Weshallapplyt}teelementaryrowtransformationson
the above augmented matrtx to reduce it to the redueed row
ccheloa fom.
The reduced form of the augmented matrix will either give
a solution of the given system or will indicate that the system
is inconsistent.
Theorem 8.12 If A is an invertible n x.n matrix, the system
of linear equations Ax = b has a unique solution given by
x= A-lb.
Proof : SinceA (A-tb) = (Au{-t)b=Ib=b,
,
i
.'. x=^&lb is a solution.
=l
*'o srnce FrA = I'
Herrcex=A-Ibistheuniquesolutlonofthegtvensystemof
linear equatlons Ax = b.
-/
.fuilct.
Worted out eramliles
G1!*,
-il *r=E 'r-il
'"
Etnd the matrices 2A. A + B and A* B'
I
arr,, : l-J
Conversely, suPPose -
"" Ii ]=
f,rl rc *t,,uor,,
L*J
"
i
r
..''.. ''
,,;;=-[l tl *"=fi i-rl
- fr
-Zl'
^-I3:.!, 3:1 ",,T,l [
i
' .
i
A+B=
,',i{
Il -? -fl.fi i il I
i*r'];',*,'.t-i,l . [3 5
[ll
r, t
" f L-z 3l ,f-2 -.3 -5I
e-B-LE l-1Jl[-,1 -4 2l
*fl:'3 - f,], .'"''
= 1i..[3ir?l iilfil';t'l
=
,,
a/rirE* t; ?l ,, , ,, .,
*"-* z.rrtar [l 8ln""te.es3n4
:
Compute the rnatfipt
BA"
,:;
'
;
MATRX AI,GEBRA
126
COLLEGE LINEAR AICEBRA
r-8
sorution,*= [; 3] I; ?J =[3]?r%..%]=[?. 3J
uo=
Sowe see thatAB * BA
1
and B
-',
r-8
*= [; |'Jlii _+ sJ
r-2
2;
I
i
[-8+ll+42-45
=l 38+4 -42-O
=l O O 0 l= O. HenceAa + A2 -21A-451=O.
LoooJ
5 3']
-2 1l
"=l:,,1 i]* "=l-i o -31
then prove that (ABP =E[rAr.
2 -31
r-2 2 -sl Ir-2
2 I
62=al{=l 2 I
-61
L-r -2 oJ, L-r -2 oJ
f 4+4+3 4+2+6 6-12+O I f lf 4 -6:'l
-14+2+6 4+t+t2 -6-6+0 l=l 4 t7 -r2l
B+ 12+o J L-z -4
r lI 4 -61r-2 2 13']
6o=62f,=l 4 tz -tzl| 2 I -6I
L-2 -4 ts J L-l -2 oJ
7-22+8+6 22+4+12 -33-%l +O I
=f -8+Stl+12 8+17+24 *12-tO2+O I
L +-8-15 -4-4-go
6+24 +o J
I
rO O O-l
(
L-r -2
I z-++o -2-2+o
38+4-42-O
-574+ 63-O I
49+ 17 -21-45 -ll4- 12+ 126-O
L-rg-2+2t-o -88-4+424 so+ts-o-4s J
,
ProvethatAs +N41A -54I=O
where I is the tdfnUty matrix of brder 3 x 3.
r-2 2 -31
Pr,oof:A=l 2 I -61
oJ
-61
4
L-zt 42 o I Lo o 4s)
It is to be noted rhat the pr@uct BA ig not delined.
a.ra=12 I -3-l
-61
a r ll
r -42 42 -63 -'l f 45 0 01
-l 42 2t -1261-l o 4s o
3+ t0+9 I
6+ 15+ 15 J
,,.,ii
38
-57
-61
4s -l ra l+l 4 n -nl
=l s8 ss
30 J l-z -4 rs J
L-rg -
sl
'
11 0 0-r
-3-'l
l -6l-+slo
r ol
-2t12
o_J
o
L-t -z
Lo rJ
3-'l
.
N -21A- 451
r-8 38 -57 1 rll 4 -61
4s -rr4 l+l 4 tz -nl
=l B8 38
30 J L-2 -4 15 J
L-le -
r-2 2
=li Zr ?l
lz -r s.l
Calculate tl:e product AB.
2+6+S -l +4-3
-_Jr+z+o
12+ 3+ lO 4+9+5 -2 +6-5
t9 n o 2zl
=lrs
t8 -t 261
38 -57 1
-1141
=l 38 4s
[-rs - ss so J
.'. A3 +
[; ?] t6 3l trtS Slgl = [i g]
*^rgfi*=[;3 3]
127
l5J
*:Gtuen
f l2 3'r
rl-221
*=13 _?
tl
"=L-r, s ?J,.
r-l 5 31
t-=l 7
2'l
. Lz o-sJ
Ls I
sl
GtuenB=l 7 -2 tl....gr=l 5 -z
r-l 7
21 rl
ollz -2b sl
L3 r -3J13 -l 4J
NowBrAr=l 5 -2
21
129
I28
COLLEGE UNEAR AI.GEBRA
MATRIX AI,GEBRA
(-1)'F2l +7'5 +2'l-Ll
5. 1-21+ F21.5 + o '(-U
3. {-2) + 1'5 + t-31 '{-1)
fFl)'r+7'2+2'3
=l s.t +(-21'2+o3
Ls.t*1'2 +(-3)'3
(-1)'2 + 7'3 +2'4
Sotutlo I GtrrcnA=
I
The sYmmetric Part
5.2+F2l'3+04
3'2+t's+(-.3}'al
I
[s?I "{iqil
of o =
l' 14 +
'{)
=*[[l ?il.li q?]
r-1+14+6 2+#-2 -2 +21+8 1
=l s-4+o -lO- lO+O 1o-6+O I
4il
I s*z-s -6+5+3 6+3-12 I
rtg 35 271
(u
=l r -2o +l
=*[iii ?i?i?l =+Vt fl=L; :;l
2 3'r r-1 5 3'l
tl
Asain* =[-i s-rllzz -z
s al L o -sl
Ttre skevr-symmetrtc
L+ z -sl
1.5+2'F-2)+3'O
(_2).5+E.t-2)+Fl).p
2
(-1)
+
7
+
5'
l-Ll:
=l {-zl '
2:S+S.F2)+4.0
L2' (-l) + 3' 7 + 4.2
rl' {-l) + 2' 7 + 3.2
r.3+2'l+3'(-9) I
2.3+3'l+4't-31 J
3+2-9 1 '' i"'
r-1+14+6 5-4 +O
|
-6+5+8
IO+0
-l z+BS-2
'lO'
6+3
1O-6+O
-12
J
*s
[-e+zt
r19 I -4 I
=l ss -zo zl
(-2)' 3 + 5' I + (-l)' (-3) |
Ln4
-sl
i
4
_lrtL
-z1Le ?
s
part of A = | A- St
qil
fl[i
-]l=
i
=;[l ii
i-|.=;[-?
Lil{
-2,
-
P::",;1""Ti*'n*trix
i j ilr"o'tr'oso""l
"=ruli
Ls 1-b
i j il
nr=nli
i j i]
Proor:G'en^=*11
n'we*=*
Unear Atspbra'-g
j
[i'i i] [l i
3 3l
1 ll
I -5 I
-5 1-l
COLI,EGE IJNEAR AI'EBRA
130
.I [9+9+f+9 9-15+3+3 9+3+g-15 9+3_15+3r
_
+I
I9:I5+.3+39+25+I+t 9-5+t- 5 9_b_5
-3619*3+3-159-5+l-5
9+t+t+25 g +I-b-S
L9+3-15+39-5-5+l 9+l-S- b 9+t+25+l )
2-31 3+5i I
il
3
5
-i
-l
I
I
l-360 0 0t
r lo s6 o o
-3610 0 36 0
Lo o o s6l
r1 0 0,ot
I
I
lo l o ol
=lo o l'ol=I .'. AAr = I.
Lo o, o- rJ
3 s 3l[3 3 3
"'"
1 rlls-5 I ?r
-5
similarry,
=*
1 l-5 ll,s I I
[j I -5 rlls r -5 ?l
t
ls Hermitlan.
2+3i 3-5il
12
3
pnoof: [=lZ-Si
fi
J
[3+5i
12 2-3i 3+5i1
*, Ao==l 2+3i
i
3
:j
. AF
5 l=.S ..
I
i,.i
'-'. A* :'A. Herice A is Hermitlan'
Exampti io.'prov6 ueii'tt"
Hence A is orthoponal.
No* aet =,1
*r"B.rrA= [],.r, '.iJ ,r,as =l?;t tI
then prove that e+B =a.+ g.
I= [l * r, 'lJ and e= [?;1, -;J
a*e=[3]ri l:'il i,, (1)
Pr,oor:
*it iiA = # [:, , il
.
proof : Given
[l
=lo
I
-i
Ls-si
:
o o ot
l o ol
o I o l=I.
Lo o o rJ
.Theiefore, aalgra=:.
,
131
MATRD( ALGEBRA
o*=
=+ [-il
^= i
is unit4ry.
[], -l]
t [1,
' g,u
.L
_l]"
trl
[],, _ll, [], :iI
i i;i] =;Y, gl
= [3 ?l= i
Similarly. A*A = I.
: 'i:
Thus AA*.=AilA=I
Hence A is'an unitary niatrx'
"1
i:
AsairrA+B =
3 5 O
Exampleff.Showthatthernatrx4=l
""--'*-''
7"'2''j'lil"'
Thus from,{1) and (2), we;get i
, :r ' "" inwlutol-v" t '
0l f-5 -8 ol
[-53 -B
o l"l 3"5 o i
5
Froof .A2=axA=i
[3;rl li'lJ,
riJ et
A*rr =
[3 ] 4, i:
-.,\+R = A+B.
I
L
.
'rvvr
Ll 2 -,tl Lt 2 -tJ
I
1
COLLEGE LINEAR AICEBRA
182
l-r
A21
r25 -24+O 4O-4O+O O-+Q+Q1
Q+Q+o
=l-i; *ii*o -24+25+o
=L-i"*-4"-'i
-e*ro -2 o+o+lJ
rt O 01
I ?l='
A2=I
^
I r=
r, . rrynd the lnverse of the rnatrrx" = u ?l
Slolutlon
I
the matrtxA is non-sin$ular and
hence ,t-l dsts'
Nourthe cofactors of D are
Atr=4,
I9t=-2,
;. x,
lz
,n,o*= \?
-3r\ =
l'l
L
o=I-t -?f = [-3 -?l
={eay,=} Li
1-1
141.
e-'=*AdjA=*,Fie ld +)
-11=i
ffi -',1
!
tl
- tl -? ',
'"*'"- 16 3 i [=o+11+!fi=s3*o''
-te?:"'
tfrerrp = 14 O -t I
lio the matrin A ls non-singular
|,
r!
li.
t.'
4,, = (-U l3 2
=
3l
" [?
- Il 3 i I I H:r*l'E'J"ii:H"H'3{""#j"*
*[o-t 3 ii1o -2I I X,".H:X'lI,i'#&u.il?"i'lJ'?'-'
the matrlx;
^,,
la- t
of the matrix
solutron'I^'r=[? 3 i 3 ? l::::1"]ffS:'*rstand
Eolutlnn I Ict D be the determlnant of
lz -I 3 l =2(O+3)+ 1(8+3)+3(12-O)
ohctors orD are
r7
t"zz= L
*-fv/6(b)' zurd *".i;*T ".T rnatrix
a=14 o -t2)l.
Ls 3
.i
-l\=n'
11 tl
r 3-11 L2 Tfr3
141
I
-5
=l-tt
-'^
A
4J
Adj
=
rherefore,
I Ltz-e
rao,plg/G. Frnd the lnverse
A12=-3
=-u'
=-n,o., = \-l -',\ = ''
matrDq then
: Irt D be the determinant of the
] ?l = 4 - 6 =' 2 *o' so
rhenAdJ
^r.
= t-I) [a
e.s = Fl) la
Hence the given matrix A is involutory'
_@r]
=(-1)l 3 1\ =,,' *=\'rB\
lz -l\
I
=LB
lp?
MATRD( AI,GEBRA
=
i3
-11
and A-l eJdsts' Now tlre
=t'
l+ ol
T
rr.e'. = l6 5l = tz'
(-1)'
second row 'bv
*"*nlv
-Zlw;
=g'A-rr
and o'' = [-?
Hence A is invertible
-?l
COLLEGE LINEAR ALGEBRA
134
[-r 2t -3r
O lana
l+ -2 sl
e-, =*
Ls 2 -sl
rhus A-,u=L
f2 I -l't
n=l O 2 t I nrra a-,n. p.U. p. rgg5l
solutron: Giveno=
-B
[-i i s_lI
l+ -2
l_r 2 -sl=- 1(5+0)-2(ro-o)- g(4-41
rrtD=hl=l 2 I oi
l4 -2 5l=-5-2o+24=-l*o
So A is non-singular and hence A-l odsts.
Cof;actorsof- I =or, =
g =$
l-;
2=A,=Fl)
lz rl
-3=e13=14 _tl =-t
2 = Azr=
r-5
g
l
4 -31f2 1 -ll
i
g
3o
lL8 3 _i J
r-IO+O-15 - 5 + 8 - 6 5 +4+ LP
9I
10-14+ 12 -10-7 =l 2o+o+3o
L to* o+25 8- 12+ 1o -8-6- 15 I
r-25 -3 18 I
-sE I.
=l so 8
L+r 615.-2e)
Ft;d tne invS;7f the following matrix by
exampte
I
rer,r=[r
2
O=A23=pl) -14 -2 =6
lz -3
4=Asr = I o =$
I
-3
- -2=4o=(-l) l_r
| 2 q =-6
I
-ll
iu -+)
t
3 :,
3 I !]#;;aro#s.
rl O 3 : O I OlWemultiplyfrstrow-by3.3nd
; I o o lz"nathensubtractfromthe
-lA 4; -i
i
: o o i lsecona and third rows respectlvely'
l;
.' . "fl O 3 : O I 9 l'srUtr"ctthird rowfrom
-Lg 3:13 i L ?1,;;;;;r;;;
-3
. rl O 3 iO r Ol
o i r -t -r lnautuplv the second row bv (-t)'
- I o -iE-loio-2
IJ
Lo
Tl o 3 : o I 9lw"*uliiptvsecondrowbv5and
" b=Agg= -l2 ?l =-u
-Lgl_?,;-J-;11,ffi'ffiHJ,il;;'il,iiii,i.*r
4
3l
.*ro=[-? -107 -8-Tr
ol=l-ro z -ol
-6
i
i.=L_lo '" -g l= L lt
13 4 -1 i 1 9 9ltnte.ctangefirstand
Fttl-| -31=-n
r5
4 -31
4 3l r-5
Solution:
-a
t=Pw=ll-r
4 5 =!
*
=
I r5
form' o
=l i
using row canonlcallrvrrrr
"-L?
l? 3l=-ro
r35
IV{ATRIX ALCEARA
r Adr A
Elampler+.tfR=l 2
"
-;
-sl L-e 6 -sJ
- l-l
i ? lwe murtiplv *rird row t, (- +r)
! 3 : -i5-7-4)
Loo-toi
L
MATRD( AICEBRA
COLLEGE LINEAR AI.GEBRA
136
[t
:i:.,i;l] [i :i:jhl]
We multiply third row by 3 and then subtract from the
first row.
[roo' B+-g I
-Lsi?:-i
ki
.1=bA-'t
Hence A is invertibre and
i
I
Example 16. Find the lnverse of the matrix
rl-l o ot
lr 2 o O IUV ustng only row transformations
A=l-o
o l -2 lto reduceA to I.
Lg t-z rJ
f l-r
O O:lOOOl
soru,on,rar.l=l-l 3 ? -3 ; 3; ? 3
We multtply second rowtY
I
n-roo:too"l
r o o'-* +2 ot ol
_lo
t-2:4
o
ol
lo
o
L" o -z l:-s-B r_l
_lo o t-2i 4 2 t ol
lo
o
[_o o -2 1:-s-B
U
We mulflply third row by 2 and then add with the fourth row.
o, 3 ] o rl
[, oI o
o o,_++ool
-lo
lo o r-zi 4 21ol
fooo-s:st21_l
I
We subtract flrst row from sOcond row. We multiply ftrst
row by 6 and then add wtth the thiid row. Also we multiply
first row by 8 and then subtract from the fourth row.
fl -1 O O : I O 0 Ol
lo s o| o:-lrool
-lo
-6 -2: 6 o r ol
Lo e-2 li-8oorl
We multiply second row by 2 and add with the third row.
We also multiply second row by 3 and then subtract from the
fpurth row
o o:-t I ool
-lolo o | -2 i 4 2l ol
Lo o -2 I : -5 -3 o lJ
3 + o rl
f, or oo o:
o'_+*ool
L-z ilo 5J
Le t-2 l:ooolJ
rI -1
g O O i I o O Ol
We add second row with the llrst row.
f s-tt 9f
*, =l -i
tg7
We:nultiply fourth row by (-'J
o o, Z + o ol
T, o
I o o,_++ool
_lo
lo o I -2 | 4 2 r oI
lo o o ,,-i-+-3-il
We multiply fourth row by 2 and then add with the third ror'
i:,,
t-'or o o'3+o
I
.,'' !'Y
o'-|+ o ol
\-'
lo
-lo o t o,i t-i--31 =rl+A-rl
Lo o o r:-r-*-3-r"l \
139
MfiTRIX AI,GEBRA
COLLEGE LII{EARAI'GEBRA
T38
A-rA)K= A-rL
or, IX = A-r L Jstnce A-lA = I
Hence A is invertible and
or.X=A-rL tand D(=X
f135
z l o rl
I r 1 o ol
l-5
T ' ,l
z
^_,=l ,f_r-Tl
I
L-l-5;E-Ei
Eramp!9l1?' Solve the following
herp of matrice"
i:;J: ll (1) i,
of linear equations
tz'
rcte= Lr _z l.*= $l,"=El,*"'.t.*
(2)wegetN{=L B)
the matrlxA' then
D be the determinant of
r l-- 4-r=-5*o.
,=1,lz _zl=_*-L--vrv.
and
So the matrix A is non-slngular
are'
Now the cofactors of D
=-2,
A91 =- 1,
rhererore, Adr
Sccondurrcesg
, fA. ""!*."iia matrix of the given system of llnear
', i'"
tALl= [? :,
: [] Ir,"*hange flrst and second rouls.
: 3I We multlply flrst row by 2 and then
tt
- lZ -2I : lJ subtract from the second row'
- -u ." muftiplysecond -*uY |.
1|,
-?l
-ll -i i-il
Now the system is in row canonical form'
Then formtng llnear system we have r = - I and x- 2y = 3'
.'.x=2!+3=-2+3= 1.
^-r-+a
^_, etdsts'
hence A-l
A12=- I
Thus the requirgd soluflon of the system is x =1 and y = - 1'
Erampt4^/**. the following linear equations with the
herp ormatrrce" , ?Xi 7-
,fr: :E I
D.u. p. rss{
2x+fu -Sr=2O )
Ar2=2'
A:If
*o
,r]';ttr-,,1
-ll
=Li i,]
6r =fi aa1 n = - | [:?
We multiply botl: stdes
lt
i;i/ations is
t? lllXI=El Qt
A.11
t] EI ;]= t
*, [;l = Lil HenG ;= -l]
linear equations with the
Solutlon : Flrat Proccss t tt" "y"t"m
as
can be urritten in matrix -form
I-€t
=[i
rhusX=[i
of equation (S) by A-t'
Eolutlon I Ftrst Procerg : The given linear equatlons can
be written in matrix-forrn as
lzi -,i\[i]{;al
u,,
14O
COLIEGE LINEARAI,CEBRA
suppose,*,^=[!
3 _rl" [i]*,r"=[;A ]
=
then the equation gven by (I) reduces to A)( = L
Let D be the determinant of the matrix A' then
(2)
5 - 7 I =s(-s + lo8)- s(- tz +24r-z(s6-2r
p= ls
14 t -t2l=irs_60_ 2#=tZ*o.
lz 9 - 3l-ote-w--w
So the matrix A ls non-singular and hence A-l odstsWe multply both sides of equation no (2) by A-t on the left,
ThenweBetA-t A)K=A-rL
or, D( = A-l L
or,X.=frL (3)
{sinceA+A
= I and
^,,
l3
lt -'?l - rr,
=
o,,=lt li =*,
ar, =
Qg,,
r*=13 -Zl=u,
...,.
lq-ql
ls-zl
(-r)
As
ss,
=
=
li-- iri =s.
l7 - irl
ls sl
o*=ln
il=-tz.
Therefore,
Now from equation no (3) we get
]l[;]
[;l=+[-tr ]i
=+[-ffi i,B]ffi1=+[ 'tr ]ff1
"" [i]=+t{l=[i]n** r=LJ
matr[r of the gtven linear
eqqaUons ts
: l'l = r*. Arz = Fr)
&r = Fr) ls : Il =-
- r 1O5 4€! -53 I
and A-r =$
# f 17 -rtr -r? l
^o ^=
ryiotaProccs. : The augmented
Now the cofactors of D are
=
t4l
MATRIX AI.CEBRA
r lo5 -r2 34 T I lo5 -485 -538'lI
a8 5 -r7 I =l-12
AdiA=l" L-ss
I -r7) L 34 -t7 -17 )
rs s -7 : IQr #:,"ffi:f*:H['ffit*subtract
rAIr =
Ll I -!r' irZ J
fryg",i#f*
ffi
-[i -,i *'i.JAl *ruf*ifHt3',f"T**
,r? i i :s'n-lw..aa second rowwith ttrtrd ror'"
- l-l
Lo rz 5 : gaj
rI-44
'l -n't]*" ","tuoly oeond row bY f +)
-lo-tz-o
Lo o-I
and thtrd run' bY {-11'
rr 4 4 . -71
-Ls
It.
aJ
I42
Now the system is in- row canonical form, Then forming
the linear system, we have z = O,
6
y+ Wz=2,x-4y-42=-7
[b) Construct the products AB and BA'
ro 1I Ol
where"=ll33llu=l:l 3 3 1l
roll0-1
Lorlo] Lo-t-t ol
ot,z=O,y =2, x=8-7 = l.
Thus x= l,y = 2, z = O ls a solutlon of the gtuen equations.
EXERCISESI - 3
-r -2 5l
Ilnd the matrices, SA, A + B, A- B, 3A - 2E}.
3] :.
Ansners,[,333131 ,IZ-L-1
[33 3l
B B ?l
r15 15-2lr-3 8 -rll
4 11 I,l 4 - I 22
Ansrwers: tall
"l-z25 --15
2JLg t2 tz )
f-2'o o'21 ': l2 oo 21
b)AB=l 3 3 S 3l*=l 3 3 3 3l
l-z o o z) l'z o o -2)
3l
I I -I Il
l-l 2
n
2
+. Ifa=l-3
-rlande=12 u_]
I
ol
LI23
L-z
l,
:
t-l I 3 .i I*o hB 3 '3"-? I: , ,;,,,. i,:
1:
(bl Sho.w tllat the six matrices
and also show that
AB_BA=_4c,where"=LB
1.(dro= [i 33i J,"ou=[? o -3 4I
,=[3 ?],^=lA -91
"+[* f]
then show that AB * BA.
"=j I aj *
.A2
]
satisff the reration"'
=W =.C2 =I, AB ! D, AC: BA= E.
z.trA= [i -B],u= [3 -1] *0"= [-i -i]
i fh"ri show that [i) A (B + C)
. r. i:1.,,
,
(ii) (A + B) CrAC + BC.
r-l 3 2]
ft -2
3.{alltA=l 4 -2 5 land e=12 3 Lo r -3-j
Ls 2
'
r2-3
-51
lR.u,P_.
\w4l
T-l 3 5l
5,rfa=l-l 4 sl,s=l I-3-slana
3 5l
Lr-s-+)
L-l
r 2 -2 -4l|
ltnen
-1 3
; shoWthat
C.=l
,
t tt2 -3i
,
AR = BA,= O, AC = A and CA =tC. i
= AB + AC
find the matrices AB and BA.
r48
MATRIX AI,GEBRA
COLLEGE LINEARAI,GEBRA
[L U.S Lg7sl
[R.U. S. I:fftl
, f :4 6
6l
z.tra=l r B _3 *"u. that A3 - 4A2 -A+'41='o'
[-r -4 I
I44
12. (i) Find the lnverse -lof the matrlx
r I -4 -r -41
I z o s-+l
e.IfA=l-r
t-Z el
L-r 4-t 6j
12 -l -l
e=l t-z tl
(il) Flnd the lnverse of the matrix
r5 4 41
e=17 4 -s
lz I 6J
11 I l-1
ro I -l 1
2
s
e=l
a=lz
^-LiZ lard -3 2 4l
t L -L;l
eJ*- Li i oJ
AoswcrlB:
r1 -1 1-'t
ro.ffa=|2 -i o ltn ttshowthat
A3 = A2 . A =,{,. .$ = I and hence ffnd A-l p. u. H. leE& R- U. S. L#ill
rt.
r0 o I 1
o -l 2
Lr -l lJ
i ? tl
!
@ ^=[?
-ll .,B=f_i i il
(ii) *'=#[-?i
-,?, e)
-?
-" -u, l
,,,0 ,
,r. oo =l
I
L-s ;
|
matrices and the lnverse matrlces of
6r.\*"O the adJotnt
eathpf the followlng matrices :
Find the matrlces aB, en' A-r' B-r'
results by veriffing the relatlons'
iagf-l lfY a-t, AA-r =I, EriB=I.
(aB)-t' Check your
i
-r 7 19 I1
-* [j I I
*"=[_i i ,i]*,+[i i i]
,*l Adrc=[i.' l-, -?]-'{i
=i-i5 L 2)
-1
-;J-'""-1
I
ADsrGr,,;=|-'i
iBl,*=1":3
rsj
--li -i-sl'
It 4? ill
L o I ol
(Iil) c=l o t
'i!
tilrr=l
L-6 -6 zJ
Lr o oJ
rrngrtrs : (i) AdJ A = [ -: ] I, o-'
p.u.P. 19791
I
9. Veriff that (AB)r =S,$ wtrere
r-l O o-r
Lo o ?-,J
p. u. IL 1s751
Ll -l 2)
Prorre that A4 - 5A3 + 9A2 *7A+ 2I= O.
Ansrcr : A-r =l
r*5
UATRIX ALGEBEA
COLLEGE LINEARAI,GEBRA
i +]
"-'=#f 11 '+ ,3 ]'
rf-2 3 ql
B-r =#l'g -'J _;
l
GBrr=ig[ ,,$ -;* -i}]
o rl
rg l ll
rt
14. rf
"=Ll I ?l*'au=ll I ll
Show that A3 - N +A-28=O'
!,t{F?r ahebrq-lq
Ic. U. P. 1fr161
2o.Findtheinverseofeachofthefollowin$matriceslby
elementary row operations (row equivalent
f2 3 41
r1 2 3l
s 4l
2larrrdl=lz
3
U a=la
t5. *'-'-12
IJ
t 3J
""i.rg-""t-the
canonical matrix)
Ls+
Show thai (A + B)2 * A2 + 2AB +* '
r-l 2 3l
l-l t ll
Lrs;zl
Lt4
r-1 2 o1
I
(iv1 l2
14 -2 *
slr"ae=l 12 9l
16. re=l l3
eJ
Find the values ofA-r B'and AB-r
rz. ffa=l1
o or
Lr I ? I
tD. U. P. 19761
o-o i n-' * al,
$.nj.{i !4.
-6
o t O ,l= l.
,r.
i
-i il
I
o 121
[_i t 2 tl
e=l I o I 1l
Ls
'tr;Fti
::1,=t
rT 'r l::il: au,
€e 6' 6
{
the inverse of the matrix '
lr,rr;1.U" !
Answer: R-r= l ,-2:
:l'-; rl
ffi
8_l
rZ
l- 3t
z -2
I
I s
li]*
lr'-2
#
:: :\ :,.-
=?i,,',,! ;l i,
6 -6
o J',':,
L I' rrQ -t
It
lrlrl
Lo
-a-4 -4a I
tu,l I -. , ,'i'
c
I
r
, !l
I
;i
:
I
fB I
4 '-4t' 16J'
:
i
,t
i
'\
I r'i s ?,1 tn.r,rinla a3 ,o, - +a -'21' iru
21.tr^=l? "o ?,1 #;;;ffiiar."- ,
Ll -r ]J.---i
ii.,., i i, i
i
i
L-i _; i s.l
3 5 l_1
ra
45 4
I '5 I
-4
I
tl*rlr'
:l
l-z !
r^
l-r
l,
l
'
l',I
5,
I 0
,l
l;r'i{,:i .ri..i.f 1}i
3
-21
T-5 4 -31 To
1
l_z
I
-1
(iv)
I ro -z gEJ
lM [_i i _i o I
'-'La
)
l*"
f2 5 2 3l
11
I
:t-3
-3 41
4 I P.orr" ttrat A3 = A-1-3
IfA=12
18.
.Find
2
[1 -l
(i) l-3 -;],r[-',;
ID. tt.P' ts77t
.LO 0 rJ
19.
L-r 5 Il
ADswers:
11 0 Ol
Arrswer: * fa-, *ay =l
Lo
i -rl Is
r 1'3 4'1u"''i
41
g ltut
-r ol
3 tl
j'l? : 'l ?1,13
rl l+ rz'o
[r-r-il
[o roi
Anscrers,e-'s=l o I ? l*-'=l'? 3l
tLo o ;5JI l-o-zz-l
r1
:
u) l3 ll'',[i
P'U'H' 196'C'U'tI.lg77l
-t ::: ;,
r47
MATRX ALGEBRA
COLI,EGE LINEAR .AI,GEBRA
146
Answers,
{*0"+,r;[ ii -i
t,
I:
, .. l
i
Er.u.H.f.+p,ft
_:l ' ,
i
148
COLLEGE LINEAR AI,GEBRA
MATRIX AT.GEBRA
rt 2 3-t
22. Find the inverse of the matrtx e=12 E s
Lr o 8J
I
l
-G {51
f -4O
Answer:A-l=l 13 -5 _31
Ls -2 -rl
e=l-sine
--Lo coso oltJ
o
rI I l rt
rlr
-rlli"orthogonal.
A=rlt -rt _lr _f
Lr -l -r 'rl
Prove that
t-cos0 -sln0 O-1
Ancwer: frr -l sin 0 cos e O
I
tJ
rl I l'l
the matrix
sl
z+.ffe=ll 3 Slanan=l I 2
Lr s LxJ Lt 4 5J
are hrro matrices, then lind A'n
r [r r+rI
e=
li-i
-il is unitary.
6
a e g'.
"t u. H. T. lsl, rs?l
p.
show that the matrix o=l-?
-Z
[3
-l
idempotent.
-l l" ------r
L-q 4 -sJ:1 ," involutory.
31. show that the matrix
= [6 i
" Ls
-l -sJI
-3 -il
3I
E' =* [-B
l,
"Lo o -zJ " "L-12 r7-s)
rl t 1-'t
12 5 31
zs. tr
^=Ll ? B.]*"=L? L ?l
p. U. tI. T. 1$3, fS4
then prove that (AB)-r = B'. A'
Aoswers , A'B =*
ll
__l
4:g,t and.SA are both symmetrie matrices.
28. Show that the matrix
is invertlble for all values of 0 and flnd A-l .
r1 2 3-]
I
G\i5J
-il t]ren show t]rat
27. oo =
[3 i
[*
23. Show that the matrlx
l-cos0 sin0 Ol
o
l-r
{o
2 .I5l
r"l
16
Lo
_t
I49
r.
26. Prove that the followlng matrlces are orthogonal :
;-cos0 -sln0 Ol
-":L%
(i) Itr3
r!
?
l-3 3
(,,)
] fiflLST' T'?J
2_1
r I 2_ Z1
I 3 3 -3 I
3l
3l
Ig-l
lz 2 t
'' | 1*, ?
Ls s -BI
I
.4
I
L-s 5 5J
:l
):
1
,ri
,fl
il
(r) iJJrrt= ,. )
x+y+z=6)
(ii) x-y+z=21
2x+y -z= I )
[D. U. p. ts77l
t..
'3x+2Y-z=2Ol
(iii) 2x+ iY - 3z = 7 |
x-Y + 6z= 4l )
3x-Y + z=-2 I
(iv)
,
*:rilr:'::8|
rc. u. P. Lsnl
x+2Y+z= 2
(v) ?x-_ lr:32__: l. II
Answers:(i)x=2iy=3,(ii)..tr='=,t=2'z=3'4'
(iii) x= 5,Y = $, 7=l' .0{)' x= -2:Y =O'z=
(v) x=-2'Y =l'z=2'
with
36. Solve the following systems of linear equations
the helP of matrices :
+
5x-.6Y + 4z= 15 )
p'
u'P' LsTs'J' u' rr l98ol
(i) 7x + 4y - i; = Ls I
A3swer:x=3,Y=4,2=6.
2x-3Y+42= 1l
(ri)
t
rrr' 3x+ 4y -t, = rc
;;-i;+22= 3 l
p'u'P' 1966' J' u' rL 19781
Ic. u.,P. 19841
x+2Y- z---l )
;;*si- z=14)
//""'
tD.u.H. 1s761
'rq+3x3=1f
{v)
\v' 2xi+
5r, * 2i2 + trs = 4)
2174
Aaswer: xl =O' &=T, xs =-0'
rwsl
AEsrertX= 1.y=S,z=-3.
2x- Y- z=6 )
3x-Y-52'ii
(vii) x+3Y*?"=! I
ID.U.P. 19851
Angr€rix=3, Y=-2,2=2'
x+ y+ z=3 )
(viii) x+2Y+22=41
x+ 4Y +92=6 )
A[ss.tE: x=2,Y=l,z=O'
2x+ Y+ z=3)
(ix) 3x-iY+22= 9l
[D.u. H. 1*]31
4x+3Y-z=-l )
Aos;wer : x= 1, Y = - I' z = 2'
x+zy+32+ + =O I
iiiiv*a'* 1o=o
Aqgref i x=-3, Y = 1, z=- l'
(x)
.^' 2x+iY +52+ Z =O I
tD.u.P. 19861
by
37. Solve the following syst6ms of linear equations
(by elementary row
using row equlvalent canonical matrix
2x-Y+z=1.1
(i) x+4Y-sr=-?l
3x+2Y- z= O
tD.u. H. 19781
(ii)
x+2y + z=2 )
3x+ Y-22=l I
4x-3y- z=3
I
2x+4Y+22=4 )
J
x+2Y+32+t=3
Ansmtr. 2 x=-26,Y = 13,z= l'
x1 +3x2 +Zx, =?1
lD. u.P.
transformations) :
Aoswef i X= 1, Y =3,2=5.
Aivl 3x+ 8Y * 22 = 28 |
x+ Y+z =1'l
(vi) 2x+5Y+52 = 2l
4x+9Y + l2z=3 )
ll
J
Ainsg3r 3X=2, Y=1, z=o,
x+!*z=91
(iii) 2x+ iY + Zz = 52 I
2x+Y-z=v)
151
MATRIX AI,GEBRA
COLLEGE LINEARALGEBRA
150
.)
v* z-t=S t
(iii) x+
x+ y- z+t=4 I
x- Y+ z+t= 2 )
AosurcrB: (l)x=o,y= 1' 7=2' 01)x= L'y=o'z=L'
(llt} x= 1, Y = - l, z =2' t= -2'
COLLECE LINEAR AI.CEBRA
t52
x+ Y+ z+ t=5 I
2x+
v+32- t=14 I
(iv)
3x+ By - 2z + 2t= |
t
4lt-2Y+ z-3!=6 )
Aosrcr:.x= 1, Y=2,2=r',i;,*,
2
ls
o
z
38. FindtheinverseofA=12 I
Ltol
ro I o -21
1l
-t l-z l -l
Answer: A =l-r I -1 ol
Lr -2 I 3l
-l
rl
p.u.P. 1S761
11 2 t 2 3l
lz 3 l o I
e=lz.zI II oI oI l
It
Lo-z o 2 -2)
I
39. Find the inverse of the matrix
I
by using onlY the elementary row operations {row
equivalent canonical matrix)'
Angwer:
2 4 5-3 I
2
-1 3-3 I
-1
[= -2 2 -3 4
0
l-l
o
7 -2 2-I
L
1
T
o
I
T
I
T
40. lf A and B are two s5rmmetric matrices of the same
and sufliclent condition for
*a.i'ur"";i;& th;i; n"i""'"u.y
= BA'
tt. *"t i, AB to be symmetric ls that ABidempotent
matrix is
41. Show that eiery non-sin$ular
matrix.
an -;i:
-' identitY
nJAis an idempotent matrix and A + B = I' then prove
B is arl idempotent matrix q"d AB = BA = O'
that
- 4g.
If A and-B are n-rowed'unitary matrices' then prove
that BA is also unitarSr matrix'
CIIAFTTR Sffi
VECTOR SPACES
6.1 Blnary opcradon (or composldoa) on a sct
It ts a rule by whlch two elements can be combined
together to glve a new element. Such a rule may be addition'
sribtractton, multiplication and so on. The most fundaqeptal
concept fo; studylng algebraic structures is that of btnary
operatlon on a set.
Defrnltlon : Irt S be a non-empty set' Then
is a
S xS = {(a, b) : a eS, b eS}. AbFary olrcratlon on a set S
funcfion (or mappin$ from S xS into S: i' e lf f : S xS -+S' then
the set S'
-f is satd to be a bhary olrcratlon on
* , X, "' etc to denote the
o,
symUots
*rJ
*,
use
we
Often
binary operations on set. Thus 'r'wlll be a binary operation
on S if and only if a *b.e S for every a, b eS and a*b is unique'
A btnary operatlon on a set S is also sometimes called a
in the set S. If a*b e S for every a' b e,S'
blnary
"ompo"ltion
then we also say that S is closed with respect to the
composition denoted bY '*'.
A set having one or more binary operations is known as
algebratc stnrcJure:
Nowwewillde{ineSome.,algebraicstructuressuchas
Groups, Rings, Fields and Vector Spaces using btnary
operations.
orgroup wlth examples
9.2 Detrnltion
a group C is a non-empty set of elements for which a
binary operation * is defined. This operation satisfi5s the
following axioms :
(i) Closure,
IfgL g-p implies that a * b eG;
(ll)Assoctattvity. If d, b, c e G implies that
(a*b-)*e=a*[h*n]
214
tr
l.
VECTORSPACES
COLLEGE LINEARAICEBRA
Eamnlc 2. Conslder the set Z = {O. 1. 2. 3, 45}
Z is ring under the binary operations of addition and
multiplicatton modulo 6.
6.4 Dcf,nttlon of field with cxamptes
ring with urut eiement rn which
A fieme
every non-zero element has a multiplicative imrerse.
Examples of fields are the ring of rational numbers, the
ring of real numbers and the ring of complex numbers.
(ui) Identlty,.There odsts a unique element e e G (called the
i
2I5
ldentityelemcnt) suchthata t e i e * a= afora[a eG'
(iv) Inversc. For eve4r a e G there exists an element a'e G
(called the inversc of a) such ft121 a,* a'= a'* a = e'
When the binary operation is addltlot G is called an
addltive group and when the binary operatlon ls
e gro.rp G is called abellan (or commutatlve) if for every
!
;i
For every pair of real numbers a and b there corresponds a
real number called their sum and is denoted by a + b. The
addition composition has the'following properties :
.SlLa + b = b + a for every a, b in R(comsnrtattve).
b) + c = a + (b + c) for wery a, b, c in R (Assocfatke)
-S]pL{6+
AJglThere odsts a real number, vD O such that
a + O = O + a = a (cdstcnse of addtttv,c fdenfity)
1p$a*b=b*a
,i
t
p:ample 1. The set {1, - l} is a group with respect to the
' i,..
*
,}
t,
blnary operation of tnirltiplication.
Example 2. fhs,sef, {1' -1, i, - i} where i = a/Jis a group
with respect to the blnary operation of multiplication'
p-xagde 3- The set oJall integers, i. e' {"" -3' -2' -2' -L' A'
L, 2,'3,...1 iq-a group #ih respect to the binary operation of
,'f
addition. :
.l
tl
tl
ll"
i
:l
ri
ol
!,.
A t4l To each real number a, there corresponds a real
number, vlz (-al such that (-a) + a = a + (-a) = o
(edstence of addttise luverse)
il
I
J
!r
I
I
,f
.'i
I
i
t
i
t
*
I
t
il.
:.
i
:
il
Also for every palr of real numbers a and b there
A rtng R is an addltlve abelian bo'p with the following
additional ProPerties :
closed with respect to the binary
. (i) The group R is
operation multiplication. i.e a, beR + abeR
corresponds a real'number called their product and is denoted
by ab. The multiplication composition has the following
properties :
tr4-OJ ab = ba for every a, b in R (Coqmutattvel
U fZ1(ab) c = a (bc) for anery a, b, c in R (Asgpciatfw)
that for every
IVI t3) There exists a real humber, viz 1 such
(e.rdstence
of multlplicetive identity')
ae R la = al'= aM (4) Io each real number a * O, there corresponds
(ii) Multiplication is assoclatlve i. e'
"
",
- thb)g =gQ*}for all4. b. c eR
(iii) Muluplication is dlstrlbutlve with respect to addiuon
on both ttre left and the right' that is'
a(b+c)=ab+acl'--^
all a' b' c eR
6 . "r"J = ;;; !l| ror
Example 1. The setZ = {O, t 1, t2, .'.!is a ring under the
binary operations of ordinary addition and multiplication.
,,lf
,,ti
i$,
** (]) . =. ("f = ,
",r"r,
(odsteace of muttlpllcatlve lnverse)
AM (1) ah + c) = atr + ac for every a, b, c in R
fDistributive law)
another, lri, |
rt{
ti
i
t
I
ii,
lt
spage over an arbitrary field F is a non-empty set
V, whose elements are called vectors for which two operauons
&fe, pr€scfibed. The first operation, called vector addltlon,
assigns to each pair ofvectors u and v in V a vector denoted by
11 + v, called their sum. The second operation, called scalar
multlplication assings.to each vector v in V and each scalar a
in F a vector denoted by a v which is in V. The two operations
are required to satisff the following adoms :
4ljrAddttioa is nfrmmutstfue.
For all vectors u' v € V, u + v = v + u.
A t2)Additlon fs eggegfaggg
For all vectors u. v, w eV, (u + v) + w = u + (v + w)
A (3) Ddstence of O (zeno vector).
There exists a vector O eV such that for all v eV
v+O =O+v=v.
(4) Hsiercc of neEatlv'e.
,A
For each v e V there is a vector
- v eV forwhich v + (-v) = (- v) * v = 0
][lUForanyscalaroe Fand anY
t'
vectors u, v e V, cr (u + v) = c[u + 0v
M(21 For any scalars a,P e F and anY
vector v eV, (ct, + P)v = anr + Pv.
!!p[For any scalars o,P e F and anY
''"' M (E Fo, each v eV, lv = v
where I is the unit scalar and I e F.
For some applications it is necessary to consider vector
spaces where the scalars are complex numbers rather than the
real numbers. such vector spaces are called complex vector
spaces.
Vector spaces are also sometimes called linear spaccs'
217
VECTOR SPACES
COLLEGE LINEAR AICEBRA
6.7 tranplcs dtcctot sPacc
nrnrrllrlc 1. IR2= (a, b) la, b e lR } is a vector space over the
Iield F = IR rvlth respect to the operations of vector addition in
IBPand scalar multiplicatio., on IR?.
dr"pr.""rrts the sei of all points in the plane'
E*unFle 2. Ep = {@' b, c) |.a, b, c e IR} is a vector space over
the field F = IR with respect to the operations of vector
addition in IR3 and scalar multiplication on IRS '
IBP asplssgnts the set of all points in space.
Example 3. kt V be any plane through the origin in R3 '
Then the points tn V form a vector space under the standard
addition and scalar multiplication operations for vectors in
IRP .
Erampte 4. For any arbitrary fie1d F and any integer n, the
set of all n-tuples (ur, u2, ..., ur,) of elements of F is a vector
space over F under vector addition and scalar multiplication
given by (u1 , u2; ..., uJ + (vl ,vz, ..', vJ = (u1 +v1 , 1r2*Y2, "', urr,+ vn )
o(u1, u2, ..., uJ = (cru1 , c[r2, ..., ou,r) where u1, v1, a eF'
Thls vector space is generally denotet by F". Th.e zeto
vector in F is tlle n-tuple of zxro i. e. O = {o, o, ..., o).
Example 6. Let M be the set of all m xn matrices with
entries from an arbitrary field F' Then M is a vector space
over F with respect to the operations of matrix addition and
scalar multiplication given bY
br" I
aln I [b, r btz
l-ar r e,tz
"r, dzz ^r! I*l o, bzz
|I ...
I
I ...
La-, a-z ... a-r, i Lb-, b",,
bz,
I
I
b-r
--l
219
VECTOR SPACES
COLLEGE LINEAR AI,GEBRA
218
=[
Ettt + brr dtz
ozt + bzt dzz
8lt
a2n
*bt, I
+bz,
+ b-r, J
I
I
3ml
""
+ btz
+ bzz
-[
+
bmt amz
?tr r
?ttz
4ln
dzz
dzn
ozt
2ml ano
I [oar t
cl?r2
I I o"r, clAzz
l=1...
o-n I Lo"-, @n2
Oaln I
oJa,zn I
.1
cra-r, I
where &1, bu and a e F'
Erample 6. f€t V be the set of all continuous real valued
f' g e v
functions defined on the closed interval Io, u. For'imy
(x)
(d
g
(.ld
+
(f
f
+
d =
and cre IR, deffne f + g and odby
(cd (x) = cd (d for every xin [o' U'
over IR '
Under these operations V becomes a vector space
of
multtple
(sirace the sum of continuous functions and scalar
any continuous function are continuous)
Example 7.I-etV be the set of all polynomials
co-eflicients ai
ao .lF + a1 ;F-r+ az /'-2 + ... + a*r x + a,, with
over F with
from ap arbitrary fteld F. Then V ls a vector space
polynomials and
respect to the usual operations of addition of
multiplicatton bY a constant
real
Example 8. L,et S denote the set of all pairs of positive
numbers. ir = (ur, uz), v = (v1 , v2),
(uto, uzo) (o is any scalar)
Deffne u + v = (ur vr, tTzvz)and am =
real numbers
where u1v1 and u2v2 are the usual prodqcts of
this
and u1s and u2q are the cr.th powers' S together with
addition and scalar multiplication is a vector
prescription for
space which we denote bY M2'
real numbers
E)rample 9. kt m', Ue the set of all positive
xywhere the
and define forx, Y = fr, a vector sum by x*y =
If a is
product on the right is the usual product of numbers'
is' the number x
any number and ,a R, define d x = f ' tllat
raised to the a Po\Yer.
vector addition
these definitions of
We claim that wtth
IR o becomes a vector space'
and scalar multipltcation
of all continuous functions
EsasPlG 1O' The set
differential equation
oo satisffin8
fH,
y=
- "' ' "':*
frre
of this
space' tln fact any solution
vector
a
is
2y
y" -y' - =O
y e-x and
a linear combination of =
equation-is
differential
y=e\.
_^_^
be a vector sPace
Theorem 6'1 l,et V
F. Then,
\
gver an arbitrary
n
and 0 eV' cro = O'
(i) For any scalar cr eF
v eV' o1
(ii) For o eF and anY vector
(-c) v = cr(-v) = (iii) For creF and v eV'
O'
v eV' then er - o orv =
(id If arr = O' *h"" aeF and
i*"t:
(r) 0O = tt'(O + 0) = oO
crV'
+cto
we get
Adding- uO to both sldes'
(40) + cro = (-oo) + (c[o] Cto)^
=(-cro+aO)+600=O+fl'O
or,O=O+q0:00 "'oO=O:
(o+ o) v + (-ov)
(u) O = ov + [ov) =
+ (-ov)
= (ov + ov)
ov
(Ov + - ov) = ov + O =
= (ov +
oll=O'
crrr + cr(-v)
(iii) O = sQ = cl(v + (-v) )=
sides' we get
Adding - cff to both
(cn'r + cr (-v))
- ca/ + Q = - Gv +
(-v)
. = (- cnr + crv) + cr
=O+o(-v)
(-v) = - ot.
or, - ay =ct(--v) .'' cr
(o+ t-cr)) v = crt/ + (- ct)v
Again, Q = ov =
-,1
fteld
22O
Adding - crv to both sides, we get
-crv+O=-crv+(mr+(-c)v)
= (- cry + crv) + (- alv
= (- cl) + a)v + (- alv
=ov+(-mr)=O+(-a)v'
or, - cn/ = (- a)v .'. (- s) Y = - ov.
(iv) Suppose that crv = O and cr* o, then, there odsts a scalar
a-leFsuch thato-rs= l, hence
v = lv = (a-l o)v = a-l (crv) = a-l (O) = O.
Ilefinition of Cartesian or Euclideaa space
Let n be a positive integer. The cartesian n-space denoted
by IRn is the set of all sequences (u1 , :u2, ..., u,r) of n real
numbers together with two operations
(ur, uz, ..., uo) + fu1 , v2, ..., vJ = (u1 + v1 , Ll2 *Y2, ..., u,, + v,r)
and cr (ur, uz, ..., uJ 1(aur, mrz, ..., mrJ'
ln particut"r, ql = IR is the set of all real numbers with
and multiplication.
their usual
2 For each positive integer n, Euclidean space
IRN
ofa vector space.
(i) Le.tu = (u1, u2, ..., u.,) and v = (v1, v2, .'., vr,) 5s in IRnthen
t
I
t.
i\
(iii) t€t O = (o, o'..., o) be lrr IRn ' Then for any
u=(ur. u2, ..', uJ m ff we wlll have
u + b = (u1, u2, .... uJ + (o,o,"',o)
= (u1 + o, u2 + o, '... ur,
u + v = (ur, uz, ..., uJ + fu1 , v2, ..., vr,)
=(ut *v1 ,u2 *Y2,..., ur+vJ
(l) is true.
= (v1 +ur, v2 +uz, ..., v, + u,r) = v + u. So axiomA
(ii) l€t u = (ur ,t)2, ..., uJ, v = (vr, vz, ..., vJ and
w = (wr ,w2, ...,wJ be in Rn. Then
{u +v) +w=(ur *v1 ,u2 *Y2,..., u'r+vJ + (wr,w2,..., wn)
=(ur *v1 *w1 , i.jr2+Y2*w2,..., ur+vrr+w,.,)
= (u1 , u2, ..., uJ + fu1 + wr, vz *Y,12, ..., vr, + wJ
= u * (v + w). So axiom A (2) holds.
:
+ o)
= (u1 . u2, ..., r-r,J = s.
IRn such
Moreover, if v = (vr, vz, .'., vJ is any vector 6
i (ur' uz' "" uJ
ttratu +v= u' then (ur *V1 , u2 *Y2, ""un +vJ
Therefore, u1 * v1 = ut implies vr =^o I
*
l? "' :.* 'il]"" ::.=
o
t
un + vn = un impues vr, =o )
i. e. v = (o, o,...,o) = O, Thus O ts the unlt vector witli
(3)
hotds'
property that u + O = u and so tlre a:dom A
(iv) Ietu = (u1, tl2, ..-' u'J and set
-u = Fur, -uz, ...' -uJ, Then
u + (-ut (ul, u2. ..', uJ + (<rr ' - uz' "" - uJ
=
=(u1 -u1, :Uq--u2,.-.' un:uJ
space.
: We shall have to show t]:at IRn satislies all axioms
22I
VECTORSPACES
COLLEGE LINEARAICEBRA
(o'o'""o)=O' I
'
ff such
Moreover, lf w = (w1 ,w2' ..., wrr) is any vector ln
(o'
and
thatu+w=O, then (u1 *w1 .u2 *w2, .''r urr+wJ = o' "" o)
therefore, u1 *w1 = o implles v/l =-ul
u2 * w2 = O imPlies w2 =-ttz
.
un +wn = 0 imPlieswn =-lln
'
property
i. e. w = - u. Thus - u is the uniqu€ vector with the
that u + (-u) = O and so axiirm A (4) holds'
and
(v) kt cr be a real number (scalar) and u = (ul' u2' "" u")
v (v1 ; vfi, ...,vr) be vectors tn IRn, Then
=
VECTORSPACES
COLLEGE LINEARALGEBRA
/
ct(u + v) = 61 (ul + v1 ,t)2 * Y2,..', ur, + vJ
+vr,))
= (cr(u1 +v1), cr(u2 +vz), ..., cr(u,
+ s/n)
= (du1 * crV1 , c[u2 + Gu2, "', clun
(ov1
,
.
ov2,
crtd
,
"',
= (cru1 , c[r2, ..., ouJ +
vJ
Y2,
"" = ou + crv'
= cr(ur, u2, ..., uJ + o(vr ,
So that a:dom M(1) holds'
(vi) I,et cr, p be the real numbers (scatars) and
lJ=(u1,u2;.-., uJbein ni, Then
(cr+
- '0 u = ((a+ 0 ur, (cr+ 0)uz' "', (s+ 0 uJ
= (ctu1 + Fur, clu2 * 0r:2, "" c[tln+ BuJ
= d{ul, t72r ..., uJ + P(u1, uz: "" uJ '
(2) is satisfied'
= cru * Bu So axiom M
(ut' !:z' "" uJ
(vii) L;et a, p be real numbers (scalars)'and u =
(apu1' "'' a0uJ
be in i*. T'tren (op) u 1 aP fur, u2, "', uJ =
= a(Bur,puz, .., FuJ
= o (0(ur, uz, "','uJ) =
a (0u)
,, $ adom M (3) holds
(viii) Let I be the unit scalar and
' ; = (ur. uz, ..., uJ be in IRn ' Then
lu = i(rr, *, ..., uJ - (Iur, lu2, ";, luJ
(u1,,u2, ..,. uJ = u
-=
So a$9rl M (4) is satisfied'
Theorem6.S.I-€tVbethdset.of.allfunctionsfromanonfunctions f' I eV
emptV set S into an arbitrary lield F' For any
'f
V
arri any scalar oe F let f + g and cr be the functions in
(x)
every
for
(cr0 (x) = af
defined by (f + g ) (x) = f(x) +,8(x) and
t'
I
i'i
xe S.
Prove that V is a vector space over F'
223
Proof : Since S is non-empty, V is also non-empty. Now we
have to show that all the axioms of a vector space hold.
t0 t€t f, geV, Then
(f + g 1:d = fl.{ + g($ = gH + IId
= (g + 0 [rd for anery xe S.
Thus f + I = g + f. So adomA (l) holds.
(ii) t€t f, g, treV, To show that (f + g) + h = f + {g + h),.It is
required to show that function (f + g) + h and thi function
f + (g + h)) both assign the same value to each x e S. Now
(f+ g + h) 19 (f + g B + h (, (flr)+ g(rd) + h E.
- (f + + h)) (d = (S + g + (.d =
h)
=f
= flS + €(d + h(d).
G
for every xe S.
But f (x), g(x) and h (x) are scalars in the f.ield F where
addition of scalars is associative.
Hence (fld + g0d) + h(x) = f(d + €H + h(rd)
Accordin$y, 6+ I + h = f + 19 + h).
So axiom A (2) is satisfied.
(iii)"I.et O denote the zero function.
O(xl = O for every xeS.
(f+ O) (d = f{d + O E = f (.rf + O = f (d
for every.xeS.
Thus f + O = f and O is the zero vector in V.
So axiom A (3) holds
tiv) For any function f e V,' let - f be the fr.Inction defirted
by (-0 x= - f (1.
Then (f+ (-0) (x) = flx) + (-0 A =f(fi -f(x) = O = O (d
for every xe S. Hence f + (-fl = I
So axiom A (4) is satisfied.
;.Y',
69 nrnrtrflcrlofrutap*Cc
V'
(v) lrt ce F and f' g e
((f + d (d)'= s 614 + gH)
ttren tatf + d) H = a
=;;;d;
H
= (ol)
+ (ad (d = (of + ugf
Exeoplcl.I,etVbeanyvectorspaoeoverthefieldF.Then
the set lol conststlng of the zero vector alone and also the
(d forevery xe s'
t:e field F where
(x) are scalars.tl
lslnce cr' f(x)' 6
over addttionl
multiplication is distrtbutlve
frerrcea[f+d=qf+og'
So a:dom M (1) is Pattsfied'
Then
=(od)(d+00(d
mP=11a,b,c) la,b,ce IRI
(d for every xeS'
Hence(cr+9)f=cf+0f'
so a:dom M (2) holds
(vll) l*ts"PeF and f e
'
V'
(Ffk))
planes respectivelY
Pol
unit scalar leF
(vlil) I-et f eV' tf'"" Ot the
"=iiot* lf (d = f['$ for every xe s'
=
M t4} rs
Hence lf = f'
vector space
satlsfled' so V ts a
are
tne-Jims
Stnce alt
J 1:*-
#;;
[:n T:
:' ]1.:: :::Ti L:il:"H;
ffiTT#;;;':n:'"::,H;ifr
'",#.
li'lJ;* :il
::ffiffiffiffi;
:$ f:
1
I ;;
oJn"* *':.1'^'T':i'*Yi
implies that
:Sil#tl'#H].";il':;;-e
crwl +$r4 eW'
\
Then W1 = {@, b. 0 la, be IR}, Wz = {0, b, c) lb, c e IR}'
and w3 = {(a, o' c} la, c e IR} are subspaces of IRP ' w'' w'
andW3 represent the sets of all points in the X-Y,Y-Z and'Z-X
(aS) fl$ = a
;;"
---=;(CIP)0 d = (o
[d for wery ,as'
dlo m =
Example 2' consider tlte vector space EP= {(a' b) la' b € R}
ttrenWl ={(a, 0 laeIR}, Wz ={(0,b) lbeIR} and',
W3 =(a,b) la=bandabe IRlaresubspacesof d'
Wr andWz represent the sets of aU points on the x-a:ds and
on the y-a;ds resPectivelY
Also W3 represents tfre set of all points on the line-5r = '16'
Uamptc 3. Consider thevectorspace
(vi) Let o, 0 eF and f eV'
+ pf 0d
(a + g) 0 (d = (cr + 0 f(d = 06(.d
= (af + F0
22:t
vECToRspicps
I
COLLEGE LINEARN'GEBRA
224
I
w' cr' 0 eF
fxanple 4. Consider tlie vector space
E9={h.b,d la,b,ce tsP1.
ThenWl ={@,O,O) IaeIR},W2 ={(O,b'q lbeIR} and W3 = {(0,0,c) lc e IR} are subspaces of IBP ' W1, W2 and
W3 represent the sets of all points on the x-axls, y-axis and
z -a:ds respectivelY.
fiample 5. I-et M22 be the vector sP*1jr of all squate 2 x 2
matrices. Then the set wof all2 x,2 matrices having zeroes on
the main diagonal is a subspace of the vector space M22'
Example 6. Let V be the vector space of all square n xn
matrices. Then the-set consisting of those matrices A = [aq] tbr
which au = aJr called synrmetric
is a subspace of v'
Linear,tlgebra-t6
StricSs
,#,
226
. VESIORSPACES
COLLEGE LINEAR AI,GEBRA
of all functt1nl of a
Erample 7.i-etV be the vector space
set of all functions f eV for
real variable x and let W be the
geW and let h = f+ g1' Then (5) = O'.
which f(5) = O. Suppose that f'
+ g[5) = O
g(O-r,
= O and so h(5) = fl5)
-'
+ O = O'
follows that heW and so W is closed ""1"j, "Oli*"^
k(5) = ofls) = ct'O = O
Again let k = crf where cr is any scalar'
multipllcation'
*i.rr"" ke W. Clearly, W is closed under scalar
Thus W is a subsPace of V'
is a
'--- ft"-ptu A. Tht set of all continuous functions
valued functions'
of the vector space of all real
'
subspace
complex numbers' Then C
Example 9. IJt C be the set of all
IR'
is a vector space over the real field
subspace of C'
l€tW = fiU lu e R i = '/eflt' ThenW is a
tlreorcm.6.4WisasubspaceofVifandorrlyif
(i) W is non-emPt5r
implies
(ii) W is closed t"tiut vector addition i' e' v"w eW
thatv +we W'
t W
(iii) W is colsed under,scalar multiplication i"":'Y
impliesthatove Wforevery cre F'
., ..,"
Proof:Supposethattheglventlrreeconditions.(i)(ii)(iiil
' etnd closed under vector
hord in w; then * o
1o".:*::1 Since the vectors in w
addition and scalar multiplicationM(2)' M(3) ana Uf4 hold in
reiong to V axioms A(1)' A(2)' M(1)'
By
to show that A(3)' A(4) hold rn W'
Y?.
r,
W. Hence we have only
ueW' then by condition (iii)
condition (i) W is non-empty' say
A(3) holds in w'
;;= ; .; *rd v + o = v for every vew' Hence
and v + (-v) = Q
Again. if veW then (-1) v - - veW
of v'
#;"; A(4) holds in w' Thus w is a subspacectearlv
the given
;;;;;;;;iv, if w is a subspace of V thentheorem is proved'
-
conditions (i)', {iil, (iii) hotd in W'
Hence the
227
Tlreorqn 6.6 A non-empty subset W of a vector spg.ceV over
the fleld F is a subspace of V if and only if
(l) u, v e W+u -v € W (ii) oe F, u e W+au q W.
Pr,oof : The conditlons are neeessar5r I j,''i ,,, . : , :, ;
If W is a subspace of V, then W is an abelian group.with
respect to vector addition.
,,. .
:
, ,
,
i, e, u, veW+u, -v€W
+u+(-v)eW
=+U -V eW.
' Also W must be closed under scalar multiplication, i, e
creF, ue W + crueW. Thus conditions (i) and (O are necessary.
The condltlons are suffcient
I€t w be a non-empty subset of V satisffing the two given
conditions, From condition (i), we have
ueW=+u-u.eW= OeW.
Ttrus zero vector of V belongs to W and it will also be the
zero vector of W.
NorOe W, ueW+ O-ueW=+-ueW.
Thus additive inverse of each element of W is at6b.in W.
Again, ueW, veW=+ueW, -veW:
+u-(-v)eW
=u+veW.
:
'"i : tl
Hence W is closed with respect to vector addition.
Since the elements of W are atlso the elernents .of V.
therefore vector addition will be cor-nmutative as well as
associative in W.
Hence W is an abelian group under vector addition. Also
fibm condition (ii) W is closed under scalar multiplication.
The remaining postulates of a vector space.will hold in W
since they hold in V of which W is a su.bset. Thus W is a
subspace ofV.
,?,
yECToRSPACES
COLLEGE LINEARAIrEBRA
Theorem 6.6 A non-ernpt5r subset woof a vertoi spac€ v over
the lield F is a subspace of V if and only tf
d, 0€F and u, v e W=oru + pveW;
hoof : ltc condltlontcneceseary
If w ls a subspace of v. then wmust be closed undervector
addltion and scalar multlpltcation, Therefore,
6ge IR, and ueW=+oUeW
peFand veW=+fl,veW.
remalnlng postulates of a vector space will
The
hold jn W sirlce
v.
Hence w is a subspace of
I
Dellnltlon : kt S and T be two sets.,By S n T uib
rnean the
lntereeectlon of S and T, the set of atl element"
and T, By S U T we mean the unlon of
to
"orr_o, S
S and T, the set of all
Theorem 6.Z.Ihe intersection of two subspaces
vector space V is also a subspace ofV.
S and T of a
Proof : Since S and T are subspaces of V,
they are nonempty and clearly O e S and 0eT. Therefore,
OeS fl Tand
henceSnT*@.
Now let u, v eS O Tthen u, v e S and
u, v eT. Snce S rra f
are subspaces of V, u, v e S implies that
ou + pv e S where
o,0 eF. Similarly, u, v eT tmplies that au + pveT
where
a, p eF.
Hence u, v eS fi T implies that au + pv.e
S O Tfor,o, F eF.
Therefore, S nT is subspace of the veetor
space V.
It is to be.noted that the union of S and T is
not a subspace
of V unless S CT orT C S,
[the sSrmbol "C" tneans contai:eed ir:l
.
Now ueWand - ue'W.
.'. u + (-u) eW=+Oe W.
.l'
zero vector of W.
elements of V,
therefore, vector addition will be associative as well as
cornrnutative in w, Thelefore, w is an abelian group with
rcspect to vector addition,
and ueW, then
they hold in v of which w is a subset.
Thus the condition ls necessary.
trhe condltloa fsgrficacmt
. suppose that w is a non-empty subset of v satisfirlng the
given conditlon,.i.e o, pEFand u, t eW= cru + pveW.
Irt o = 0 = I, then leFand u, veW+ lu + lveW
1.e,u+veW
ueW=+ueVand
[since
lu = u€V.l
Thus W is closed un&r vector addiflon,
Now takingo = - I, F = o, wesee that if ueW,
then (-1) u + o ueW+ - {lu) + O eW=+ - ueW.
Therefore, the additirre imrerse of each element of W isralso
Since the elements of .W arE also the
Now taking v = O, we see that if o, peF
cru+poewl.eorr+oew =+cruew.
i
Thus w ts closed under scalar multiplication.
Now cueW and pveW+ou +Sv€W.
Thus zero vector of V berongs to w and it udll arso.be the
22g.
{
''|
$
T
I
,#
,"1$l
,ro
tr,
ffi
Theorem'5'g The intersection of any fam,y of subspaees
ofa vector space is a subspace.
koof : kt {S, liet} be an5r family of subspaces oli
space V over the fleld F. Let g
= O 51.
tel
vector
28t
VECTORSPACES
COLLEGE LINEAR AI.CEBRA
230
We have au +
non,Since 0 eS1for every i e I, we have OeS' Hence S is a
empty subset of V. Now u, v eS.and o, pe F impltes that u €S1'
= (oa, clb,
ig
tr
I eI, since each 51 is a subsPace ofV'
.'. Gn + 0v eS.
WORTED OI,T E:TAMPLES
Show that S = {(a o, c) : a, c e IR} is a subspace
F IR'
over tlee real fleld =
matrices
^'^ofvwhere :
;;;;thatw ls a subsPace
pr,oof : For 0 e R3, O = (o, o, o) eS
Since the second comPonent of O is o,
Hence S is non-empty. For any vectors u = (a, o, c) and v = (a" o' c') in S and any
scalars (relrl numbers) s, P, we have
t"tttces' that is' all
svmml:T
all
(i) w consi*t"-or
ij = 1' 2' 3' """" n'
for which at1 = a1t'
matrices e = {all
(ii) W consoo
= (oa, o, uc) + (9d,o'Pc')
= (ca + Pa, o, ctc + Pc')
that A = [ag]
equal. Now suPPose
b''
that is, 4t= Etandbg =
'. Since second comPonent is zero'
Thus cu + pve S and so S ls a subspace o{.'*''
. Then
rx^r"r'fle 2. Shorythat
ij
IRa'
T= (a, t, Oe IR4t 2a-3b+ 5c-d) is a subspace of
",
pnoof : Foroe R4,0 = (o, o, o, o)eT
Sir:ce 2. o - 3-o+ 5.o - o = o
Hence T is non-emPtY.
Suppose ttrat-u = (a, b, c' d) and v = (a" b-' c"
then 2a - 3b + 5c - d = o and 2a' - 3b' + Sc'd' = o'
Now for any scalars (real nurnbers) cr' B
with a given
tii'-"t"""" which commute
19sol
tD' u' H' T
: AT = TA)'
LV
*':
matrix T; that
'"' ve
(0
OEW since all entries "j ":: ::trj:T:to W.
Proof i: (u
anc19 = [bi,l belong
,cu + $v's (a, o, c) + 0h'. o' cl
.
d
IR anda-'o*3=5]
' ra'e'r"lilc'3' w=-;';:; i"'u'ce
of d'
is not a subsPace
a 3'Q = o * 5'
o) €W' Since o-2'o
(o'
o'
o
d,
xn
=
For O.
of all square n
be the vector space
l-et'V
4'
Exanple
of the vector sPace IR3.
,.,
ac' qd) + 0a" 0b- Fc., Pd')
+ Pc-'dd.^f''
cr,b + Pb-' cc
and so
Thus cru + FveT
Therefore, S is a subsPace of V'
ffil.
b" i' d')
g"') - (qd + Pd')
= (oa + Pa,'
+ 9bl + S(cc +
3(ob
pa1
+
Also we have 2(aa
po = o'
+ 5c- - d') = co +
-3b'
A)'+
FtZa
5c
+
3b
= a(Yaof
'
T is a subsPlce
.Jl
v€'q for every i e I as S = nst and hence cru + pv eS1 for every
;.;
pv = cr(a' b' c' d) + P(a-'
d-) are in T
,
il
fl
""*T,
e F' oA +
fo' t"y."""tt'" a' I
*
aA
i"JJ,:i r",, + pbli'rhus
pB is the matrix whose
+ pB is also a svmmetric
aA + FB eW'
matrix. Therefore'
of V'
Hence W is a subsPace
B e W;
suppose that A'
Now
TO'
O
(ii) O eW since OT = =
BT = TB'
that is, 61 = TA and
41'
.232
COLLEGE LINEARALGEBRA
For any scalars cr,peF
VECTORSPACES
(CI Let,
=0(AT)+BGrO
=;, S. T ew and o = Ge IR= F.
rhen * = G @,s.2)= (sfi, s.15, z13 e,
)
=aEA)+pEB)
since its components are not rational.
(aA + pB) 1= (cA) T+ (pB) T
T(0A) +T(FB)
=T(uA+pB).
Therefore, oA + pB commutes with T. Thus aA + pB eW.
Hence Wis a subspace ofV.
eample 5.I€tV= IRP , show thatW is not a subspace of V
where:
lD. U. P. t9;n2l
(i) W = {(a, b, c) I a2O} i.e W consists
of
those
vectors of
^
IRP whose first component is non-negative.
(ii) w = {(a, b, d la2 + b2 + cz s l} i.e W consists
of those
vectors of IRP whose lengfhs do not exceed l.
233
HenceW ts not a subspace ofV.
Erample O. Let V be a vector space of alt 2x2 matrices over
the real field IR. Show that W is not a subspace of V
(i) W consists of all matrlces with zero
where :
determinant;
,{2 =A
(ii) W consists of all matrices A for which
noor: (il l-.tA=
[3 3J andB= [? 3 I
ThenA, Be[ since det (A) = O and det @)
ButA.'= [3 il .
= g.
[ 3l= [? 3] **
sincedet(A+B)=-l*O.
(iii) w = {(a, b, c) I a, b, c e
O} i.e w
""i"?",1J.:;'tr:,:l
of IR9 whose components are rational numbers.
koof : (i) ktv = (2, g,S) eWand c = - Se IR= F.
,
Then ov = -3 (2,5,5) = (-9, - g,-f E) GW, since _6 is negative.
Hence W is not a subspace of V.
(ii)
'
kt u = (O, l, O) and v = (O, O, t)
Then u(W since 0P + 12 +G = I < I
and veW since @ +G + 12 = l < l.
Now for any two scalars 0 = p le IR F,
=
=
ocu+0v= I (Ol,O)+I(O,0, t)
=(O;l,O)+(O,O, t)
=(O,1,l) 6y
Sirrce o2 + 12 * 12 =2 <1.
' Hence W is not a subspace of V.
Hence Wis not a subspace ofV.
(iil The unit matrix , =
,,=[3 ?]
But nr=lt ?
[; !J ororg" w, since
[ ?] B ?l
=
=r
Jo*" not belong to w
since (4t)2=tfi
.!]
tf;. lJ=[|u,!J*nr
Hence W is not a subspa.ce of V.
6.lO f,inear combiration of vectors
Let.V be a vector spaie over the field F and Iet
v1,. .., vre V
then any vector v e v is cared a *near
"o-uioatiootor
y1,y2,.'.., v,, if and only if there
exist gpalars d1, d,2, ...,*.;;
such that v = Gr v.t + %,yz + ... + okrVn
=l?",.
VECf,ORSPACES
23i-
COLLEGE LINEARAI-GEBRA
(2' L' 41'v2 = (1' -1' 3)
Examplc 7. Consider the vectors v1 =
(5' 9' 5) is a linear
and v3 = (3,2,5) ; UP. sno* that v =
combination of v1 , v2 and v3..
of
Proof : In order io sho- that y is a linear combination
that
such
F
in
oP
ez and
v1 v2 and v3, there must be scalars o,t'
,
v=0lVl +azvz +C[3v3
*
i, e, (5, 9, 5) = or (2' L' 4l + wi (1, I' 3) + oe
+ (3",'-2%'
=.{2c,1, sr, 4or) + @2, -a'2, Wl
g, + 2oa, 4rl,1+ 3a'2 * 1'u)
= (2ouy'+ a"2 +3ca, o1(3' 2' 5)
1'l
Equating corresponding components and
forming linear
cr1
4o1 +3a,2+fus=5
J
by elementary
Reduce the system (1) to echelon form
Ttren we
op.r"Uorr". Interchange first and second equations
hlve the equivalent system
o1 -
o'2 + 2'c4
=9)
2a1+ ca+3o.=5;
4 and then subtract
third equations respectively' Then we
We multiply first equation by 2 artdby
from the second and
have the equivalent sYstem
- A,2+2a3= ?^l
3W - cr3 =- t3.i
7g'2-343=-31 )
(9
are unknown scalars'
*3, 2) + wz 12' 4'-1) + cls' {l' - 5'n
i.e (2, -5, 3) = crr (1,
Sos' 7%)
+ (W' - 44a' - o'zl +(oe' = (or, -3crr' 2sr)
and forming the
linear sYstem we get
a1 +2a2 + cls = ?l
-g", - 4az - 5cr3 = -5 |
2ot- *2+7o4= 3)
form by the elementary
Reduce the system to echelon
equation O'
operations- We multiplv lst
respectlvely' Then we have
subtract from 2nd & 3rd equations
.'.1Y::
then subtract from
We multiply second equation ty I ana
syptem
the third equation. Then.we have the equivalent
dr- dz+20s = 9l
W- 2crs=-I3!
2l
_
{t=_ i )
Sothesolutionofthesystemisctl=3,aa=_4'o3=1.
Hencev=Svt -4v2+vg'
of vl ' v2 and v3'
Therefore, v is a linear combination
5' 3) in Ef ls a linear
Exemple 8' Is tl:e vector v = {2' -
Equating correspondtng components
Vl
4o,1 +Saa+5ctr=51
C[1
getcr, + 4+2=9 "'0'1 =t
the vectors
(1)
- W+2as=9 I
we have Gg = 1' Substituting
From the third equation'
12
we get 3CI2 - | = - 13 or' 3c2 =o3 = 1 in the second "qt'"uo"
Of d'2 =-4'
'
equation' we
first
the
in
I
4' uB =
Agatn substltuttn g W = -
I
' combinatlon of
and vs - ll' -5' n'
4'-1)
{2'
(1,
3,21,v2
=
vl
' = + cts v€ where o't' oQ' and o3
Solutlsn : trt v = o1 v1 + wv?
systemwe get
2o1 + c2 +3cr3 =51
:
{4
"
235
the equivalent sYstem
dt + 2W+ os = 2-1
2W -2a3= I f
'5a'2 + Sttg = -L )
tn*
236 i
COLLEGE I,INEAR AI,GEBRA
I
I zsz
I
VECTORSPACES
We multiply 2nd equation by
! ana then add with the 3rd
,l
We multrfly 3rd equatton bY-i. Then we have the
equlvalent system
equation. Then we have the equivalent system
.
a1 +2a2 *
0,3=2] or +2o,2+ 0e=21
2a,2 -2qs =^l
2e, -2o, = 1 |
l=
o+ o ='u)-
3
i.e it has no solution. Thus the vector v is not a linear
combination of the given vectors vr, v2 and v3.
Example 9. For which value of l. will be the vector
v = (1, 1",5) tn IR9 is a linear combination of the vectors
v1 = (1, -3, 2) and v2 = (2,-I, l).
Solutlon : kt v = o(l vt + aq vq where o, 1 and a2 arrunknourn scalars. i.e (1, l, 51 = sr (1, * B, 2l + a2 @, -1, Ll
={or, - Scl1, 2o1) + (zae, -o"2, ad
= (c1 + 2*j, -3q - o,2,2a1 + o2)
Equating corresponding components and forming the
linear system, we have
cr1 +2a2=11
(*)
Reduce the system to echelon forrri by the elementary
operatlons. We muldply Ist equation by -S and 2 and then
subtract from 2nd and Srd equations respectively. Then we
have the equivalent system
a1 +2fr2=l
I
fue ='l'+31
-3o,
=3 !
1
:
)
This system. has solutlon for l. = - 8 and the qolutlon is
Ot =3, U2=- 1
whtch is not true. Hence the above system is inconsistent
-3ar- ez'=)\l
2o4+ %=5 )
fu l"+3I
e2 =-|
o =rul
This system has an equation of the for* O =
ct1 +2a2=l
=
t.e th-e systern (*) has solutiori for'}, = - 8.
Hence v ls a linear combinatio[ v1 and v2 tf l" =:8..
ffiplc lo:writethematrix"= [?-] I*"
Itnear comblnatlon of the matrlces
=
^, [; -i I,
E U.E T. 1S4
=,Li ,1tr *o o, =[] I I
^,
Sokrttoa : Set A as a linear combinatlon of .$1, A2 and A3
usqg the unknourns rr,r, (b and os.
A1c1A1 +'tbzAz ttbAs.
rlratis, [?:i I *o,
[i-\ I .* [-l I I*". Ii ;l
. [-ff 3'l . tr' ;"'l
+oa+gu; 0,t+b-ael
='l,o-CIz+o -or+o+o I
[?: il = [3'-;'l
"'''Ja,
Equating co,mesponding components and. rrrming the
0,t+&Z*d'3= 3l
C[1 +g]2- OS =-1
[
-o"2 = tol I
-c[1
=-2 )
ttrence the ttre solution of the system is ct1 = 2, wz = -1, n, =2
Therefore, A = 2Ar -A2 +2Asi
that lis, A'is.a linear comb{nation of ,{1, A3 q$ 43.
.
Exemple ;ll. write the matrix
combination of the matrices
" = [?'- il ""
o,={l 3] ,"=[? ?] andAs=[3-?]
a lirrear
i
I
I
,.J
239
VBSTONSPACES
238
COLLEGE LINEARALGEBRA
Solution : Set A as a linear combination of ,{1' A2 and Ag
using the unknown scalars dt, h and cr3.
A=Ctr Ar + de rA,2 + caAg
[?-tl= *[l 3J .*t? ?l .*[3-?l
=[xl f'].[3, LI .[8-ff]
=[Il
.*:l
r?,:']
Equating corresponding eomponents and forming the
linear system, we get
01
01
C[1 +
-B
+2A3 =
0,2
=
W- 03=-
i]
Hence the solution of the systern is ct'1 = 3, wz = - 2, ac = - |
Therefcire-A = SrAr - 2Az - AA :
Ttr:ur4 is a linear combination of A1, ,{2 4nd A3.
flf f:near qran of a shbcet of avrcetor spacc
If S is a non-empty subset of a vector gPace V, then L (S)'
the linear span of S, is the set of all linear combinations of
finite sets of elements of S.
Example 12. The vectors e1 = (1, o, o), .e2 = (o, 1, o) and
e3 = {o, o, l) generate thevector space mt..fo, any vector
(vr, vz, v3) € mP i" a linear combination of e1, €2 8rrd €3,
specifically (vr , vz, vs) = vr (1, o, o) + v2 (o, 1, o) + ve (o, o, l)
= Vl€l +vz e'z + v3 €3'
Theorem 6.9 I€t S be a non-empty subset of a vectOr space
V, ttren L (S) is a subspace of V containiqg S. Furthermore' if W
is anrother subspace of V containlng S,.then L (S) c W.
S is a subset of L (S)'
Proof : If ueS, then lu = u€Sl hence
Also L (S) ls non-empty' slnce S is non'empty'
(S)' then
Now suPPose that u, veL
u = ctlul + ot2v+ "' + cl"u-and
v = prvr +52v2 + ..' + Prrvr, where
F' Thus for )" peF
ui, v1 eS and oi, ft are scalars in
l"u + pv = l,(cttut + %!z + "' + ot"uJ
t
+ F $rvr +fuv2 + "' + P"vJ
= (l,or) u1 +
(la2)u2 * "r * (l'otJ u*+
+
ftrpr)vr + U02) Yz + "' 00Jv"
elements of S and so
which is a linear combination of the
of V'
is again in L (S). Hence L (S) is a subspace
of V containing S and.
Now suppose that W is a subspace
all multipl€s cr1u1 ' azutz' "" cq"u*eW
u1 ,1r2, ..., u.eSCW' then
+ + oqnum€w; that
*
where cqeF and hence the sum c[1u1 buz "'
combinations of elements of S'
is; W contains all linear
theorem is proved'
Consequently, L (S) C W' Hence the
vector space V
Theoreq 6.1O : If S and T are subsets of a
over the field F' then'
(i) sGT+L(S)GLrD
(ii) L(SLrI)=L(S)+L[I)
(iii) S is a subspace of V if and only if L (S) = S
(iv) L(L(S))=L(S)
Pr,oof : (i) Lrt u = c[Iul + %rJz + "'
+ a"u" eL{S)
of S and
where {ur, uz, "" ur} is a finite subset
gf' therefore'
c^t, se,..., c&r€F' Since S
T'
u2, ..., ur,} is also a linite subset of
{u1,
So u = (t1 u1 * uavz + "' + crru,'el- fI)4'
e,
frn
,,t.
l.,-F -r-
24t
VECTOh.SPACES
,14O
COLLEGE LINEARAISEBRA
thusuel.is):+u€LfD. .'. t($ ELfD
HenceS e:r+L(s) cL(I)
(ii) I-et uel, (SU'I) then
+ +
g =orur + &zW+..., +CIru" + Frvr + 01,.vz "' F-v$ibset of SUT
where {ur, trz, ..., lrn' vt, v2. "1' vm} ts a flnite
vJ Ef' '
such that frr, rr, ..., uJ G S and tvr ' vzr ""
Now ct1u1 + o,2W + "' + qtu"eL (S)
and brvr.+ gzvz + "' + P,rrv,rreL[I]'
Therefore, uel. (S) + L (I).
LG) (A)'
Let w be any element of L (S) + Ifl)
Eudently USUT) EL (S) +
* v for ueL (S) and v eI{t)'
6$'.
?
then w =.1t
a linite ndmber of
Now slnce u is a llnear comblnatlon of
of a linite number
elements of S and v is a linear combinatlon
a linear combrration
ofT. Ttrerefore, t'*
;;;"*s
l9o"
Thus u + vel. (St].D'
of a finite numbers of elements of SUT'
Therefore, from (A) and (B)' we get
.
USIjD = US) + L(D'
to
(iii) Suppose that S is a subspace of V' Then we have
prrye that US) = S. Irt ueUS)'
cr2""' on€F
Then g r o,1u1 + e2W + "' + onuowhere c[1 '
ofv
and ur, u2 , ... u,'eS. But S is a sublace
scalar
Therefore, it is closed with respect to
multiplication' and vector addition'
Hence u = 0lul + d2W + "' + crru,,eS'
Therefore, L{S} E S' Atrso'S g -Iu,{S}i
i
'
,
L --r.
,!:
..
-,t
Conversely, suppose tJrat L (S ) = S. then we hayg to prove
Now we lmov that US) iq a subspac" of t., ,
.- : r,
Slnce S = US), Therefore, S is also a subspace of V.
,
, ,
r
(M We kr:aw that L (S) is a subspace of V.
thefiore, by part (iii), it follows th,tt L 1q511 = L (S).
gle 12 (a) Show tlrat tJ:e vectors u i (1, 2,'B), rv:s (0' 1,
d.
E;U"P. 19?51
Proof : We must determine whetl:er an qrbitrary veei<ii
2) andyT= (0, O, 1) generate
v'= (d, b, c) in dc"n be expressed as a linear combinati#
v'= xu + yv + zvr of the veetors u, v and w. Eryressing this
equation in terms of components gives.
(ab.c)=x(1,2,31+y(O, 1,2)+z(O, O 1)
- k 2x 3{ + (o, y, 25r) + (O, O, z)
= {x"2x+y, 3x+
;
4 + z)
i
Equating corresponding components and forming the
x
=a
2x+ y=b
)
i+
3x+2y +z=c )
z+2y+3rcc1
y+2seb I
ra )
ri : :l
.
The above system is in echelon form and is consistent. In
fact, tlre sys@ has the sohrtion x= 4, Y =b' 2a, z = c - 2b + a
rAu6 vand w generate (spanl mF.
ple 13, Determine whettrerrreeto.rs Yr =(2,, -1" 3J
yz=(4,1.2)andve={8,-l,S)span d.
I
whether
arbitrar5rvector
an
Sokrtloa : We must determine
v = (a, b, c) in d,W be orpressed as a Unear combinati6n
v = orvr * o'2v2 * ct3vs of the vectors v1 , v2 and vg.
&rprqssing ttlis equation in terms of components, we get
Ltnear Ngebra-I6
COLLECE LINEARALGEBRA
w?
r',r..:
(ab, c) = at (2,- 1, 3) + oe 14, t,2l +% (8,.i,:- 1, 8)
,i
j,...,
I
,
.. ':r.t
'-.
VECTOR SPACES
i
Now form the system of llnear equations by equating the
eorresporldlng components.
i
= (2a1, - cr,1, kr) + @o"2, aa,zoel + (t3os, - as, Scts)
= (2o4 + 4uz + S%, - or + q2 - oa, Sar + fyq.+ *6)
Equati+g. cg:IfFPold[g.3opp11nent3 and formin$ the
2a1 + 4uz + 8cr3 = 41
we
havC'd
llnear system,
I + 'W - os = b i
+ Zoq + 8o3 =9.,
3O1
r
ir-i r ir."r: i'
: ?hts pftiblem thus reou*eesito deterrrliirihg whether or not
Wrsystspl$ consistent for all values of a' b and c. No-w this
ssstefn y{!l F-e iaorlsiqtg$t,fo{ all a, .b and e, if and only if the
matrix qf co-efficients.
i,:,i
ii{):l'.rr1rri',r
..
r2 4 81
'' '"' A=l'-1 ll -1 lis invertibte.
Ls
.
2 Bl
il
:.
4",
lz 4 8 = 2(B + ii - a1-g+ 3) + S(4 - 3)
hl=l-r I -l8
=2O+20-40=O.
ls2
l'
{
I
"orr".qycftly,
(spud
ri, - (-2,-4,-8) and u3 = (3,9,27) generate
d.
Solution : We must determine whether an arbitrary vector
be expressed as a'linear combination
u = (a, b, c) e tri3
"r,
' u=xrur +huz*x3u3 of thevectorsrr,u2,zlnd u3'
Bxpiessing this equation in teil of compbnentS, we S
,
11!,")="('' l'') .*F2'4'-+r)+ xs(3'e'2v
x1-2x2+ 3x3=2i
ixt -4h + 9x, =b I
ixt-8;h,+27a5=c)
or, equivalently
x1 -2x2 +
3x3 = a1"'"' '-
''
x1- 8x2+ 18xs=2bf (1)
x1 - 32x2.+ loSxs =4c J
Now we reduce tHe system to eeHelon ,,form by the
elementaty transformations. We subtract ftrst equatiori frorr$l
the second and thtrd equattons respectively,'Then we haveE
.e
.: r r
theequivalentsystem , ,,
EI
!)
xt-2xz + 3x3 = 3 I
-.6:q +15x3=2b-al (21
-3Oxz+lo5xs;4c-a)
Hence the co-efficie nt matrix A is'not invertible and
vr, v2 urnd v3 do not span IR3.
u1 = (1, i,b,
2A&
:
tr-,zn' -4xi',8*,) + ts*' s" z z*11 '
: (*''gu;' ]"1),f,
E
"9
We multiply second equation by 5 and then subtra"t fro*fl
the third equatlon. Thus we'get tlib' equiValent system
x1-2x2 + 3x3 =4
-6xz+15x3=2b-a
I
: i6t
which is in echelon form
From the third equation, we get xs =i (2a- 5b + 2cl
Putting the value of x3 in the second equation,
we get - 6xz + 2a- 5b + 2c = 2b * a
or, 6x2 = 2a - 5b + 2c - 2b +a = 3a - 7b + 2*
.'. n=*,(sa-zb +2c).
'
)ii
Again, putting the values of x2 and xs ilr.the first equation
we get r, -l ts" -zb +2c). i {2"- 5b + 2c) =-r'
.
,
:
I
i
,l
.{
I
I
"{
{
''l
tia?_
coLLEGE t INBanALcEBRA
"oi, iS"i - rsaigbu- loc+6a- tsb+6c= 15a
Agaln, the columns of A are
or, 15x1 =%La-2'Ob+4c
xr =* {%La-zr)b+ 4cl.
c, =l
Therefore. (a, b, c) =
fi
[arr I
larz I
[ar" I
; , :; ,, ,.i
.L;;,J-L;;,J
::' o=l?.'.' I "" ::"
L;;"]
\Z+^-ZOA+ 4c) u1 +
* tg" - zb + ?.cl uz +* {zr - 5b
Hence ur, u2, u3' sPan trf .
Yerifioatlon of tlrc result
?lp
.YECTORSPACES :.
+ 2c) us.
(i) *@q"-20b + 4")-i $a-7b+2nl +*tz"-5b + 2s)
=]utz+^-%)b +4c- 15a + 35b- loc + 6a- 15b + 6c)
,, .: *$ {SO"* 15a + 35b- 35b + l0c - loc = fi {tSu') = *
", (ti;, $ .6tza- lob + 2c)-i tou- 14b+4c) +i 6a- 15;b+6c)
'':'" ''ofigza- lob +?,c-Soa+7ob-20c+ t8a-45b+ 1&)
*
ZAd
= S tsO^ 3oa + 7Ob - 55b + ?,oc -
=ft {rsu) =u.
(iii) * to"- 5b + c) -ittz"-2{3b + 8d "* (18a-45b + 1&)
I
=fi tOa- SU.+ c- 6Oa + 14Ob *4Oc + 54a- 135b + 54c)
=StOO"-6Oa+ 14Ob - 14Ob+55c-40c)
=* (15c) = c.
I
.,
=l
{
I
;
These n columns of A viewed as vectors ln Rm' span a
subspace of Rm called tlle cohrmn slnce ofA
,
'a
vbctor
of
T
subspaces'S'"rrd
trvb
any
Defialtioa : For
'l
space V, thelr sum S + T is the set given by {u+v ue S and veT}'
Thdorem 6.11 For any two subspaces S and Tofavector
i:,
1.,:.
spaceV, S +Tls a subsPace ofV.
Proof : CIearlY, 0 = 0 + OeS +T : So S +T * o
,
i,e.S+Tisnon-emPtY.
IJtrcye S+T, ot, peF.Thenx=ul +vr'Y=rrz*Y2
i,
where ur, uz eS and v1 , v2€T.
Thirs ax+ ff = o (u1 +rlr'1 + Pfu2 + v2)
= o(ur * cnrl + fr:s + pv2
=coltl*Frz+cr,ir'+frz'(l)
Now stnce S and T'are subspaces of V'
':
crul *frrze S and cnrl + Pv2e T. '
":'
'i
i
Hence tlee result is verifled.
6.12 Row qnae end aolumn sllEcc of e ua'rtx
Consequenfly, (1) Cfives that ax+ pyeS + T.
Irt A be an arbitrary m xn matrix over the real field IR:
6.19 Dir.ect sum of nftcPaccs
The vector spaee V is sald to be the dlreat gufoirbf'lts
l-arr dt2
a=1"" 4zz
I ...
L"*, an2
Hence S +Tts a subsPace ofV.
arn I
subspaces S and T, denoted by V = S @ T if every vector veV can
be written in one andonly one way as v ,.'u + w, where ueS and
a.zn
I
I
a-n I
The m rows of A are Rq= (a11, a,12,,"', arr,)
Rz = (azr, dzz, ..., aar). .,., Rj, = (a*rr 462,"',a6j'
- i.
,:1,:
' These
m rours viewed as vectors in Rn span e subspace of Rn
j
I
called the tos sPacc of A"
:
weT.
?heorem 5.12 The vector sPace V is the direct sum of lts
subspaces S and Tif and only if (i) V = S + T and (CI SnT = {O}:
Proof : Suppose that V = S (DT. Then any ve! 9,9 be
uniquely written in the form v = u * w, where ueS and weT.
"*fl\1l,)
*
:il
,iii
246
.
COLLEGE LINEARET,bBARA
2e7
VECTORSPACES
Thus in particular ; V = S + T.
Now suppose that veS O T, Then
{1) v = v + O, where veS and
:
l
2 . {0 flrrd xe d cuch that u +x=v,
$rherc u = (2, -1, O, 3. CI andv = (O'l' 2, -L,-2) D' U' P' 19ml
(lt) Solve the vector equatlon u + x = v for the following
o' r) :' '!'t''':
d:u=(|
(21v = O + v, urhere OeS and 'eT.
veT.
s' l-
palrofvectorsuandvm
S-ince gugh a srrrry for v mlpt be gnique.
.'. v = o; Accordingly SOT = {O}.
01 the other hand, suppose that V = S + T and SftT = {O}
kt ve V since V = S + T, there odst ueS and weT such that
v = u + w. Now we have to show that such a sum is unique. If ttre
sum ls not unique, letv = u' + rV' rvhere u'e S and r/ e T. Then
u + ur = u'+ w- 45rd so u - p'= w'- rr. qut u- u-€$ and w'-weT,
hence by SnT = {O}, u-u' = O and w' - W_ = 0.
Theref,ore, u = p and w = w'. TtU6 FuFh a sum for ve V is
uniqueasdV=S(ET.
ffpnte tF. Let S endTbe suFsp4cep of na 4"g4ea ry
S = {a, b, o, o} la, beF} ana T = f(O. O, c, d} h, a .F}. One can
easlty verts that both fhese subgets qre subspaces of IRa.
The zpro vector of Ba !s e, o, o, e. TtpF z ta, b, O, S
"
= {O, O, c, d) for sorne ai b, c. {e!lifpp[s tha]
ir = b = c = O. Flenqe z.* O, &th4t S OT= pf.
Therefore, S + T = S (ET, Npry pver.y {et,b;r pr, 4r} #"un
be expressed as {a1, br,O. o} * @..P, cr,dll urtth (ar, F,,. "O, Ol e S
::'
,
pfpB0l6pff-ff {4,
l. [,.etu = (1. 9, l, -Z),v= (3, O, 4, U a&C
'',,"* :. {6, 3, - 3. 0 be thp veciors in Uf . Compute"the foilowing
retons: {02u +v-w(il) 5u-3v-w.
,ffaryrerll: {i) (-1, f,9. -3}
{rt} F to, 7,-4.- r3).
I
Anmr: (i) x= (-2,2,2,4,4)
(ulx=(0,
t
fl
i
,,{
f
I
'}.
r'-t'-tz'o). d: ,
3. Let u1 = (-1, 3, 2, O), u2 = {2,9, 4., - 1)' u.a
:
f F, I' l, '*l nnd
2). Flnd scalars crr, a4, Qs ?r1d ct*.6pch that
E+ = (CI, 9,
o1u1 * &zrtz + ds us + %u* ={p' 5, 6, -3)
Ars-rEr i o1 = L,or2= 1, 06 =-1. ao = l.
,:,.:.:
l,
4. LetV be a vgctor sPace over the fleld F lrt u :m{ v any
_
I
!l
,},
Shorv that (t) (-1) v = -v
(ii) s(u - v) = cu - cw-
of
S.Verify whether the followtng sets are subspaces
.
. a','
vg (lR):
r
{i} t0q 2Y' 5) : x YelR }
i,
l
(ii) fiiq x+y,*z),& Y, z€IR l.
{nf*f*
0l t!€ eet ls rpt a subspace' '
F. U',P, l{Qtf
9f EP.
qv tlr,at lil = ltra, h, c, d) l[a" U, c' 4)
: naBn4, I ,. '.;. 1
b - 2s- d = 0, is a strhPace of B4
that lV
=
{(a, b,
"l
.
p-, F, s lpsfl,
e ts" h+b+c.: !1,!+,1,pghPfapE
that V = fia, b, c, d)
of ts4
pp:q lF{
,
eB" la +b c d 0l is
+
+
=
a
COLLEGE LINEARAICEBRA
';i..i'
'::j.
of the following are subspaces of IR3?
{k y,. z) eIR3 ly - oz = ol
[{:s{*{ .,(r),S^;,
rl
1, i.
, Fr T= !,n_\o =IR3 l**y
,
)
I
i
I
Urit W = {(a, b,
lr; U, ce IR and 2a - b+ c = 10} is
"t
ofVs 0R ).
ID. U. P. 19341
Shovs that each of tJ:e following subsets of the vector
ls a subspace of EP :
{i)w=11*b,c)la+b=6; -
D.U.II.T.1986l
t)
l
Aorrtl8 : (1, 0, q =;(1, O,- I) +0 (0, l, o) +i(t, o, U
(o, o. 1) =-)0,Q -t) + o (0, r, q +; (1, o. 1)
-4
WWrtte (8, 6, 0) as a linear combination of (-I, 2, Ol,
B,f,q.@,-t,oland(o, t,-1)
[D.u.pnel: ts3]
(5.6,0)=2(-1,2,Ol+ t (g,1,2)+ I (4,- I,O)+2(O, l._l).
-An-srer:
@. Wnrther or not the r"rror- (1, Z,;i't"; ,rr.",
cd:dination of the vectors u1 = (2, l,O), r:2 = (I, -1, 2)
and us = (0, 3, -4)
,
[C. U. p. Ig?gf
Anawer : (1, Z, 6) can not be written as a linear
(ii) W= {(a, b, c) lb + c = 0h'
(ui) W= {(a, b, c) lc + a
249
19" .Write the vectors (1, O, O) and (O, O, I) as linear
eombinattons of the veetors (1. 0, - l), (O, I, O), (1, O, l)
ID. U. P. 19851
::;
.
VECTORSPACES
, :.,';i,..
= 0}.
13. Let W = {(a, b,
U. lR } be a subset of the vector
I
",
spacetffP . Then show tirat W is not a subspace of IRp .
,:
p. u-H. T. resl
, L3',*1* = q'b' c.) la2 b) be.a subset of the vector space mP
Then show that W is not a subspace of IRF . tD. U. IL T. fggEl
.
I5. Show that W = (a, a, a) la e IR l ts a subspace of Eg .
B. U. IL 1e851
16. show that each. of the follorrrng subsets of -thq-retor
spac€ mP is a subspa.ce of IBp :
:, r," (flWl ={(ab,O) la,belR} (DWe=(O,b,c) lb,ceIR}.
coTF\)affon of u,, r.9 ard us.
ffiWnte the vectors (2,8, -Z, B) and (- 4,6, -lB, 4) as a
iirr# comblnaflon of the vectors
g), v2 - (3,
, v1 = (2, I, O,
-1,5,21, vs = pl, O,2, t)
Ansrcr 3 ti (U3, -7, 3) ;.2v1 - vz vs
:
(iil F4, 6, -13, 4) = 3vr - Bvz -vs.
rO0\fl O.turorine u,tether or not the vector (8, g, -4, -2) is a
lineAr-eombtnatlon of the vector set {(1, -2, O, A). {2,8, O, _l),
(2, -t,2, Lll
D. U. p. 1S4l D. U. II, T. 19Sl
Aorser: (3, 9, 4, -Zl= I (1, -2, O, g) + S(2, B, 0, -t)
.
l*r#"',t7.,J-e* Y be a vector space of all n-square matrices.over a
real field IR. Show that W is a subspaee of V if W consiits of
&&.*-"y""inetrie matrtces (i.e At =-A).
B.U;II. fry
''
i&
Shornr'that
w
'
is a subspace of IRa where
**iirir
q
=(a;lb. o) la, b, ceIR l
*r' ($W=
flxr, h. xs,*] la1x1 +&z xz+ssxs +%x4=eqeIR I
24. Express the potynomial p = t2 * n - . 3J1
comtrinatlon of tlrs polynomtals ..
1
pt =F -^.+5, k *h, -St and ps =tt3.
ADsmr:p=-3p1 +!p +4ps:'
I
*;?
.,
"'L*,
.
25.|n tjre veetor space tsP erpress the v = (1, - 2, S) as a
Iinear combinaUon of the vectols vectors v, (1, l, 1),
=
v2 = (1, 2, 3) and vs = (2. -1,.1).
Anccr:v={vl +S,ir2 +2vg. t
q.i
36. Determine rvhettrer (4,, 2, L, O) is a-linear combination
of ich of the foUprvi,qg sels of vectors. If oo'find s'e such
combinatio$.
(il l{L,2,-1, OJ, (1,3, l' ?)' (6: 1, O. 1}
93, I, o, u, [1,2' 3. 1), to' q'_6'-61]
(8, l' o' o)' (3' 8' -2-' -1)l
flo, -t' ?, u, (1. v,4.-21.
emrg" i (4 (4' 2, L, ol ls not e llnpar co-mbfnat{on
iiii- ti, 2, t, ol is ngf a linear corrpfnat[on'
(iii) @,2, L, O) is a linear combination and
(L' 7' 4' -21
{4, ?,1' 9: 216. -L,2, 1) +.1
-3(3,1,O,0) + O(3' 3' -2.-1)
z?.Express the matrix =
ll * a [near
{iil
iliit
" [-3
combrnauon of the matrrces
[l 8l' * = [l
^,
-i
=
I
p.u.Er. lsEN
srd AB = [-? el
f*rFGr:4=!$'*SP*tt&
l9 rl
28. Express (if popqlplel the mqtrtx A = [-1 -eI +s a
*;=[i-i[;:i-l
30.In qB3 , bt s =ll,2,l), (3, 5, o)l and
f ={1,2, l), (3, 5, o), (2, 3, -1)l
e-whether< S> * <T>.
R2.
,..3),6eterrnine whpther ol not the followtng vpptors span B3
" (il ,r, = (1, l, 21, rt2 = (1, -1, 2), us = (1, O, l)
= (-1, 1, O), v2 = F|, O, 1), v3 = (1, 1, l)
-4W,
Aemr: (i) ur, uz.ue span IR3
lfi ut,v2,vq sPgn m'3.
p9,.6noW that the space generated by the vectors
u1 I (1, 2, -1, 31, *z*{2,4, L,-9ilandtrs = {3, 6, 3, -n and tIrc
space generated by the vectors v1 Z 1L, 2, -4, I U arld
v2 = {2, 4, - 5, 14) are equal.
te.u.P. tq7g|
g),
(l;-1,
8}
l.
33.{il Show thet Ep = ffI,8, U, {&,
Q.u.Er. tgp$
ll ;;*=[l 3l
p.rr.E.t 19O$8-['s rgpq
fArppf I A parr no.t be erpreesed as a llnear conrbtrt+4qr
of
?i. Whlch of the following matrices are llnear
paq+HnaHons of t-he matrlcPa
3f'
^=[-l 3l'"=E 11,".[fi
3l *[-3 rll
',[3
tprrp, u,[3 8f =n*n*c
u,l [-3
:il "2A-,ts+c'
p. U-P. fgFq
h Show that the veetors Er = (1, U eld u2 = (1,-!f spap
ltnear combinatlon Pf the 4PfftcPa
Ar,
- Ae a4d A3.
251
VEC-TIORAPACES
COLLBGE UNEARAI'GEBRA
25O
(til FInd a condttlon on a, b, c so ttrat v r (+, b, Q
ic a llnear combinqtion of v1 = (1., -S, 8) a$
vz- lP, -1, -U, that !B , so thatvbbngB to spn [vr, vzl.
fncr: a+b+c=O.
*. WU ard W be the srrbspea of IBp deflnpd by
, P-.V. A. fffi
U@tt/'
S is a subspace
subsets
the
follourlng
that
eEch
35. 6hgv
of
ShorY that Ro =
of the lndicated space V:
{il Ye lR3. S is thp collectlsn pf all trlples {x y, al auch
r$a!x =yard z*Q,
(ii) V= Ifa. S ia tlre cpllection of all rF-tuples (x, r,z,$)
wchthatx-Y=?+L
II/ECTOR SPACES
COLLEGE- UNEAR AI.GEBRA
,,25i2
Linear dcpeodoacc md llncer lndependcncc.
36. Show tltat:e, p1aneW = (a, O, c) in R3 is generated by
Definitloa : Irt V be a vector space over the lteld F. The
vectors v1 . V2, ..., Vm eV are sald to be ltncarly dcgendcnt over
F or slmply dcpcndcnt if there exists a non-trrvial linear
(i) (1, O, l) and (2. O. -1)
tifl) (1, O,21, (2,O, 3) ard (3, O, 1).
,37.li) ProvethatS={(xy, z,tleIRa lx +y-z +t=O..and
.
combination of them equal to the zero vector 0.
Zx=yl is a subspace of IRa
(ii) Pfove thatT = {(xr, xz, .,,xJeRI lx1+x, +... +;q 1O}is
forht treast one i.
On the other hand, the vectors v1 , v2, ...., vm inV are said
38. kt Wr and W2 be the subspaces of HF deflned by
.to be llaearly independeat over F or simply in@GudGat
Wr=fia,b,O) labelRl
arldw2 = (O, O, c) b€IRI. Show that IRP =Wr@V/2. .
39. kt W1 and W2 be the subspaces of IRP aefinea ty
l
N.Ir:LS andTbe strbspaees of IBf dellned by
=
(a, 0, O O)' I a e IRI and T = (O b, c, d) I u, e, d e IRl.
Show that IHf = S(ET.
4r.tew (IR) = {[: ] J , .,b, c, d . ts']. rhen strow that
the only lrnear eombination of them egual to.o (zero vector) is
the trivial one. tleat is,
e['1v1 * ezvZ +... + oqnvm = O if hnd only tr
X@"eoElrn
(\/space
=
{[: 3 I ,". R].
6.13 The non-zero vectors v1, v2, ..., v' h a vector
v are linearly dependent if and only if one of ilre vectors
V1 is a linear combination of the preceding vectors
Vr, v2 . ..., Vk-r.
RroOf : Ifvs=GtVl +%vZ +...+Gk_rvr_r.
V(IR)=Wr Orillz where
iwz
if
dt=&Z=...=(LIr=O
N A single non-zero vector v is qecesgarily independent.
*'iffi". crv = O if and only if a = o.
Show ghat EP =Wr (il[2.
'S
\
That is, flrv1 * %vz +.., + oqnvm = O whene ar * o
asubspace of IRn.
W1 =(a,o,O) laeIRl andwz=(o,b,d lb,ceIRl.
258
fi,
\!.
:
ttrencrlvl +%vz +... +Or_rvk_r + (-I)v1 *ov1a1 *...*ovr,=e.
and hence the vectors yt,v2, ..., Vn are linearly dependent.
Conversely, suppose that the vectors vr, v2, ..., vn'are
42. I,et V be the vector space over ttre lield F.
linearly dependent. then srvl + a.2v2 *... + c'vn = O where tlle
Shour that V = Wr@ W! where Wl and Wz are the subqraces
scalars q1 * o, ttren c1v1 + %vz + ... + c1v1 + 0vL*1 +'...'+ &r,
of symmetrlc and skew-symmetrlc matrices respectively
Or, C,rvl + %vZ + ,.. + 01V1 =Q.
over F.
Now if k = l, this implies that c1v1 -O.
=0
w@nspAcEs
COLLEGE LINEARALGEBRA
B$4
r*rlth o1. + g" 90 that v1 = 0, Slying a contradiction' lrqequse
the v1 ls a non-zero vectof' Hence k > I and we may
wr,tevp=-(fJ,' (tr) Y2-..- (H)*-,
+ (d"rbr + oab2 +... + qlbp)v,
gfvlng v1 ?s s llnear combinatioh of v1' Vz' "" vL-t'
Thus the theorem ls Proved'
L.fi6r 1. L€t V be a vector space which is spanned by a
be
flnlte set of vectors, saY {vr ,Y2, ..',vr, }' Let U = {ut' u2. "" uit}
an itidelibndeflt set in V. Then k S'n.
i*Oof : Since &r, vz, ..., v,,l spans V' eactr vectoi lrt U is a
llneal combihaUon of {vr ;ur, .",v',}, i'e'
lI1 = O11V1 * d'21v2 + "' + okrvn I
U2=O12V1 *d22v2 +..'+Clm2vn |
(1)
{
t\ = 61r*Y, * 0.2YY2 + "' + a4vt
'l
Suppose that k>n, then tl:e system
C[11X1 * A12X2 * "' + C['1XJQ. = O
CL2tXr + dZ2h + "' + CZtXf. * O
'
(2t
'.:
ClntXl + A19X2 + "' + Ar1X1 - O
has at least one non-trivial solution, say fbr ' b""" h')
c[11b1 + apb2 +... + cr11b1 = Ol
02rb1 + o.22b2+...+cr2kbk =Of
(3)
t
,cnrbr * a",.2b2+ ... + ohkbk = 0
Thpn b1u1 +t:1.ttz +'.. +Lrr.gr.
,: .
... + oktv.,). +
= !, (cr11v1 + ctafz.+
0v1 +0v2+...+Ovr,=Q.
r.
Thus tf k>n, we can chooge b1, b2; ..., bk not all zero sueh
that b1u1 + ... + hur = O t. e. {ur, uz, ..., ur.} ts a dependent aet
whlch ls a contradiction to the given eorldition, go we
conclude that k < n. Hence tlte lerrtrna is proved.
IfAAa 2.lx:t {u1,;Ir2, ...,u.} be an tndeireUdent set lii d
vector space V. Let W be the sub3pac€ spanned by {ur, uz, ..., rlr}.
[.et v be a vector ruhich is in V, but not in W then
{ur, uz, ..., u, v} is an ttrdependent set.
Proof : Letclul + o4uz+ ;.. +ol rtr + cv= O (l)
We will show that et = % - ... + c(r = c[ = o.
If o * o, solve for v. This gives. i
Gr
A.o0-
v=-EUr-o=u2-...- Ar+
r::)
(2)
,:
The right hand s!d9 of fhip expressin (2) is a linear
comblnation of u1 , t)2, ,..,, F"
1!fl |re4ee in W,,b,ut the left hand
sl<le of (2) is not in Wt which is q eontradiction..Bewe eonelirde
lltirt a = o:,Tlri€ $ives ixiui + buz,+ ... + qk{rr,iQ.;,; i': /,.,, r,
:
Since {u1, rr2,-,..,.;iu. },is arr,independent s8t, 'rilU dttrisit,trave
'
't\all *(X2 =..,=0(r=O.
Hencewe have stfown thdt Ul =&) =... =cq. ='g=6
'lhercfore, {ur , uz . .J:,"u;,'v} is an iirdependent set.
,
.,,:-
.
So the lemma is .rt
proved.
'' :.: :ii;1
',
,
.,
j:
,i
r.' ri
.
r:,
i;
_l-
i r: I
:t 1 . ... t'
,
'
Thcorem 6. 14 The non-zero
rows of a matrix in echelon
,
;
i _i
hrrrrr ilre llnearly independent.
:
+ bg(c1r.vr *,&2YY2 + "' + q4vrr) .
8s.6
= (orrbr + o,12b2 + ... + ag brlvl +
(aerbr + o.4bz + ... + trertt) vz +
%of. , Suppose that the
set of flon-zero rows say
dependent rhen:l:*:11''*"
&o-r, ...,Rr) is linearlv
of the preceding rows i: e'
say R- is a linear combination
+ +
R- = ocrn+t &n+r + oLn+z &n+z "' *t ^ll ia
*c.Brs
flrst nonorR{n i8 its
rtr'
"o*ponent
the ktll
matrix is 'echelon form'
the
slnce
erltsy.Then
zero
-ln
kth
Rn are all zero and so the
of
"'
components
";.; o + cr"'*2 o + ao*2o + "' + crno = 0' But
cornponent of (r) is ctt-l
R*
that the kth comn3ne3t of
this contradicts tf'"1"1'*puon
Rn are linearly independent'
is not zero. Ttrus Rr'Rz' ""
il;;$#;;;;
prored'
. r^-^linearly independent
rn
be
Vm
t''
""
ttcorgm 6'15;;'''
Vm
Hence tlre theorem is
vi' v2' ""
a ltrpar combination of
vectors and a vector u is
Ttren the
+ clmvm where €t1 dfe scalars'
+"'
*
cl1v1
hvz
u
i.e. =
u is unique'
above representation of
{I}
+
Proof : Giveng=cl1v1 tCL1v2 "'+o(mvm
of u is not unique'
Suppose tl.at the representation
' Ttren let u = Frvr + P2v2 + "' + F-vwhere Fr are scalars'
(2) from (1)' we get
Nowby subtractin$
o=u-u = (crr-Fr)*ta, - Fzlvz*-'-'*(cr',,-FJvlinearly independent'
are
But the,t"to*-', ' w"""vm
o' :"' oqn- F* = o
So that this implies or - 0t =
i. e, ol =Ft'
* c,2Y2 + "' + c*vt is the unique
This proves that u = Glvl
of v1' v2' ""' vrr'
a lineai comblrration
as
u
of
representation
lireearly
the set {vr' vz' ':" vrn} is
Thcorem O'fe Supose ttrat
'r
II
ii
i
t
F
dependent' Then w is a
!t'Y2" "' vm
linear combinatign of the
Ibdof : $foicc tXre cet {vr ,Y2, ..., v*. \il} is llnearly depeqdent
there exist scalafs o1, oa' ,.'. q* F not all o (zero), such that
o1v1 f hvz +... + (&nvm + Prv = O'
If p . o thert one of ttre o{ ls not zero and
c1V1 * c.2v2 * "' + o'mvm = O' But this contradicts the
hypothesls that {vr,v2, ..., v6 } is linearly independent'
Accordingly P* o and so
o,* (-?) ",.(-?}v2rr..+( r")
*
That ls. w is linear combination of the vectors v1' V2, ;', V6'
G.1? I,etur, uz, ..., un be any n linearly,'
fi\"or"
lndependent vectois in a vector sPace V. Then any (n + l)
vectors y1,Y2,..., Vn+I, each of whiCh is a linear cornbination
of u1, u2, ..., un are llneally dependent, /
Proof : We prove the resuft by inducton on n. If any of the
v,' s ls zero, then trivialty given (n + 1) vectors are linearly
dependemt. So we suppose that none of the v1' s is zero. Now as
v1 and v2 are both ltnear combinations of u1, :u2, ..', "', u' with
n = I, We $€t vl = a{rr ' vz =- Garrt with a1 * o' o'2 * o'
This gives,r, =*,*, =fi
or,' \d1-Y=ur
-ur= o.
g-z
or, vr - argi-l v2 =O
Flence v1 and v2 are linearly dependent.
grn'
"', cqrr=
f
257
ITECT,OR, SPACES
COLLEGE LINEAR AI'GEBRA
vectors
To apply itxduction suppose that the result holds for an5r k
linearly independent vectors rvhere k < n. We can now write
V1 =Ct11u1 +Cl.lzU'z +"'+0lnut
vz = GztUt +ePP2 +."' + 02run
Vn+l = On+llUl * Ctn+tZU2 + ... + OQr+lnUn
fut sume sealars ary in F.
Llnear Algpbra-I7
r$l , t,
,1, .
'. iri
zSE,
COLLEGE LINEARAICEBRA
I[qr1,:f o forevery i=1,2,..., n + I t]ren each of vl is allnear
combination of (n - 1) vectors u1,Ir2, ..., urr-t and the induction
hypothesis will yield that v1 , Y2, ..., vn. are also linearly
dependent and consequenflyVl , v2 ... , vn+l are also linearly
dependent
So we suppose tl:at at least one of a6 * o.
Irt q, + o, ttren for each I = 2, 3,.., n + I
vl - 0rr, 0irr-lv1 = (air - Cftrr(l1101r-l) u1 *
(0;2 - C[inC[,12gln-l) ttz * ... + (cti'rFr - 06C[1'n-1C[1rr-l) Un*l
So by induction h5rpothesis the vectors
wr = Vr - ornorn- lv1 (2<i <n + 1) are linearly dependent.
Conseguently, there gdst 92, fu, ..., Bn*r in F; not all zero, such
that fow2 + fuws + ... * p,r+twn*l = O i. e.
(vg - d3ncllrr- rv1 ) +
fu2 (v2 -rlu2nalr,-lvr ) + Fr
. + &,*r (v,,+r - 0ea1'11C[1rr-lv1 ] =Q
i. e. - {fu20,.2notn-l + foo3rrcrlrr-l + ... + pnal crn'1'1o1rr-l) vr
+ Fzvz +ftva + ... + p.,*r vn+t = O.
This gives that v1 , Y2, ..., Vn.'1 ore linearly dependent'
Hence the theorem is Proved.
Theorem 6.18 Every set of vectors containing a dependent
subset is dependent and every subset of an independent set is
independent.
Proof : Eirst Portion
Suppose that the set S = {vr. vz, ....v-}
constains a dependent subset' say {vi, v2.. ..., vr}
Since {vr, vz, ...,vr} is linearly dependent, there exist
scalars G.t, a'J,..., oq not all o, such that
O(1V1 * UAVZ +.'. + O(V. =Q
25g
vEcToRSpACES
Hence there exlst scalars c,t, o.2,..., o. o, .., o not all o
(zero), cuchthatc,lvl * o,2Y2+...*orVr *ov111 +... + ov-=Q'
Accordtngly, S ls dependent.
Eccond portlon : kt T = {v1, v2, ..., v,,} be an independent
set then Prvt + Fzvz +... + pr,v- = O implies that
0r=b=...=Br=O
Now let {vr, vz, ...,vr} k < m be a subset of T, then there exist
scalars 9r, fu,..., Pk whereeach p1=o, i= T,2,..., ksuchthat
prvr + hvz +... + fuv1 = Q
the subset {vr, vz , ..., v1} is
lndependent.
OUT EXAMPI,ES
Prove that the set ofvectors
lQ, 1,4, @, t,-l), (4, 3, 3)) is linearly dependent.
Proof : First Procesa
Set a linear combination of the given vectors equal to zero
by using unknown scalars &y, z:
x(2, 1,2) + y (O, l, - 1) + z(4,3,3) = (O, O, O)
,
or, (2x, x. %d + @, y, * y) + (42, 32,32) = (0, O, O)
or, (2x+ O + 42, x+y + 32, 2x-y + 3zl = (O, O, O)
.
.
Equating corresponding components and forming the
linear system, we get
2x+O+42=O)
x+y+{"=6 I
2x-y +32=O
trl
)
Reduce the system to echelon form by the elementary
transformations. Interchange first and second equations.
Then we get the equivalent system
x+y+32=O)
2x+O+42=O I
2x-y+32=O
J
el
",!
4
26t
VECTORSPACES
we subtract second row
coLLEGE LINEARAIcEBRA
2@
the
by 2 and then subtract from
We multiply first equation
get the
respectively. Ttren we
second and third equations
equivalent sYstem
x+
--ivY+32=O I
-zr=9 I
-3i-32=o )
(3)
by-2 and-3 respecttvely'
Divide second and thlrd equations
get the equivalent sYstem
y+ ,=ol
(4)
Y+ z=O )
we can
equatlons are identlcal'
the equivalent system
of tJ:em' Then we have
Since second and third
disergard one
x+y+rz=3
y+
1
(5)
form and has only two n.orr-zero
This system ls in echelon
has non-zero
unknowns' hence the system
three
in
*u"*n"
vectors are linqarly dependent'
solution. Thus the original
matrix whose rows are the
Sccond Proccss : Form tlte
to row echelon form by
mltr{
the
reduce
and
vectors
given
Gfttg the elementary row operations'
12 | 21
lo I -1
I
2 and then subtract
we multiply first row by
row.
'
t.
l,
T
J
,tt
12 |
o ol
I
form and has a zero row:
Thls matrlx is tn row echelon
linearly deperrdent'
hence the giverllectors are
of vectors
r.xatnPle 17. Show that ttre set
is linearly dependent'
1'
(3, o, 1, - 1)' Q, -l'o' 1)' (1' 1' -2))
ID. U. S. 19841
to the
of the glven vectors equal
Set a linear combination
scalars x: y' z t
zero vector ,ltttg unknown
(O' O' O' O)
Ll + z (l' L' l' -2)=
x(3, O,'1, - f) + y (2' - 1' O'
o' o)
o' yl + (z'z'z'-2zl =(o' o'
or, (3x o' x -d + (2y'-y'
(O' O' O' O)
x + O + z' - x + y - 2zl =
or, (3x+ 2y + z,O - y + z'
equations by equating the
Fornt the homogeneous linear
corresPondin$ comPonents'
3x+2Y+z=Ol
-!+z=O L
x+b+z=o I
trt
-x+Y-22=O )
form by elementary
Reduce the system to echelon
first and third equations'
transformations' Interchange
Then we have the equivalent
L; 3 3J
.
21
-16 I -l
-lo
Proof : Ftrst lEocesg
Thenwe
x+Y+32=O I
12 I
from the third row'
21
II -r
-lo
Lo -1J
I
from the third
x+O+z--O)
-Y+z=q I
3x + 2Y+ z,= l)
sYstem
el
|
-x+Y.-22=O )
we multiPlY second equation
bY -1'
,
262
263
COLLEGE LINEARAI,GEBRA
We multiply first equation by 3 and then subtract from the
third equation. We also add first equation with the fourth
y- z=O
I
the second and thtrd rows
respectively'
,l"i
,:i
rr I I -21
-2 5l
-lo -3
-3 -2 5J
equation. Then we get the equivalent system.
,1,
Lo
(3)
iY -22=o |
VECTORSPACES
frorn
by 2 arrd 3 and then subtract
We multiply flrst row
y- z=O )
We subtract second row
from third row'
Again, divide third equation by 2.Then we get the
equlvalent system
:
x+0+ z=0 )
e,
;-1,=|f
y- z=O
{n u..torc are linearlY dePendent'
(2' - l'4) (3' 6' 2) and
18. Show ttrat the vectors
)
'
Since second, thtrd and fourth equations are identical, we
can disregard any two of them. Thus the system (4) reduces to
*.
9:
z:31
(s)
This system is in echelon form and has only two non-zero
equaUons in three unknowns ; hence the system has a non-
zero solution. Thus the original vectors are linearly
dependent.
Sccond Process : Form the matrix whose rows are the
glven vectors and reduce the matrix to row echelon form by
usirlg elementary row oPerations
lz -l o rl
Lt I I -2)
:t
{,
Ls
-ll
Proof : First ltocesg
to the
of the glven vectors equal
Set a linear combination
scalars )c'y' z "
zero vector usin$ unknown
10' - 4) = (O' O' O)
x(2,-1, 4) + y (3' 6' 2l + z(2'
(0' o' o)
zyl + t2z' ltz' - 42) =
or, @x - x. +$ +(3y' 6y'
(O' O' 0)
x+ 6y + LOz' 4x+ 2y -4zJ =
oy,(2x+ W + Zz'of linear equations equating
Form a homogeneous system
:
tJre corresPonding comPonents
2x+3Y+ 2z=O I
- x+6Y+ loz=0 |
'(1)
to echelon form by the elementary
equations'
Interchange first and second
Reduce the systenr
Interchange lirst and third rows'
t
are linearlY indePendent'
4x+fu- 4z=O )
:
13 0 I -lt
l-1 I I -21
0 1l
-lz -1
oI
ir, *,:4)
-
transfromations.
sYstem
Then we have the equivalent
-x+6Y+1Oz=Ol
' Zx+5v + 2z=O I
4x+ iY - 4z=O )
{21
264
VECTORSPACES
COLLEGE LTNEARATIEBIA
We multiply first equation by -l and we divide third
We multtply flrst row by 3 and 2 and then subtract fi.orn
equation by 2. Then we have the equiValent system
x-6y- loz=O
2x+3y+22=Ol
2x+y - 2z=O )
the second and thtrd rows respectively.
rl
)
We multiply second row by $ and then subtract from the
second and third equations. Then the system reduces to
'
-
third row.
"[rO -95 -2 1
-|
t4)
1
I
|f .ra then subtract
from the third equation. Then we have tfre equivalent system
x-6y-l0z=Ol
,ur.*
o^l
?r:=
-16z=O)
)1
{
1
(5)
which is in echelon form.
In echelon form there are exactly three equations in flrree
unknowns; hence the system has only the zero solutJon x = O,
y = O, z = O. Accordingly, the vectors are linearty independent.
Second Pnocess
Form the matrix whose rows are the given vectors and
reduce the matrix to row echelon
_form by elementar5r row
operations
f2
I
.!
ti
i
i
I
1l
{
t
Since the echelon matrix has no zero-row, the veetors are
linearly independent. fi
Exanple 19. Let rf v ana w are independent vectors, Show
that u + v, u - v, u - 2v + w are also independent.
p. u. s, 198(), rsgl
Proof : Set a linear combination of the given vectors equal
to tlre zero vector using three unknown scalars x, y, z:
x(u + v) +y (u -v) + z (u-2v + w) = e
or,,rr.r + x/ + yu -)n/ + zt - ?,nr + zut = O
or, (x+ y + z) u + (x-y- 2z)v + ant =A (l)
Sfuece u, v and w are linearly independent, the co-efflcients
in the above relation (l) are each O (znro), tlrat is,
x+y+ z=O )
x-.y-22=O I
41
ls -l6 2l
Lz lo -4)
We divide third row by 2 and then interchAnge with the
first row.
rI 5 -21
-ls 6 2l
Lz -r 4)
9^ lwhich is in row echeton form.
Lo o -;t -l
J
We multiply second equation by
5 -2]'
-e 8l
-lo
Lo -lr 8_J
(3)
We multiply first equation by 2 and then subtract from the
x- 6y-loz=Ol
t5y +222=0 |
l3y+l8z=O
265
z=O )
The only solution to the above system is x = O, y = O, z O.
=
Hence the given vectors u + v, u - v and u - 2v + w are
independent.
F=rrnplG 20. Test the dependency of the following sets :
(r, 2. - 3), (2, O, - U, r, 6. - rl))
(ii) {(2; 0, -1), (1, 1, o), (o, I, l)}
(r)
ID. U. P. 19841
266
VECTORSPACES
COLLEGE LINEARAI,GEBRA
Solution : (i) Set a linear combination of the glven vectors
equal to zero vector using three unknown scalars )c y, z
{.L,2, -3) + y (2, O, -t) + z (7,6, -11) = (O, O, O)
1
or, (x 2x - 3,{ + {2y, O,-y) + {72,6z,, - llz) = 19, 6, 6;
or, (x+ 2y + 72,2x+ A + 62, -3x-y - llzl = (O, O, O)
Equattng the corresponding components and forming tJ:e
linear system, we get
x+Zy +72=O )
2x+ O+ 6z=0 I
3x- y-llz=O
(t)
)
Reduce the system to echelon forrn by the elementqrlr
operations. We multiply llrst equaflon by 2 and then subract
from tJ:e second equation. We also multiply tfre flrst equaflon
by 3 and then add with the third equation. Then we have the
equivalent system
x+2y +72=O 1
-4y- 8z=O I
5y + LOz=O )
@)
(3)
disregard one of them, Then the system (3) reduces to
I
(4)
The system, in echelon fornt, has only two non-zero
equations in the three unknowns; hence the system has a
non-zero solution. Thus the original vectors are linearly
dependent.
/i2,O,- l) + y(1, l. 0) + zl0 - t, 1) = (O' O' 0)
',
or, (2x 0, -,S + (y, y, O) + (0' -2, zl = (O' O' O)
or, (2x+Y + O, O +Y - z,-x+ O + z) = ( O' O' O)
Equattng corresponding components, and forming the
linear system we get
2x+Y+O=O I
O+y- z=O I
-x+O+z=O )
(I)
Reduce the system to echelon form by the elementary
operations. We multiply third equation by - 1 and then
interchange with the first equation. Then we get the
equivalent system
x+O-z=O I
O+y-z=O l
(21
We multiply flrst equationby 2 and then subtract from tlle
third equation. Then we get the equivalent system
x+O- z=O l
y_ z=O |
(S)
Y +22=O )
Since second and third equations are ldentical, we can
x.'{iU:8
(ii) Set a linear comblnatlon of the given vectors equal to
the zero vector using unknown scalars x, y, z :
2x+Y +O=O J
We divide second equation by - 4 and the ttrird equation by
5. then we have tlle equivalent system
x+2y +72=O )
y +22=O I
y+22= O J
267
r We subtract second equation from the third equation'
Then the system (3) reduces to
,;"=31
x+0-z=O1
(4)
From.tlte third equation we get z = O.
Putting z = 0 is the second and lirst equations we get y -- O.
x = 0 respectively. Thus x= O, Y = O, z=O, Hence the given
vectors are linearly independent.
268
COLI,EGE LINEARALGEBRA
rx-mple 21. I-et V be the vector space of all 2 x 3 matrices
over the real lield IR. ShowthatA, B, C eV:
"= [;
,.,a c =
*n
il'"= fi -l -;l
269.
vEcroR$PACps
6th equation by -1.
We rnultiply 2nd equation bV i
"tr.a
Then we have the equlvalent system
x+Y+32=O )
[! fr -T I *" rnearly a.p"r,a".,t.
[D.U.H.T. 19S61
hoo,f : IrtxA+yB + zC = Owhere xy, ze 8..
*.",[i -'^il.r[i -l -tl.,,E;3 -l]=[333]
." [rx -X'x] . [d u{ 1'rr]
.l3z;i: -'z] = [B B BJ
-T; {r; ?&
-I
I
7,
",,lrXl
'i: t{ :': I = IS S S I
.Equating corresponding components and forming the
linear system, we get
x+ y+ 3z=O l
2x+ 4y + 2z=O I
2x- y- 8z=O L
4x+ 5y +lbz= O I
3x+ 4y + 7z=O I
x-2y- z=O )
Reduce the system to echelon form by the elementary
operations. We muiUply lst equation,by 2, - 2,4, 3 and,,l and
then subtract from 2nd, 3rd';'4th sth and 6th equations
respectlvely. Then we have the iquivalent system.
x+ y+32=O'1
2y -42=O
I
{:"2=3 I
-{;3:=3 J
:
y-22=O I
y-22=O L
y-22=O t
y-22=O I
Y-A=O )
Now 2nd, 3rd, 4th, 5th & 6th equations are identical, we
can disregard any four of thegr.
Thus we have the equivalent
flstem
x+y+32=OI
y-22=O I
.This system is in echelon, fgrnr lraviag two equations in
three unknonrns. So the syslem has 3 - 2 ='I.free variable
which is z. Thus the system has non-z€rci:Bolutions.
Letz= l,theny =2artdx=-5.
'r
.'.-5A+28+C=0
Henethe$lvenmatrices A" B and'C are tinearly dependent.
fampte 22. ,etP(t) be the vector
"*".
of all polynornials
of degree < 3 over. the realtfi.Id m.. Determine whether the
followlng polynomials in P[t) are linearly dependent or
independent:
u=f + 4t2 -2t+3, y=13 a61z 1L+'4and ''
w= 3ts +8€ * 8t+7.
Solution : Set a linear combination of the gjven
polynomials u, v and w equal to the zero polynomial using the
rrnknown scalars x y and z: thatis, xr + yv + zttr = O.
, -,1..:*:
27t
This system le ln echelon form harring two equations in
three unknowns. So the system has 3 - ) = I frree variable
VECTORSPACES
27O
COLLEGE LINEARAI-CEBRA
Thus xl1+4t2-2t+ 3) +y (ts+6t2-t + 4l +z {3t9+
+ 4y + 3zts + 8zt2
or, xt3 + 4xt2 -2t +3x+yts + 6yP -fr
-8zL+72=O
+ 3x + 4y +72 = O
t
+
or, (x+y+32) 1s'1 (zlx+6y+ 8zl t2 l-2x-y-&zl
8t2-8t+Z) = O
of t each equal to 0
Setting the coefllcients of the powers
:
(zerol,*Jg.i the following homogeneous llnear system
4x+6Y +82=O L
J
loar"" this system to echelon form by the elementary
by 4' -2 and 3 and then
operatlons. We multipty lst equation
subtractfrom2nd,3idand4thequationsrespectively'Then
we have the equivalent sYstem
x+ Y+32=Ol
.
2Y-42=Ol
Y -22=O I
Y-22=O )
Then we have the
Interchange 2nd and 4th equatlons'
equivalent sYstem
x+ Y+32=O I
Y-22=O L
Y-22=O
I
2Y-42=O )
2 arrd then subtract
We multiply 2nd equatiolt by 1 and
from the Srd & 4th equations respectively'
equivalent sYstem
x+Y+32=O I
Then we have the
-i;*Bl= ".Ily=Bl
O+ O= 0
J
Hcnce the system has non -zcro solutions' that ls'
ru + )w + zN,t =O does not imply that x= Y = z = O'
Thus the given polynomials u, v and w are llnearly
dependent.
PXERCISESB(B}'
l. Show that the vectors (l' O, O)' (O, 1' 0) and (1' 1' 0) in
ID'U'P' 19801
Vs (IR) a2\earlV dependent'
x+ Y+32=O l
-7.; *\;?',=3
whlch ls z.
- Z. rttticn of the following sets of vectors in lR
rnffindent?
3 are linearly
(i) (1, O, 1), (-3, 2.6),14,5' -2)l
(ii)(1,4,2),(3,-5, LI,p,7,8), (-1' 1' 1)l
-/'
: (i) Independent (ii) Dependent'
nnsrf
./
.,{Vrou. that'the followlng sets of vectors in Rs are
llnearly independent;
^4
(i) (1, O, 2). (-r,1, O), (O,2, 3D
r (ii)(l, 1, 1), (o, l, 1,) (0, o, 1)l
:
4. (i) Prove tl:at the set of vectors (2' 1, 1); (3, - 4' 6)
and (4, -9, 11) is linearly dependent in m'3' tD' U' S' I#tl
(ll) Test the set {1, O, 1)r (O' 2,2), (,3,7, 1)} for linear
rk'lrcnclence on IR3.
Anewer : (ii) the set is linearly independent'
!-r. Sl:ow that the set of vectors
ri = l(1, o, O), (O, 1,0), (o, o, 1)' (1' 1' 1)lr" R1 is linearlv
rlr.;rt.rrtlcnt brrl that any set of three of thern is linearly
Inr ['llcnrlt'trt.
4l*
,:,*S,;::C
,{
272-oA**r"*er
VECTORSPACES
COLLEGE I,INEAR AI,GEBRA
ttre foltowir4g sets are rinearry
" -tuuo, o, 1), (1, r.-21 p' a' lD
lndependent :
(iil t(l, -1, s'(1' 4, r,l' @ -3'nl'
p'U'A' 1S1' &g' P' 1S1l
(tB ttlA-1)' (1' 2' 3)' (4' 5' -3$
({iif'dependent'
(0 lrdependent (ii) Independent
;};,
fotlowing two
the llnear dependence of the
nZ/Wrrrine
----n R3:(1, O; ts3' 2' 61' (4' 5' -2)l
"udasof
-L),
til's =
'
tD'u' P' 1s8l
(1, r'rt' io o' o)' (1' r', 0)l
els;Grs : (fl S is linearly independent'
(ii) T is linearlY dePendent'
4' 21' (3' -5' 1)' (-1' 1' 1)l is a
8. (8 Decide wt*ttee* s = i(r'
p'U'P' 19&n
of IRa
lineaely independent subset
IRs'
2}' (1' :7' -81(2' 1' -1)} of
(il} Conrsrder tlte subset {(l' 3'
tD'U' P' 19?91
subset'
Test the &Pendu,r€e of8rc
Aocrctt : (il'S is lbaedt' rndepcmdent' . , '
($"The subeet ls linearly dependent'
1)
u'= (6' 2' g' 4l'v = (O' 5' -3'
9. $ftrow ** L veetors
s..198o
llnear{y independerrL ID'u'
and w = (O o, 7, -?tare
prove that
v2' v5} is indeperrdent
IO. If'ffre vector set {tr,
vt + v3) is independent
(a) the set tvr +vl-2trs' vr -Y2 -v3'
+ 3u2 - vg' vz + v3} is dependent'
{v1 + v2 - 3vs' v1
a$d (b) therct
u' P' 1s4l
'/
.
G0 T =
P'
'*t/ru,t'*ine whether
in IR3
or not the following vectors
or hdependent :
are hnearly dependent
(2' 1' -1)' vs = ff ' 4' Ll
(i) vr $, - 2'1)' vz =
=
(fl) u =
$,-2'4' 1)' v ={2' l'O''3} dndw=
(3'
[C'U'P' 19731
-9:t-3*
dependent
Ic"U'P' 19781
Ailtr€rg: G) vectors are linearly independent'
(ii) vectors are linearly
ZZs
12. Determlnc whether each of the following sets are
ilncarly depenrlent or linearly independent :
0l l(0, r, o, 1) (r,2,3, -u(8, 4,3,21. (o, 3, 2, O)l
(u)(1.3,2), (1, -7,-8), (2, 1, -l)l
D. U. Pr,ql" 19831
(i)
AnasGrs: vectors are linearly independent.
13. (l) Show that the set {(4, 4, O, O), (0, O, 6, 6) (-5, O, S, S)} is
';
Ilnearly independent in IRa.
(ii) kove that the set {(1, O,2,41. (0, l. 9, 2), (-5,2, 8, ,.1,0}
is linearly dependent tn IR4.
. 14. Determine whether the following sets of vectors in IRa
are linearly independent :
(i)fi3, o,4, l), (6, 2,-r,2), (-1, s, 5, l), (-3, 7.8,3))
(ii) (4 - 4,8, O), {2,2,4, O), (6, O, O, 2), (6, 3, - 3; O)l
Angwers : (i)The set ls linearly independent.
(ii) The set is Unearly independent.
A5.
Show that the followlng sets of vectors in II{3 are
dependent :
ffiury (i)
i {(1, l, -r), {1,2. sI, {3, 5, -3)} i
(ii) {(2, r, l), (3, -4, 6) (4, - e, l l)}
i'
ID.u q. rfflll
tD. u. s. re83l
:
16.,Show that the lbllowing sets of vectclrs irr [R3'arre
linearly independent :
(l) (1, -r, 3), (1, 4,5), {2, --3,7)}
lD. v.H. Lg/,4ll
(ii) lu. -2. Ll.(2. t. -t),(7 4. ri)
lc. u. P. 19731
17. Detennine whether the f<tlkrwing sets ol vectors in IR3
rure linearly depenclerlt or inclcpendent :
(0 {(r, -2, 1). (2, 1, -Ll.t7, -4, 1)}
rc. u. P. L97SI
(ii) {(--1, l, l), (1, ',i,2) (3, -5, l)}
lD.u. P. 1982I
Ansrpers : (i) Lineirrll' rlc:pt'rrtlcrrt
(ii) I.irrcrrrly intleperrdent.
[.lnear Algebra- lI3
27.4 .
r ',18.
CoLLEGE UNEARAT-GEtsRA
'koverthatithe following set of vectors in &.3 is linearly
dependent:
{(s, o,'-a),r(-r; l, D, @.2,-2),(2, t, t)}.
19. Which"of"Ihe following sets of vectors in IR3 is linearly
(r l(2, -I, 4) (3, 6, 2t, Q, t0, -4)l
(u) (1, 3, 3), (o, l, 4), (5, 6, 3), (7, 2,
-\l
Aadwcrs : (i) Linearly independent (ii) Lirrearly dependent
2O. Determine whether the following sets of vectors in IRa
are linearly dependent or independent :
(, {{1, 2, t, -2), p, -2,-2, O) (o, 2, s, 1), (3, O, -3, 6)}
(CI {(3, O,4, l), (6, 2, -1, 21, F1,3, 5; 1), (=3, 7, 8,3)}
Ansmcrs:
(i) Linearly independent (ii) Ltinearly. independent.
2I. For which real values of l. do the following vectors
YECTORSPACES
275
24. Show that the vectors i i'r"-' u
(1. |,2. 4). (2,
-t, -5,2). (1, _1, _4, O) and (2, t, l, 6)
25. L€t V be the vector space of all 2 x
2 matrleeso$s;the
roal flcrd IR. show that the n'-ratriees
A, B, c e V are rtnearly
ln<lependent where
o=
[i i],r= [d ?l *"= [ ;l R.u.M.sc.p. reesl
26.|n vector space V (IR) of all2 x 2
whether the following matrices are
tJ
?l,tS -jJ
*t'
matftces determine
linearty,depeitdCriti:."-.-,". , i
rSI
Anewcr: Linearly dependent
27' ,.etv(IR) be the vector space of an 2 xs
1,,::::l]
llnearly
i.:1j::1:-jprhe
;. _ ,,.,
matrices over
fouowing qatrisq.s in v(rR)
ase
A=W lJ,
II , =l-;; -SI andc =Ii :l Zl
za. tq{v, (IR) be th_e
the vector space of
ar all
a, polynomials
n^r.-^-r-r
".;;
;,. , Ansmur: )'=-2, ),'=1.
22. Show l-hat the set of vectors in IRa
(3, -2,4. 5), (-1, 3,2,6), t4,2,5.12)) is linearly
rd w are linearly independenl vectors ina
then show that the vectors
\,.i \tr, w+u
{ril
I-I -1'\r, ll -'1r, ll
- 2v + rv
a,c lrlr;r.r linearlv independenl .
It2 +t+
+ t, 3t2 + 2t + 2] of polynor.ntals
is ltnearly
p. u. P. rs8?l
dnenen darcctorryccc
Dr0altloa : l,et V be a vector spaee and
lnllr ect of vectors ln V.
L
of
V(IR) be the vector space of all
Rolypopiah or
<3. Determine whether the
following polynoriithls are
llrrnrly dependent or Independent :
(t) t3 +Zt2 +4t:1, Zts_t2_3t+E, _4t2
(tt) t3 - 3t2 _ 2t + 3, 2t3 _ 5t2 _ 5t + 7,ts _ +A+S
t3 2t2_3t + 3
Aarpcrc: (i) Linearly tndependent
C.ft Brdl d
,,
{v1, V2 , ..., \,,,j
:.
.
276
COLLEGE LINEARAI.CEBRA
VECTOR SPACES
We call [vr, vz, .... vn] I basle for V lf and only if
(i) lvr ,y2, ..., v,rf ls linearly indepe4d-ent.
(ii) {vr ,y2, ..., vrr} spansv. ,ii :1, ",..,
. De0nitlon I let e1 = (1,0, 0, ,.., O), 6, = (O; l, O, ..., O), ...,
;.rgn Fr(O, O, .,.,.O, U. Then {e1 ,r,2,..., en} ts'a ltnearly
independent set tn IRn. Since any vector v = (vr , !2, ..., vr'! ln
ffipq4,be-$fiSt€Las v. = v1Q1 + rt1az t ... + v,re,r, {e1, e2', ..., es}
sp.qs, IRn Ther-e{ore, {e1 , e4, .... qJ is a basis. It is called the
.,
stsndard basls or usual basls for g1n.
Dcflnltlon : A non-zero vector sp3ce V ls called flalte
dtmcaslonal if it contains flnite set of vectors [v1, v2i...., vn]
whleh forr-ns a basls;for V. If no'such set exists, V ls called
lnflnttc dlmenclond.
g The dlrrFaclsn of a linite dirnensional vector
D;Bnttlon
f:.
1-
i.
i
spaceris the nudrber of vebtors in any basis of it.
. . ;.'
Or, equivalently, the dlmcnslon of a vector space is equal'
to.,{he, maximum nurnber of linearly lndependent vectors
conhrred Dt It.
".Difinltloir :'If v1 , v2, ...,'vrare vectors of a vector sPace
V such that every vector v e V,ean be written in the form
v = olVl:*, Qv2 +.,... * OnVp where'ot are sealars, then vg, v2,
'r...
,,';].:
:
..., v. b called a gcneratlng system of tlte vector space V.
Theorem 6.19 If V is a vector sBace of dir,lrension I1, every
generating system:pfeV.'epr1{airas'o;.1: bt{t; r.ho$r:61616]"{-hdir "tr,
linearlyindepender(.ypctors. i
:".i
i::i:'i';r.1-:-- '
277
hod: lct v1. v2, .... v,n be any generating system of V. Let r
be the ltrclterl number of llnearly lndependent vectors that
ean bc ch<rcn from v1, v2, ..., v-. Then we may assume that
vr,Vl, ..,, vr are llnearty independent but that v1 .y2, ..., v1, v11j
erc llnearly dependent forl = 1,2, 3,..., m - r, Hence we have a
tron-lrlvlal relatlon o1v1 + ezvz +... + okvr + o!.dvr*J = O ... (l) in
whlch cr+; # O, since othetwlse the above equaUon (l) would be
a non-trlval relatlon between vt, ve, ..., vr contrary to t&e
llnear lndependence of these vectors. Henee the, above
equation (t) deftnes vsay os a llnear combination of v1 , v2, ..., v.y
forJ = l, 2, ..., m - r. Therefore, lt follows that v1., v2, ..., v1, is
also a generailng system of V.
Now any set of more than r vectors of V is ltnearp''.dependent. Since the dlrnenslon of V is n, V.eertalnly emrtains n
llnearly tndependent vectors. We must have n Sr. But elnbe V
eontains r linearly tndependent vectors ul, uz, ..., ur we also
have r sn. Hence r = rr dnd th? theorem ts proved.
Theorem 6.20 lf V ts a vector space of dimenslcin n,.erery
basls of V eontalns exactly n Hnearly lndopendent vec.to,rsl;
conversely. any n llnearly lndependent vectors of v consutute'
n basis of V.
Proof : Since every basis is a generafing systern.: It
contains n but,not more than n linearly independent vectors.
Slnce the vectors of a basis are llnearly independent,
llrcrefore, it contains exactly n vectors tn all.
Conversely, let v1..v2, ..., v, be any n linearly indepe.nflent
ver:lors of V and let v be any other vector of V. Slnce n is the
rllrrrcnsion of V, the (n + I) vectors vr, Vz, ..., vn, v are linearly
rlrpcrrdent and there exist a non-trival ielation of the form
,, 1J \
278
II
CoLLEGE LINEAR AI,GEBRA
prvl + azYz + ... t dnvr, + cv = O. Moreover 0, * 0.. Since
otherwise we would have a non-trival relation among
vr, vz, ..., vn. Hence we have
,, =
-# ", -*y2 - ...-f *
It follows that every vector veV is a linear combination of
vt, yz, ..., vn and hence these n vectors form a generating
system. StncC they.are linearly independent, tfrey also form a
basis of V. Hence the theorem is proved.
Theorem 6.21 Let V be a vector space of dimenslon n and
let v1, v2, ...,'v1, (r <n) be any r linearly independent vectors of
V. Then there exist n - rvectors v...1, v1a2, ..., Vn of V which
together with v1 ,y2, ..., v, constitute a bAsts of V.
' Proof : Slnce r < n. the vectors vy, v2,..., v. do not generate
the whole space V. Hence there exists a vector V111, of V that ts
not In the space genereted by v1 ,v2 , . ,vr.The vectors v1 ,V2 , .. . v1,
ve..1 zlr€ therefore linearly independent, for tf
0,1V1 * bVZ +... * Ofvr * OfalVsal =O.
W'e must have d,r*1 = O, Slnce otherwlse vr;1 would belong
to ttre space generated by vr, y2, ...;.., v,. Ttlen tt follorvs that
dr - &z = ... of = 0 sirrce vr, vz. ..., vr are linearly independent.
Now if r + I <n, we can repeat this argument to obtain a
vector v1.,2 such that v1 , v2, ..., v1 , v1a1. V.12 or€'ltriearly
independent and so on until we have n linearly independent
vectors vi, vz, ..., v., Vr*1 , yr+2,..., vn. These n vectors constitute
a basis of V. Hence the theorem is pioved.
V, then every vector ve V can ,be expressed uniquely in the
form v = CIIVI + Aevz + .,. + Otnvn,
,..d
iI
i4l
..tlr
,il
r1
Proof : Sihce {vr, vz; ..., vr} is a basis of V, any vector
v e V can be wrltten as a linear combination of the vectors
Vt, vz, .... vn i. e, v = cllvl + oevz + ... + cr,vo where cq are. scalars.
Therefore. we have only to show that the co-efficients d,1. cr,2,
..., c,n are uniquely determined by v. Suppose that
v = CllVr + %vz*. ., * OkrVn = p1v1 + fov2 + ... +- P"h
Then (crr - 0r) vr + @e-fizl vz + ... + (oQ, - 0.,) v, =O
Since Vr, Vz. .... , vn are linearly independent, it follows
that a1 -gr = O. w2-fu =O, ot -gn= 0, thatis, G, = Ft, az =P2, ...,
0. = F.,. Hence the theorem is proved,
Theorem 6.23 l.et W be a subspace of an n-dimenslonal
'rector space V. Then dim W <" n, [n particular, if dim W = n
then W = V.
Proof : Since V is of dimension n, any n + 1 or more
vectors are linearly dependent. t
Furthermore, since a basis of W eonsists of linearly
independent vectors, it can not contain more than n elements.
Accordingly dim W S n. In particular;'if {w1 ,w2, ..., w,} is a
basis of W, then slnee it is an independent set with n elernents,
it is also a basis of V. Thus W = V, whendim W = n. Hence the
theorenr is proved.
Theorgm 6.24 lf S and T are subspaces of a finite
dimensional vector space V over the field F then
clim (S + 1) = dim S + dim T- dim (Sn11
Proof : kt dim S = s, dim T = t and dim (Sffi) = r.
L,et {u1. u2, ..., ur} be a basis of SfiT. Slnce SOT is a subspace
of S, We can extend the above basis to a basis of S, say {ur, uz, ,
u", vr, ... , Vs- . }. Thls basis has s elements since dim S = s
Similarly, we can extend the basis {ur , uz .... uJ to a basis of t;
say (ur, u2, ..., u,-, w1 ,wz, ..., wt *)
lrtA = {us, tr2, ..., u. v1 , ..., V"_1, w1 ; ..., wt_o}.
;280
VECToRSPACES
COLLEGE UNEARAI-GEBRA
Clearly, A has exactly s + t - r elements. Thus the theorem
ls proved if we can show that A is a basis of S + T. Since {ur, rt}
generates S and {ui, wkl generates T, the union A = {ur, vl, wr,}
generates S + T. Now we have only to show that A ls ltnearly
independent.
Suppose that orul + oery + ... + ohur + grul + ... + fi*r.v"-r.
+ Yrwr + ... + Yt-, Wr-u = O.
(l)
where ca, F:, Tr are scalars iu F.
Letv=cl1u1 *...+o!.ur+prvr +...+p"* v"-,
l2l
Then form (l) we get
I
v = - ylwr - lzwz
yt-r wt-r,
(3)
Since {u1, r1}e S, v e S by (2) and
Since {wu} cT, veT, W (3}
Accordingly veSflT.
Corollery : If S and T are two subspaces of a flnite
dlmenslonal vector space V such that SflT = l0l then
dkt, (S + T) = dim S + dim T,
Pr:oof : Since St..lT = {O}, dim (SnT) = g
Thus dim (S + n = dfin S + dtm T-dim (SnT) implies that
dim (S + T) = dirn S + dtm T, Hence the corollary is proved.
6.16 Quofrcrtspsec
Definltlon : Lrt W be any subspace ofa vector space V over
the lield F. I.et v be any element of V. Then the set W + v lru + v
=
: o€rM is called a rlgfit ccet of W in V generated by v.
Obviously, W + v and v + W are both subsets of V. Slnce
addition tn V ls commutative. therefore, we have W + v = v * W.
Hence we shall call lv + v as simply a coset of w in v generated
But {uy, w1} is a basis of T and so is independent: Hence the
above equation forces Yr = O. ..., Tt-r = O. Substitufing thts into
(l), we get o.i ul + ... + ohur + ptvr + ... + fu,. vs-r = O.
But {u;, vr} is a basis of S and so is independent. Henc,e the
ab6ve equation forces Gr = O. ...i o(r = O, 0l = O, ..., Fr- = 0.
So the equation (l) implies that q, 01 and ft are all O (zero),
So A = {u;, vJ,wr} is linearly independent and form a basis of
S+T.
'flrus dinr (S + n = s * t - r = dirn S + dimT- dim (SnT.
IIence the theorem ls proved.
'
byv.
Pnopcrtiec :
Now since lurl is a basts of SnT, there exlst scalirrs 6r, ..., 6,
such that v = 6r ur + &uz + ... + 6rur. Thus by (3), we have
6rur + oary+... +$u, + ylwr + ... + Tt-,. wt-, =O,
Zgl
(i) aV=+W+o=W
(it) oleW=+W*to=W
(fll) IfW + v1 arrdW+v2 are two.cosets of
W inV. thenW+v1 =W+v2 *vr -v2eW
fhcorcm€.Z5 If W is any subspace of a vector space V over
the field F, then the set V/W of all cosets W + v1 where v1 ls any
arbitrary element of V, is a vector space over F for the vector
addition and scalar mulilplication composttions deflned as
(W +vr) + [W+vz) =W+ (v1 +v2) for every vr. v2 eV.
and q fW + vr) - W + c.u1 where ae. F and v, eV.
Prloof : Let v1 . v2 eV then v1 + v2e V and also
ne F, vt e V =+ cnzl eV. Therefore, W + (v1 +.v2) e V/W and also
W + crvl eVlW.
Thus V/W is closed with respect to addjlion of cosets and
scalar lnultiplt"cation as defined above.
VECT0RqPACES
CoLLEGE LINEARAIEEBRA
:282
, ktW+v1 ='W+vt'wherevl ,v1 '€V
(lll) Hstcnce of additlve ldgqttyIf O is the zero vector of V, then
and W *Y2 =W +v2'where v2,v2'€Y'
vr -Vr'eW,v2-v2'eW 9vr - V1tY2-v2'€W
(vl + vz) - fur' + v2]ew
==+
W + (-v1) - W - v1 dVlw. Also we have
[W + vr) + [W -vr) = W + (vr-vr ) =W+ O =W.
.'. W-v1 is the addiflve inverse of W + v1.
Thus V/W is an abellan group wtth respect to addition
composition.
Therefore, addltlon of coeets in V/W is well-defined.
A.gain, PeF, v1 - v1'eW =+ pfur - v1')eW
Further for scalar multiplication. we observe that if
+W+gvr =W+pvr'Therefore,'scatar multlpltcatlon in V/W.'ie, also-.well
;
a, pe FandlV+v1,W+v2eVl[ then
(v) o ltw+ v1) + (W+vz)l
defined.
*vr)
(vi) (q + $ (w + vr)
'
,
,
l,et W + v1, W + v2 and W + v3 fe any'"tnree elements of V/W.
Then [W + v1) + [(W + vz] + (W + vs]l
+v3]l
=W+[(vr +v2)+v3l
= [W + (v1 + v2)] + (!V +,v3)
= fiW + v,) + (W+ v2)l + (W + v3)
=W
= w+ (ct + Flvr
=W+(crv1 +6r,;
= (W+anrr)+ 0M+ pvr)
=0{W+v1)+9(W+v,;
(vii) a0 [W + v,; = W + (oB)v,
= W + o(pv1)
= o, (W + pvr) = c{p(W + v,)l
=(W+vr)+[W+(v2 +v3]l
=W + [v1 + fu2
= s [W+ (Vr + v2]l
= W+ c (v1 + v2) Sir:ce crW
=W+(mr, +av2)
=(W+ffir)+(W+avz)
=a(W+v1)+o(lV+v2).
(0 Addftfon b conmutatbe
. (tttlAddtttoo ts Gsoclatlve
,:
If W + v1 is any element of V/W, then
+W + (vr + vz) = W + (vr' + vzJ
= W +vr) + (W I v2) =(W+ vr') + (W + v2')
1v1]=W+'tvj
=(W+v2 )+(W+vr)
: j :{',r.j: :. r,,
... W + O = W is the additive identity;
(iv) Hbtence of addidve lnvcree.
Since W ts a subspace of V, we have
Then W+vr)+[W+rr) =Y_* fur
Zgg
W+ O = WeV/W IfW + vr is aqy element ofv /W,
NowwehaveW+vl =W*v1 ':1v1 -v1 '€W
and W * Y2 =W+v2' I Y2 -v2'€W.
I.et W + v1, W + v2 be any two ebments of V/W.
i,
1
{viil) 1(W+vr)=W+ lv1 =1ry+v1 where I e F.
Thus v/w is a vector space over the lield F fcrr the addition
cosets and scalar murtiprication. The vector space V/lV
is
'f
ctrlled the $uotieat space of V relative to W.
'ftre coset W + O W is tlre zero
veetor of this vector space.
=
284
COLLEGE LINEARAIiCEBRA
6.17 Dtmendm of gsoticnt $Pacc.
285
YECTOR SPACES
,.:
Theorem6^26lfWisasubspaceofafinitedimensional
vector space V over the lleld F' then drm V/W = dim V - drn W'
Proof:lrtmbethedimensionofthesubspaceWofthe
vector space V. l€t S = {wr ,}r2, ...,wJ be a basis of W' Slnce S is
whert w - ylwr + yzwz + ... + y,nw.eW.
*lW + v.W + (w+61v1 +6,vz +... + 6.v.|
= (W + w,| + 61v1 + q2"vz + ..., + 6.v.
=W+61v1 + 62v2 +... + 6.v,
linearly lndependent subset of V, therefore. tt can be extended
SinceweW ...W+w=W
to a basis of V.
IJt S' = fi r, w2, ..., wn1 . v1 'Y2. ...,vrl be a basis of V'
= (lV + 61v1) + (W + &vz) + .:. + (W + ev.)
ThendtmV=fll*r.
.'. dimV - dhn W = (m * rt - m = r.
So we have to Pr6/e thatdim VII/= r.
Now we claim that the set of r cosets
Sr = {W + v1, W tY2, .-,W + v.} is a basls of V/W'
frrst we have to shorr that Slls llnearly tndependent'
'Ihe r.ero vector of V/W ts W- '
kt cr1 (W + vr) + ob (W + v2) + ... + o.r (W + vr) =\fi
+ orvr)=W
=* (W + crlv1) + (W + nzvz) + ... +
(cr1vl + %vz +..- + ot'vr) =\[
=+ W +
(W
:*(}1V1 +%vz+"'+OtvreW
:* o1v1 + (hvz + ... + ckvr = ptwt +fuwl + "' + $rtl1'tn
Since {wr , wz, ..., w-} ts a basls of W'
:+ (11v1 t oavz + ... + oq-vr - Flwl- Frwz - "' - Bnlw'n= O
:* (I1 = O, 02 = O, ..., U, =O
Since the vectors vr, V2, ..., v, wt, wz, "', wm are linearly
independetrt. 'lJrrrs the set Sl = [W + v1, W * vz' "" W + v'] is
linearly indePendent.
Now we have to shorv thal l, (Sr) =VIw'i'e 51 spansvftar'
v/w''Ihen veV catr be expressed as
Ir1 W + v be at1' clelrtent of
y = 111V1 * 12w2* .:. + Y,nWrr * 61v1 +lbvz + "' +0'v,
=w *6rvl +&vz + ... +6,-v,
= (6, (tV+ vr) + & (W+vz) + ... + 6" (W+v,)
fhus any element W + v ofv /w can be expressed as linear
combination of 31 = {W+v1, W+ v2, ...,W +vr}
: v/w=L(Sr)
So Sl is a basis ofv /w
Thus dlm Y /W =r = dtm V- dim 1ry'
Hence the theorem is proved.
6.18 Coodlnatcs of rvectsrelative to ebasls
,
Let [v1 , y2, ..., vr,] be a basis of an n dimensional vector
space V over the field F. Then any vector veV can Ue expressea
where xr €F' (fSi <n) and (-r1 , x2, ...,xn) are called the
co-oldiaateg of v relative to the given basis. There is clearly
ve| apd orlered
rr-tuples (xt, x2,..., ,6) with elements in F'.
Oqe
,cjt! easily vertfy thart if v, w h4ve co ordinates
(rr, xz. .,.d,Ur,,yZ, ..., y,,) rcspr:c:tively and oe I,',thenv+w
Iras co-ordinateg (xr + yr ,x2 * yr^ ..., x,, + yr) and sv has
<'o-ordinates {ux1 , u.rc2, ..., a.4j.
VECTOR SPACES
COLLEGE LINEAR ALGEBRA
186
system of llaear
6.19 solutiotr specc of a homogcncous
rquatlons
Sttxt + atzXz + ... + 8tr,&, = Oo
+ a{zi + ... + tznxn=
m linear equations in n
is a hotnogcneous systcm of
(1)
IR ' Ttie solution set W of
iieta
the'ieat
over
unknowns
the
and this subsPace is called
constitutes a subspace of IBf
(i) The solution set W = l(Za' a)
equatlons'
: ae IR) of x-
? ='
subspace of IRz
is a solution space and W is a
q IRI of the linear
(ii) The solution set W = (2a' -a' 4)' 'a
I
solution 'space ard w is
system of equafiion , * -tr
"
:
i l:3}.'"
e: subsPace
ht W dcnote the fletternl solutlon of a homogeneous linear
rxprrrncd rrttlrluely as a linear combination of u1, u2', ..., u".
'llrc rrrrnrber s of such basls vectors is called the dlmenslon of
4mlXt * A1y,X2 + "'+ a-rr:6 =O
solutlon sPacc of the system of linear
As for examPles
l5a} rrtt.ltL
nyalcm, 'ltte ttrtn zertt solutlon vcctors ut, u2, ..', us'are sald to
lbrm r |rrlr of W lf every solution vector oe W can be
(1)
?:'"'
287
ud dlmcnrloa for thG Spnctal solutloa of a
of lR3 '
the non-homogeneous
Rcmart : The solution set W of
n
(i 1' 2' "" m)
linear system il.uu n = bi =
o'er,n" ,"'i, o"ld lRdoes not constitute
: t6orcm O.ZZ (Imthout Fod)
Q)'
a sttbspace of EP '
are equivalent :
The follorvingithree statements
equations AX = B has a solutlon
{i) The system of linear
' (it) B'ls a linear combination of the columns of A
augmented matrix
(iii) The'coefflcient matrix A and thc
(A, B) have the same rank
W. wrlttcn as dim W = s. UfW - {O}, we define dim W = Ol
6.21 Procedure of findtag the basls a1{ dirnelslon of the
rolutlm Blrac.e of e homogpnm llncar EIEtcm.
Let W be the general solution of a homogeneous linear
nynlcm and suppose an echelon form of the system has s free
vrrrlables. kt u1, u2, ..., u" be the solutions obtained by setting
rnrr <rf the free variables equal to one (or any non-zero
lrrrrst:rnt) and the remalning free variables equal to zero.
't'hen dim W = s and u1, uz, ..., u" form a basis of W.
Theorem 6.26 The dimension of the solution space W of
llrc homogeneous'systeql of linear equations AX = O.ds, p-r
wlrrrc n is the number cif unknowns and r is the rank of the
loelllclr:nt matrix A:
Proof : Supposeul, u2, ..., u, lbrm a basis for the eolumn
rlnr('(' of A (There are r such vectors since rank of A is r). By
lltrrrrtrrr 6.27, each system AX = ui has a solulJon say v,. Hence
A\rr =tli, Av2 =t)2, ...,Av, =u, (l)
:,l rl)l)osc dim W = $ zDd w1 . w2. .... w. fotttl a l;;rsis of W.
lll ll - {tr1 . V2. ..., Vy, W1 , W2, ..., Ur"}
\\ r claim that B is a trasis of lR" .
llrrrs nc lrr:ed to prove that B spans JR" arid that B is
[ntru lt' lrrrle;rndent.
288
COLLBGE UNEARAIJGEBRA
(0 Proof of 'B.strnns-
VECTORSPACES
El:'
Suppose ve IRr and Av = u. Then u = Av belongs, to the
column space of A and hence Av ls a linear cornbination of the
ur, sayAv = clrur + o"zq+.,. +okur {2)
l-etf =v - o1v1 - %vz - ... -: ohvr. Then using (l) artd P)
we have A(r/) = A (v - GrVr - hvz -... - okur)
=il-;;';*yti ;*,fr.*'
=Av-Av=O
Thus y' belongs to the solutiqn spaee W and hence'f'ls 4,n
linear combination of w,, saY
s
y'= 0rwr +\zwz + ... + F=w, =X Ftwt.
i=[.
Grvr =
i-Ji.;.-u,*,
i=l
J=l
Thus v ls a linear combtnation of the elements ln B and
hence B spans IF .
(ll) Proof of 'B le llnearly lndelrndeaf,'
Suppose Yrvr + ^lzvz +." * Trv. + 61w1+ Qw2+ ... + 6"w" = $
since w; e w' each Aw1 = o' usin8lthis fact and
(l) and (3), we get
1ls
O=A(O)=Al I y,v,+ ) o,*,)=
\i=l
J=l
rs
E1O = y, u1
= I yr u1 + I
j=l
i=I
-t-
i ,v, er, *
i,
un*,
289
However, wr. wt, ..., wr are llnearly lnclependent. Thus
each 61 ' 0. Therefore, B rc rrnearly tndependent. Accordi,gly,
B lsa borlrof lR' , strree [t has r + g elements. we have
r + s = n.
Conroquenlly dlrn W - g - n - r. Hence the theorem is proved.
Dnnpio 23. prove that the vectors (1, 2, O), (0, b, Z) ana
(* I, l, 3) form a basls fo.IR,
Proof : The glven vectors will be a basis of IRs if and
only
lf they are linearly lndependent and every vector tn
IRs can be .
urrltten as a llnear combtnation of (1, 2, O) (0. E, 7)
ana (_1, l,gl.
I{rst wc ctrail pove that thc rtctors arc lincerly indepcndcrt.
For arbitrar5l scalars x, y, z.let
,,,1
d.1,2, O) +y (O, S,V +Z {-1, t,S) (0, O, O)
.,
=
.:
or. {x 2x o) + (O, Sy, Zyl + (-2, z,3z)
g,
6,
19.
=
or, (x- z,2x + 5y + z, Zy + 3z) (O, O, O).
=
Equating corresponding components and formihg
.system, we have
x _z=o1
. 2x*_Sy*^r=91
7y+32=O )
(t)
Reduce the system to echelon form by the
elernentary
transformations. we murtipry ,rst equation
by 2 and then
subtract from the second equation.
Thus we get the equtvalent system
x-z=O I
5y+32=o I
7y +32=O l
|2u2 + ... + Yrur.
el
or, we can write
Since ur, uz, .... ur are linearly independent. each Yr = O
Subst.itutingi this in (3), we get 6l wr * ... * 6uw" = O'
'
linear
x-z=Ot
132+_5y=O |
3z+7y=g
)
:
lJnear Algebra-lg
{Bl
2gA' '' ;
, '; CoLLEGE LINqAR ALGEBRA
We subtract second equation from the third equation'
:
Then we get the equivalent sYstem '
' ',..1 'iiil;'.';16- Z
=0.1,
@)
3z*r?==oo
::'
i
is in echelon form and has exactly three
This systqm
-l
eouations in the three unkno#ns; hehce the system has only
the zero solution i. e. x = O, y = .O, z = O' So the given veotors
are
linearlv independent.
that the
tBy the application of rnalnces we can also show
*given vectors are linearly independentl
To shqqr that the given vectors slna R9' we must show that
an arbitrary vector v'= (h; b, c) can expressed as a. linear
(-f 1' 3)
'
combinatiorr, = (]: b, c-) = x{1,2, O) + y (O, 5, n + z
g"i
Then formingf liney sYstert'r,
""
';.
il;,i::\
(s)
folm by this elementary
' Reduce this system to echelon
operations
.Wemultiplyfirstequationby2andthensubtractfrom.the
: second equation. Thus we have the equivalent system
x-z=al
(6)
i.: 5y +32=b-2a |
7y +32=c , )'
We subtract second equation liom the third equation'
Then wc get the equivalenl sYstem
x - z=a
5y +32=b-2a
l
I
A
VECTORSPACES
I.i'orl the third equation, we have y = | {" -b + 2a)
29I
Sr-rbstituting the value of y in the second equation
weget r=f,va-5c- t4a)
Again substitutin gz = L (7b - 5c - L4a.) in the first equation,
wr.gr:l
"=
-] 1zu
- 5c - Ba)
't jrcrefore, v
= (a, b, r:) = L W* ft:-Ba) (1,2,O)
* j rc-u +2al(o,b,v+it a-sc- r4a) (-"I, r,3)
Thus every vector in IR3 can be expressed as a linear
combination of the vectors (1,2, Ot, (O, 5, 7) and (-1, 1, 3). Hence
lhe vectors (1, 2, O), (O,5,7) and (-1, l, 3) form a basis of IR3 .
o.l). (1, I, 1)l to t
i) Extend
,lQ,
T":_u. "jT1^^.
P;19(**l
lD.
xtend the set 11'3., Z, I), (O, l, l)l to a basis of IR3 .
lD.u.s. 19ffi
Solution : (i) First we have to show that the given:jI of hro
vcctors is linearly independent. Set a linear combinatiorf.of
tlre two given vectors equa!,to zero by using unknown scalars
,rimd y:
I
.,,
,42: O,r) + y (1, I, t) = (0, O; O)
or, (2x O,-rd + $,y, y) = (0, O. O}
or, (2x+ y, y. x+ y) = (O, O. 0)
..:
Iiquating corresponding components and forming the
llrrrlrr system, we get
2x+Y=O1
x.*I:3 i
**.rvc have x'= o'Y: o'
-,r*'
*
llcrrt't: the given two vectors i3. ffP' are linearly
Ittrlt'3rcrrrlcrrl. So {(2, O, l), (1, I, l)},is a part of the basis of
lH'rrrrrl lrt'lrcc we can extend them to a basis of lR3 . Now we
292
COIJLEGE I.:INEAR AI,GEBRA
Solutlon : The given linear system is
x-ry+ z=0'l
(1)
2x-6iy +22=O I
3x-9Y +32=O I
Reduce the system to echelon form by the elementary
3 and
transformations- We multiply lirst equatton by 2 andby
equations
then gubtract from the second and the third
respectlvely.Thgn we have the equlvalent system
(ii) First we have to show that the given set of the vectors is
linearly tndependent, Set a linear combinatlon of the two
given vectorsequal to zeto by.using unknown sca-lars xand y t
x(3,2, 1) +Y (O' I' l) ={O' O' 0)
x-3Y' *z=O l
O=O |
O=o J
';;l&quntfixg corresponding components and forming the
Iinear systGm,.we get
3x
=O'l
2x+y = O I Thuswehavex=O, Y=0.
x+y=0
J
Hence the given two vectors in IR3 are linearly
independent. so the gtven set of vectors ls a part of the basts of
.;
,.
ii,'.
to a basls of R3' Now we
i IR' and hence we can extend them
seek three independent vectors ln IRs which lnclude the given
vgctors. Thus we can easlly veri& that {3, 2, l, (o, I, I), (1, o, o}
,l_ are hgarty independent. So they form a basts of IRs which is
arf extenslon of the glven set of vectors to a basts of IR3 '
Determtne a basis and the dtmension for the
28.
dhd**
t
i dlution space of the followlng homogerrcous system:
II :' '
z=O I
x-SY+ z=O1
x-Sy+
2.tc- 6y + ?.2=O l
{2)
,
3x-9Y+9l2=O )
i
I
+x-$l+z=O
i
'
Thls system ts in eehelon form and has o:tly one non-zem
I 2 free
equatton in three unknourns' So the systern has 3 - =
of the
dtmenslon
variables which are y and z, Henee the
solutlon sPace is 2 [two)'
set (t) y = I, z = o llllY = 0, z = I toobtain !h9 respective
solutions v1 = (3, I' O). v2 = Fl' 0' l)'
Hence the set l(3' l' O)' (-1, 0, 1)I is a basts'of thdSohrttsn
or,, .(3x. bq $.+ (O, y' y) = (0' 0' 0)
or. {3rq Lu+ Y,x* } = (0, O' O}
.-
293
VECTORSPACES
/
which include the given
in'Ih3
vectors
seek tJxree:independent
twovrctors. Thus we can easily veri$r that (2, O' l!, (1' I' l)'
tO. I, O) ar,e linearly independent. So they form a basis of
,'fu.3 which ts an extenslon of the given set of vectors to a basls
of mf
,
\ soaca,
ltffinmplc
26. Ftnd the sotution space W of the fogov{ptg
/ homogeneous system of linear equatlons :
I
x+zy- z+ 4t-O )
2x- y+Bz+qt=9 [
p.u.H.T.tg8q
4x+ Y+32+S=9 f
iJ-u.H. 19881
'
v- z+t=O I
2x+t$- z+7L=O )
-
Bolutlon : Reduee the given system to echeld'iiffi'tiy tt.
cletttcntary operaUons. Wemulttply lst eqtiatlcr WZ'4 andz
and thcn subtract from 2nd, Snd and 5t}t' eguatlqus
rerpecllvely. Then we have the equtvalent system
x+2Y- z+4t=O I
- 5y + 5z-51=O I
-7y+72-7t=O t
Y- z+ t=O t
- y+ 2- t=O J
294
T
COLLEGELINEARALGEBRA
We multiply 2nd, 3rd and 5th equations by -
ll
:
i,- + and {-l)
respectively. Then we have the equivalent system
x+2Y -z+ 4t=A )
.
y-z+ t=O
Y-z+ t=O I
Y-z+ t=O
Y-z+t=O )
I
'
I
.
Since 2nd, Srd, 4th & 5th equations are-identical, we can
disregard any three of them. Then we have the equivalent
system
"
i.L
*
295
VEgfOR SPACES
f]',./i: 'r'i
The free variables are x z andt,.Pfi!,,, i..-.,.',,...,..,;(i-,,r
i.iulc.l'luir'r
'i'\'
i i[ti'- '1 "''"t :':li'
';11''i
1'il ilI,..il r,;rrto*rtrIa) x=l,z=Q,{=O
(b) x= O,z= l't=O'
solutions
(c) x = O, z = O,t = I' to obtain the respective
t'0' t)
u1 =(1,O,O'O)ru2 =(O'2' l'O)'us =(O'
of S and dim S = 3'
The set {ur . trz, u3} is a basis
(x' y' z't) of the
(li) We seek a basis of the set of solutions
'{ -'r:-t:8 |
t,,.,
This system is in echelon forrn having two equaUons in 4
qnknowns. So the system has 4 - 2 = 2 free variables which
t-t.r,
equatlons
; -2i:3 )
Y-22+t=9'l
x_t=o I
y-22
or,
x
ry
subspaces of IRa :
T= {(x V, z, tl I x-t = O, }, -22 = Oll
find abmis and tlre dircnstron of (il S (iil T (iiil SnT'
tBof,utrm ; {i} we seek a basis of the set of respective
$rrfi*,
J
the third equation'
Y.v- 't, * ,==uo Irrom
x-t=bl
y|2z = o I wtrictr is in echelon form'
t=OJ
| to obtain the solution
The free variable is z. Set 7 =
u = (O,2, I' O).
(SO'IJ = 1'
Thus (ul ls a basis of SnT and dim
of IR3 teanryl
erampte za. (i) I€t U be the subspace
(O' - 1' O) and Q' A' 21' nnd a
(generated) by the vectors (I' 2' I)'
Then we have
W = (-a- 2b, a -b,a, U : 4 b e IRl.
@"-rr. 27. Iet s and r be ttre following
s - lbcY'z'g lY- 2z+t=al
=O
t- 9 I srutt""t second equation
I
the.r.equi@ solutlon sPaee is
O' (b) z=O'
'llre freevarlables atezandt' Set (a)z= 1' t=
u1 = (tO' 2' 1' O) and
t = I to obtain the respeCtive solutions
is a basis of T and dim T = 2'
u2 = (I, O, O' 1). The set {ur' uz}
which qtisff
(iii) SnT consists of those vectors lx' y' z' t)
T' i' e'
all conditions given in S and in
are z and t and hence it has non-zero solutions' l*t z = a and
t ='b where a and b are arbitrary reaf numbers' Puttingz -- a
an{ t = b tn the 2nd equaHon we get Y = a- b' A$airt putting the
values of y, z and t'in the lst equati<in' we $et x ='- a - 2b'
{x,y. z, t)of ttre'equafony -22 + t = O'
tft
basis and the dimension of U'
a
;:
.
296
vEc'l.oR SPACES
COLLEGELINEARALGEBRA
of tRs spanned u)r'tn. ,g"ior"
,, (lil l,et w be tfre qub"f.".
*
(2.
(o,
6' 18, 8,6).
5, 15, lO. O) and
|t, - 2, O, O, 3), 12, - 5, 3. -2,6),
l'ind a basis and the dirnension of W.
Solutioa : (i) Form the matrix whose rows are given
vectors and reduce the matrix to row-echelon form by the
Redr,rce thts matrix to row echelon form by the elementary
row operations.
we multiply Ilrst row by 2 and then subtract from the
second and fourth nows respectlvely.
rl
i J ""utt'ct
second row by 4 and
fl ? L l *. multiply
from the third rour'
L; J 5 J ""utt""t
second row by 2 and
fl -6
? I I *. multiply
add with the rlrst row'
6 Jtt""
L;
L;
':
rl -lo Oll
-lO
lrvemultiptysecondrowby-l
-------'r-J
Lo o oJ
rl o I1
t.:
''
(ii) Form the matrlx whose rows are the given vectors and
reduce the matrix to row-echelon form by the elementary rorv
0 3l
row.
rt -2 0 0 3'l
.iL
lo -l -3 -z ol
-lo
o o o ol
Loro l88oJ
Interchange third and fourth rows
rr -2 0 0 3l
lo -r -s -2 o
-lo
ro ls 8 ol
I
of U; that ls, Basis of U = l(1, O, U, (O, I, 0)l and dim U = 2.
0
We multiply second rour by 5 and then add \lrith the thtrd
ro
-lo
Lo o oJ
This matrlx ls ln fow-ecttelon form and the non-zero
rows in the matrix are (1, O. l) and (O, I, Ol. These non-zero
rows form a basis of the row space and consequently a basts
-2
-z ol
lo -l5 -B
-lo
15 lo ol
LoloIs 8oJ
elementary row operations.
*. multiply first row b1r 2 and then
fl -6
'z, I I
from the third row'
I
Lo o o o oJ
We rnultply second row by lO and then add with the thtrd
row.
rr
0
0 3l
-2
Io
-r
-3 -2 oo
-lo o -tz
-t2
Lo o o o oJ
I
I
-I and divide thlrd row by -12,
rl -2 0 0 31
We multiply second iow by
o 3l
-20
-5 -3 -261
515 10 0l
li 618 I 6J
297
lo r s 2 ol
-lo
o l r ol
Lo o o o oJ
299
VECTOR SPACES
clol'l-ltcll t',lNltAlt Al-cDI']ltA
298
This matrix is in row echelon;forrn having two nonzets .'
..'llhe'arbovt'trri'rlrixisitrrtlrvcc;ht:lotrlbrttl"lhctror*atlifor't::ir'rt
(O: '}'l}i2r':O}3 '
tttittrix'ittt'tr'{1;;+21''O.:iOl;3}:
Ir:ot\rsitgte{ots}:in tllt'trbovc
lbr thc row
llol]-zcl'o rorvs lirrtlr lr basis
arrrcl (0. 0' l' l ' O)''l-tr-sc
birsis of W' 'lhtrs belsis tt
spilce trtrd cotrseqtrt'nth' -ar
\
uncl climW = 3'
f ' S' 2' O)' (O' O' l' i' O))
tO'
0.0'3).
(1.
-2.
=
by the
29' t't't W bc tltc sttbspircc llcncrirtcd
Example
.
2t + l'
polyrronrials p1(t) = trt +212 ^
+t2 -71-7'
pz (t) =t3 +3t2 -t +4turdprr tt) =2t1t
ol W'
tiinci a basis aucl the climc.irsion
space V (F)
tl subspacc of the 1'1'6tor
is
W
Clcarlr''
:
Solution
1' 1z' 1s} is
5 3' Tlrus the set 51 = {l'
olclcgrct:
t
itl
ol polylrotnitlls
. basis of V (Ir)'
(t)' Pz (t) a.d
the givettI ncctors pr
ol
coorclitratcs
the
Now
(l' 3' -I' 4) and
51 arc (l '2' -2' I)'
p3 (t) relatirre to the basis
12. l. -7 . -7) resPectivelY'
F'orurin-g the rrratrlr
vcctors. rvc $et
rvltose rows eu'e the irbove
rows (coordinate vectors) (1, 2, -2, t151i6'1191:;1''tll"3)it'"hitfiuiill'I:
f,orm a basis of the vector space generated by the coordinate
and so the set of corresponding polynomials is
vectors
will forrn the basis of W'
*
1ts + 2t2 2t + t, t2 + t + 3l whtch
ThusdimW=2.
*lr-rn$le 30. L€t U and w be the subspaces of ts.4 generated
by the set of rrectors
(1, I, O, -1), (1,2, 3, O), (2. 3, 3, -l)l and
1O,2,2. -21, Q.3,2, -3)' (1' 3' 4' -3)l respectively'
Find (il dim (U+W) and {ii} dirn runW.
Sof,rrtloa : (0 U + W is subspace spanned (or generated) by
all given six vectors. Hence form the matrix whose rows are
the given six vectors and then reduce this matrix to row
echelon form by the elementarSr row operations :
r1 I O -1 -1
coordinate
lr 2 B ol
lz s s -I
lr 2 2-2
z
Ie
Li B
s 4 -s
-sl
I
2 -2 4Il
rr
lr 3 -I -7)
lz | -7
I
I
I
the elernerrtary
rorv echelcltr lbnn by
Recluce this nrirtrix to
lst rorv by I ancl 2 irrrd tlien
,or,.Up"rotioirs' We rrrultiply
respectivel]''
sttbtrilct ltonr 2rrcl & 3rcl rows
fr
2 -2 ll
3l
-lo -3I -3I -$l
3 arrcl t.hcn irdcl rvitlr tlre
tl 2 -2 ll
I I 3l
-lo o
o oJ
Lo
Reduce tjhis naablx to rsqr-echelon forrtr by the elementary
row op€sdions.
We subtract lst row from z!rd, 4th and 6th rows' Also we
multiply Lc* rorr by 2 and ttsr subtract from 3rd and Sth
rows.
Lo
Wc nurltiplv 2trcl torv br'
D. U.H. rgffil
llrcl raw
1 0 -l-r
tl
l3
ll
13
'.{
2 -ll
t2-rl *zJ
2 4
ir
301
VECTCIRSPACES
COLLEGE LINEAR AI,GEBRA
3OO
Also we
We subtract Znd row from 3rd' 4th & sth ro\f,rs'
6th rour'
multiplY 2nd roul by 2 and then subtract from
r o -l-1
lo l B tl
rl
-l
Lo o -z -+J
rt 1 0 -I1
t
l s tJ
-lo
I
3
Lo
We subtract 2nd row frorn third row.
11 I O -1
I
We multlply 4th row by I and Z and then subtract
and 6th rows nesPecfivelY'
I
and 3rd rows respectivetY.
I
ol
-lrlo ooo -lo -2-zl
rl
We mrdttply l"st row by 1 and 2 and then subtract from 2nd
frtrn sth
O
o ol
l"oo
-lo
o -lo -2o
o
lo
Lo o o oJ
I
I
I
I
Lt s 4 -3J
We rnultiply lst row by 2 and I and then subtract from 2nd
and 3rd lows nespectivelY.
2
r1 2
-21
I
-rt -z
-lo
z
-rJ
Lo
I
This matrix is ln row-echelon forrr havtn$ thee
a basls
(1, l, O. -l), (O,l' 3, I) and (O' O''l' -2) whtchwill form
nolr-zeno
ofU+W.ThusdimU+W)=3'
(til l,et us flrst ftnd the dirn U a4d the dtm W' Forrn the
andtlr@redtreetbe
inatrixwhose ro\ils are the generators of u
matrix to row-echelon form by the elementary rtrnr
Lz 3 s -lJ
This matri:r is ln row-echelon form having two non-zero
rows tl; l, 0, -1) ard (O. 1, 3. l) which will forrn a basis of U.
Thusdlm U =2.
Again form the matrix whose rows are the generators of W
and then reduce the matrix to row-echelon form by the
elementary row operations.
lz s 2 -s
rl r o I
lo I B
lo o -l -2o
-lo
o o
lo o o oo
Loo o
lr 2 3 ol
rs r
-lo
Lo o o oJ
11 22-21
We interchange Srd and 4th ron's'
-1
rl I o -l 1
-'l
I
-l-'l
lo I s t
operations
i
We adtl 2nd rowwith 3td tow.
2 2
-21
t-l
I
-lo -2
-lo
o
oJ
Lo
I
This matrlx is in row-echelon form havlng two non-lrero
rows which will form a basis of W' Thus dtm W = 2.
Now by theorem we have
rltm (U + W) = dtrn U + dlm w - dtm runw)
rrr, dlm runW = dlm U + dim W- dtm (U +ril =2 + 2- 3 + t'
.'. dim {UnW1o I (one).
3O2
VECTORSPACES
COLLEGE LINEARALGEBRA
Exanple 31. Ift V be the vector space of 2 x 2 rnatrices over
[1 2
.the real field lR Find a basis and the dimension of the
subspace W ofV sPanned bY
o= [-i'r],"= [? -i I,
*u"=[-, gl
tr
F'?
"=
Solution : The coordinate vectors of the given matrices
relative to the usual basis of V are as follows :
lAl = (1, 2,-r,}l,tBl = (2, 5' 1,-1). [C] = (5, 12, 1, I]and
ID I = @,4'-2,5).
Form a matrix whose rows are the coordinate vectors and
then reduce this matrix to row-echelon form by the
elementary row operations and join successive matrices by
the equivalence sign - :
T1 2 -r
3l
lz b r -rI
ls 12 l ll
Ls 4 -2 sJ
2nd, 3rd and 4th rows respectivety.
T1 2 -1
,
.
linearly independent.
Hence the corresponcling matrices
Wemultiply2ndrowby2anrl_2and'then*strbtractfrom
..rs
tl 2.. -T 3l
l'o I 3 -7 1 's
-loo
o ol
Lo o z -rBl
[_] 3 I tS _] I *o
E -rB I form a basis ofw and dim w = 3.
Example 3.2. Let V be the vector space of 2 x2 matrices over
the real field lR- Find a basis and the,dimeflsion of .the
strbspace W of V spartned by the matrices
=
U tl "= [-i iJ," = [-; fl.*o, = [_u, -il
solution : The coorcrirate vectors of the given matrices
n'lirlive to the usual birsis of V are as lbllorvs :
.
;
I
8rd and 4th rorvs resPectivetY
I
[A]=(l.,;5,1.2).tB]=(1,1.-1.5).tC]=t2'4._5,.7)
:urct [DJ= (1. -7. -5. 1)
3l
3 -7 1
lo 2I .6
-lo
-I4
Lo -2 | -4J
3t
-]
I
lo
-rl
-lo o 3
z -18
Lo o o ol
This matrix is in row-echelon form having three non-zero
rows (1, 2, -1,3). (O, L.3, -T) and (0, O. 7, -lB) which are
"
' We multiply 1st row by 2, 5 & 3 and then subtract from
303
Interchange 3rd and 4th rows.
Iiornr a lnirtrix.rr4rose rot\l$ arcl the goorclinatq riectors,4nd
llrt'rr rt-dtrce .tltis. rltatri-r to r.or,l,-echqlorr ..form.,.l:y , the
r'l('ulcltti'r11: l'()\\' opcrations irrrrl
llrt't'rlrrir-irlt-I-tt.c siett -
:
2rl
rl -5 -l
r _.t s.li'
lr
12 +
T
-'5
i r -r -r--, I
_l
-ioin succqssjt,e t4a{ficgs b1,
:, .. . ,.,,..j ,ii it I
',
I
304
COLLEGE LINEAR ALGEBRA
subtract from
We multiply lst row by l' 2' and I and then
2nd. 3rd and 4th rows respectively'
TI -5 4
i.
(1)
Reduce the system to echelon form by elementary transfor
We multiply 2rrd row by I and - | and
then subtract from
3rd & 4th rows resPectivelY
rl -5 4 21
lo 6 3 3l
matlons. We mulUply ftrst equation by 2 and then subtract
from the thlrd equation.'Thus the above system reduces to
xr-xz
=Ol
xz +2x. =g;
-lo o oo ol
Lo o oJ
2x2+3x"=61
two non-zero
Thts matrlx is in row-echelon form having
are ltnearly
rows (1, -5, - 4. 2l and (O' 6' 3' 3) whleh
l-i f I *o tS S I to'- a basis
llnear system, we get
x2 + 2x, =O |
2x1+3x3=91
3l
lo 6 3
-lo
6 3 3l
Lo -2 -L -lJ
3O5
Equattng correspondlng components and forming the
xl - rcz =Ol
21
independent. Hence the corresponding
VECToRSPACDS "
matr{ces
or\M and dim lsr = 2'
(1' O' 2)'
ErmPle 3& Pr<rre thatthevectorl i u1 =
(O' 2' 3) fornr a basls of lR3and find
u2 = (-l' 1, O) and u3 =
(l -1. l) and w = {-1'8' 'tU
the co-ordtnates of the vectors v =
relative to this basis'
Proof : Flrst Portlon
IR3if and only if they
The given vectors will be a basis of
and every vector in lR3can be
are linearly independent
u3' Flrst we shall
wrltten as a linear combination of u' ' u2 "rnd
linearly independent'
prove that the veetors ut' uz and u'' ar ''
For arbitrary scal rrs X1 ' x3 tmd x3
letx1 u1 +t81u2lJQU3=:$
(o'
or, x1 {1. o' 2l f 4 (-l l' Ol + re {O' 2' 3} = O' ol
of, {xr -lQ'x2+''1x3"'2xr +Sxs) *{O'OQ
l2l
Again we multiply second equation by 2 and ttren subtract
from the thtrd equation. Then we get the equivalent system'
xr-x2 -Ol
x2+2xs=O1
- xs =O
(3)
:
J
This system is in echelon form and has exactly three
equations in thrie unknowns, hence the system has only the
zero solution i. e. x = O, y = O, z = O. According[y' the vectors are
linearly independent.
To show that u1, u2 and u3 span IR',
*. must show that an
arbitrary vector v = (a, b, c) can be expressed as a linear
combination v = xlur +l0r& +x.gus
or, v: (a, b, c) = xr (1, 0, 2) + & (-1, 1, 0) + xs (0' 2' 3)
Forming the linear sYstem. lve get
xt'xz
-a I
)02 + 2'k =b I
j
2x'1 +
?*. =o
Linear Algebra-20
COLLEGE LINEAR ALGEBRA
3oo
VECTORSPACES
Solvlng the system, we get xt =2, h = l. xs= - l.
echelon form by the elementary
Reduce this siystem to
transforrnations'
from the
by 2 and then subtract
We multiply first eqrration
'lhus v = (1, -1, l) = 2ur + lu2 + (-1) us.
third equation, Thus we have
Similarly, let w = (-1, 8, 11) = 1l1u1 +Y1uz + Ysue
the equivalent system
=a )
xr-xz
^I
ii,+zxs=;^ |
2;2 + 3xs = c-2a )
from
by 2 and then subtract
equation
second
multiply
We
system
Then we get the equivalent
I
ii*zxs=\ ^ ^Li
)
tot
- xs =c-2a-2b
x2 =-4a-3b+2c'
x2 in the first equation'
substituting the value of
we
3b + 2c' Therefore'
$et x1 = - 3a (% +2b-c) u3
+ 4a- 3b+ 2*) uz +
v = (- 3a- 3b + 2c) ur F
+ 2c) Fl' f ' O)
+ 2c) (1'O' 2l + (4a-3b
3b
c)
(a,
F3a
b'
=
or,
3)
+ (%.+ 2b - c) (O' 2'
as a linear
x-3can be expressed
in
vector
every
Thus
1' 0) and (0' 2' 3)'
of the vectors (1' O' 2)' Fl'
'u'combinatlon
HenceU"""ttoolft'u2andu3formabasisof=x-r'
+ )c2W + x3u3
pordon : Irtv = (1. -1, 1) = xtut
Second
or, (1, -1, 1) = xl
(1'' 0' 2l +
n(-1'
1' 0) + xs (0' 2' 3)'
we have
Forming linear sYstem'
1l
*t _'l
X] * 2*"-=
I
2xr +3x3= l,l
t7)
Forming linear system, we have
--t I
Yr-Yz
Yz+2Ys= 8|
2Yr +35rs =lIJ
(8)
SoMng the system, we get Yr = 1, Yz = Z,ys =3
I
we have xs =?a+2b-e'
From ttre third equation'
we get
o{x3 in the second equatlon'
value
the
Substituung
Again,
So the vector v has co-ordinates (2, 1, -l).
or, (-1, 8, 11) = yr (1, O,2l +Yz FI, I, O) + Ys (O, 2, 3).
(5)
*" *t i ;t*n-on'
xr-X2 '-a
^r
.3O7
Thus w = (-1, 8, 11) - lur + 2u2 + 3u3
So the vector w has co-ordinates (1, 2. 3).
Bsarnple 34. Given the vectors (2, l, l), (1, 3, 2), (1, 3, -l)
and (1, -2, 3). Test whether they are lineaily independent by
Srreep out method.
Solutlon : [,et o1 (2, l, 1) + o2 (1, 3,2) + os (1, 3, -1)
* d+ (1, -2, 3) = O = (0, 0, O),
where clr, da, cL3 and o4 are scalars. then
(2o.1 +de +os +oQ, or +3a2+3o.3-2a+,u1 +2ru2 - crg + 3&)
= (0, O, 0). Equating corresponding components from both
sldes and forming linear system, we get
2a1 + o"2 + 0,e+ oa=Ol
0,1 + 302 +3a3-2o"a'Ol
a1 + 2a'2 - 0s *3uo =6J
308
COLLEGE UNEARAItrEBRA
VECTORSPACES
ctg
21
13
t2
ll
3-2
-1
3
o
O-+Rr
O-+Rz
O-+Rs
O-+R+= ^f tst pivotal ro\r,
O+Rs= Rz-Rl
O-+Ro= Rs- R*
o-+Rz=$z"a pivotal row
Equating corresponding components from both sides and
formtng the linear system, we get
2a1 +2o"2 + tlas =O)
a1 +4or2- 9as =O.l
a1 +7a,2 + llq =QJ
Ol
224
t4-g
I
O"2
o-rRr
7
ll
2
O-+Re= Re
o+Rs= S sta Pivotal row
o,z+%- qa=O
(lr)
-i an =O
(rii)
This system is Ir echelon form and has three equations in
4 unknowns and hence 4 - 3 = I free variable which is aa.
Thus the system has an inftnite number of non-zero
solutions.
kt o4 = t, where t is a scalar.
Then cr3 =tt,or=-it.and0,1 =-1
Slnce all o's are not zero, so tJre given vectors are linearly
dependent.
lst pivotal row
O+Rs= Rz
_!3
O-)Rz=f Z"O ptvotal row
From the pivotal rows, we have
lll
(i)
ar +icLz+;s3+;
a4 =O
cr.
O-+Rz
O-+Ra
O-+R+=
O-+Rs = Re-8 Rz
o
309
l
rr'*.lrr[fu 8li. ArE these vectors (2, 1, U, @, 4, n, (4, -9, l1)
dependent ?. Test by Swecp out method.
Solution:Irta1 (2, t, 1) +oq(i,4,V+43(4,-9, 11)=(O,0,O)
Or, (2q + 2a,2+ 4a4, a1 + Aoe- 9cr3, a1+ 7a2+ llo3) = (O, O, O)
O+Ri = Re -6Rz
o
O-+Rg=
fr sra pivotal row
From the pivotal rows, we have
cr +(& + ?aa=Q
(i)
o, -* as =O
(ii)
0g =O
(iii)
Therefore, the svstem has zero solution i. e o,1 =az =o,a = O.
Hence the given vectors are linearly independent.
F-orlrlllc 36. FIxd the rank and the basis of a given set of
raectors {2, -1,5, 4, (O, t, Z, gl, (4, O, 6, U, (O, - 2, 4,
n
by usingSwccp out mcthod.
Solutlon : Rank of a glven set of vectors is the number of
linearly independent vectors of that set and these linear
independent -vectors form a basis of thqt set.
3to
COLLEGE LINEARALGEBRA
2-rb4
_Rr
406l
o-247
_eRs
orz3
o
o
o
(ii) Extend {(1, 2, 0, 3), (2, *I, O, O) to
_rRz
-+ Ra
o
--+Rs =
123
2-4-7
*247
! t", pivotal row
-eRo =
ORu
*Rz &= Rs- 1Rs
-eRe=Ra-ORs
-*n Y
-8 -13
813
3Il
VEC'TOR SPACES
2nd pivotal row
-eRro= R7 -2Re
+Rrr = Rs +2Rs
+Rrz=
-$e
sta pivotal row
-+ &g= Rrr- SRrz
Now the rank of the given set of vectors is equal to the
. number of pivotal rows = 3. A basis of the given set of vectors
is {(1, - i,lorl,@, r,z,s), (0, o, t,#)}
Since cr, (t, -i, i,Z) * oa(o,lp,B) + os (O, O, t,i3)= (O, O, O, O)
implies dt = &z = ct€ = O, (7aro solution).
ETRCISES - 6 (C)
l. Prove that ttre vectors (1, 1) and (t, O) form a basis of IR2.
2. (i) prove that {(2, - i. tl, E, 2, l), (O, t, l)} is a basis of IR3.
(ii)Provethat{(1, I, l, l), (0, l, t, I), (O, O, I, t).(O, O, O, r}}is
a basis of lRr.
3. (i) Extend lQ, O,O, -l), (1, 3, -I, O)l to a basis for IRa.
ID.U.S. 1e84t
a basis of IRa.
Answers: (, (1, S, _t, O), (2, O, 0, _t), (0, O,
l, O), (0, O, O, l)l
(ii) {(1, 2, o. 3), (2. _1.0, o), (o,
o, I, O), (0, o, o, t)}
4. Decide whelher S = (2, t, l), (t, 0, O), (S,
I, t)I is a linearty
dependent subset of IR3. what is the
dimension of the
subspace spanned by S?
ID. U. S. r9g2l
of the
Answer : s'is linearry dependent and aimension
subspace ls 2.
5. The subspace U of IR{is spanned
by the r,,ectors (I, O, 2, S)
*d (O, l,:1, 2) and the subspace Vof IRais spanned by
(l,2,3,4), (-1, *1, 5, o)md (0, o, o, r).
Find the dimension of U, V, UOV and
U + V.
Ansnret$ : dim U = 2. dim V= 3.
dim (UnU = I, dim (U + V;
=4.
6' Find a basis for the subspace of lRaspanned
by the given
'
vectors:
(l) (1, l,
-4, -3), (2, O, 2. -2), (2.-1, 3, 2)
(ri) (t, 1, o, o), (0" o,
*1**:
(U)
(r) (1, r, -4,
l, l), {1, 0,2,21 (o,_3, O, 3),
-3), (o, l,-5, -2t, @,0, t, -))
(1' 1,0, O), (0, I, I, U, (O, O, l, l), (O,
O, O,t)}.
7. Pind the dtmenston of the subspace
generated by the set
{(1, 2, l), (3, l, 2) {1, -S, a)} of Vs
GR).
tD. U. p. r9z9l
Ansrer: The dimension of the subSpace
is S.
8. Let W be the subspace generated
by the polynomials
1" -[s -2P + 4t+ l.vz
=2t3 -3t2 +ft_ I,
vs = f3 + 6t- 5, va 2ts St2 + Tt+
S.
=
Find the basis and dimension of
W.
IGU. P. 19ZI),,86;.L U. H. 19g6l
Answer: Basis: {ts *!12 +4t+ l, t2 +
t_S}and dimW= 2.
i
312
IfECTOR SPACES
COLLEGE LINEAR ALGEBRA
9. F-ind a basis and the dimension of the subspace W of P(t)
spanned by the PolYnomials
(i) Pr (t) = t3 + 2t2 -3t+2,pt (t) = ts +2P -2t + 3, and
Ps (t) = 2F + 3t2 -5t - 5'
-Pr (t)
(t) = 2F + t2 + t - 4, and
{ii)
= ts + t2 - 3t + 2, P2
Pg (t)=4ts+3t2-5tt2.
Ansnrers: (0 BasisofW= {ts+ lfz -3t+2;t2 -t+9, t+ U'
dimW = 3
(ti) Basis ofW = {t3 + f2 -3t + 2' P -7t+ 8, 2}
dlmW = 3
1O. Determine whether the given set of vectors is a basls
for gl3over IR:(0(f. I'O)'(l'O, 1),(o, 1. 1)l
(li) (-1, L,21,(2' -3' U!.(10' -14'O)l
Answeis : (r) Set of vectors ie abaQis for m.3 .
(ii) Set of vectors ls not'h b$is for m,'
11. I.et {vr, va . vs} be Uasts foi a vector space V' Show that
+ v2 and
tilt = vl,
=
{ur , uz. u2} is also a basis,
v3'
',
'1
us = vt * v2 *
12. LetW be the subspace of g1o generated by the ve<:tors
(3; 8, -8, -5), Itnd a basls and the
lL, -2,5, -3), (2. 3, l, -4) and
lD.u.s. rpssl
dimenslon of W.
*S'
2)), dimW = 2'
Answer: Basls lll,-2,5, -3J. P. 7,
13. Consider the following subspaces of IRs:
U = span {(1, 3, -2, 2, 3}, U, 4, -3,4,21' (2,3, -l' -2' 9)}
W = span (1, 3, O, 2,.U' (1, 5, 4, 6, 3)' (2' 5. 3' 2' 1)l
lD.u.H. 19S71
Find a basis and the dtmeusion of
(i)U+W (iD Unw.
dlm(U+W)=3.
' (ii) P."i" qf UnW = {(1, 4, -3,4,211
dim flJnw) = 1'
14. I.et S and T be the folloulng subspaces of gla :
S=[(xY,z,t)lY+z+t=o]
T={(x y,z,t) lx+y= O,z=2t}
Fired the basts and the dimension of
'{r}
s.,liil T (tu}
'
snt
:
{al, o, o, o), (o, -t, I. o), (o, -1, o. I)}, dim s = 3.
(li) Basis : {F1, t, O, O)' (O' O' 2, l}, dimT= 2:
(lti) gasls : {(3,'3, 2, lU, aliri tSnT) = l.
15. LetW = (a, b, d la, b, ceRand 2a +b + 2* = Ol
ID. U. Prcl" rgEgl
Flnd a basis and.dlrnension of W.
Ansrer : {{I, l, -U; (-1, 2. O)} is a basis of W and dim W.= 2;
16. Frovethatin Uf tfr.vectors (1, -1, O)' (O. l, -1) forma
,,
basis for the subspace U = {(x Y, zl ets3 l x+ y + z = Ol-
-:
17, Ffnd
"
Uasis of eat!of.thefollorrrtngsubsPaces.of nR3 :
(t) U = {(x Yr z) 11'
z=Ol ",' ,
i-
'
(tt)V= llXy,zl l'x=O,andy+z=Ol ':: 'i ' '
:' .
Ariwerlg : (i) (2, l, 9r, (i, O, f)l it a basis of U.
(ii) {(O, -1. 1), (0, 1, -1)} is a basis of V.
18. I€t V be the vector space of all 2 x2 matrices orrcrlhe
real field F. Prove that V has dimension 4 by exhiblting a
basis for V which has four elements.
19. I.et V be the vector space of 2 x2 matrices over the real
fteld IR; Find a basls and the dirmenslon of the subspace W of V
'
":.lrgl;": E i J* ";
rices"=[i 31,"= qr3 I and u = [S
spanned by the matrices
[3 ?
eniwers : (i) Basis of U + W
= {(1, 3, -2,2.31. (0, I, -1, 2, -1)'
3T3
(O' O. 2' O, -2}}
AEEscfs: Basisoflf,r=
dimW=2.
tll 31, I-? l]}
I
COLLEGE LINEAR ALGEBRA
315
20. lr:tV be the vector space of 2 x 2 matrlces over the real
field IR. Find a basis and the dimension of the subspace w of V
space
314
for the following homogeneous linear equations :
x1 +2x2 -xs *4rq =0')
2x1-x2+3x3+3x+=01
4x1+x2+Sxs+9xn=O7
spanned by the matrices
o= [-l 3l,r= [?-? I*o .=l-3
tI
Answers : Basis
"* = t[-l 3l tB -il [? i,l]
dimW=3.
2L,l*tV be the vector space of 2 x 2 matiices over the real
x2-x3+xa1Ol
2x1+ 3x2 - xs * 7x4 =O ) '
Ansnrer: Basis : {(*1, I, t, O), {-2, -1, O, f}}
dimension = 2.
'L
WU. orri_a basis and the dimension of the solution space
for the following homogeneous linear.system
field IR. Detennine r.vhether
x+2y- z+3s-4t=O I
2x+4y-22- s+5t=O I
2x+4y-22+4s-2t=O )
o=[lll,u=H l],"=[? ?l*r=[3 ?l
form a basis for V.
tt
ID.U. P. 1S4l
lRU.r{. r9B5l
Basis: l@,-t, O,0, O), (1, O, I, O, O))
Answer : They form a basisforV.
22. Find the dimension and a basis of the solution space W
.6\dt of the following sYstem
p''l' ai*.rrsion = 2.
26. Find a basis and the dimension of the following
system of linear equations :
ID.U.S. lSOl
x1 +2x2-2xs+2xn-.6=O1
x1 +2x2 - xa + 34 -2x" =Ql
2x1+4x2-7xs+ &+&=OJ
: dimW= 3"
I D. U. H. 19881
lR u. H. 1988 |
Ansrers : Basis : {(-2, I, O, 0, O); (-4, O, -f , 1, 0), (3, O, f , O, l)}
dimension = 3.
27. Prove that the vectors u1 = (1, 2, | -2),
2x1 + 2xz -
X3
u2 = (0, -2, -2, O), u3 = (o, 2, 3, 1) and ue = (3, O, -3, 6)
+.,cs = O'l
-X1 -x2+2xs-3xa+rE=O I
xr *,
form a basis of tRo and find the coordinates of the vectors
v = (5, O, - 8, -1) and w = (-9, 20, g4, - 25) relative to this
* -Tr* ;,t;3,[
*
ID.U.P. 19841
Ansrrcr : Basis : {(-f ,1,0,0'0), (*l' O' -1' O' 1)l
dimension = 2.
z
lrirsls.
Answer : v and w have coordinates 12, - l,- 3. 1l and
13, -2,5. -41
respectively.
I
\
COLLEGE LINEARAI'GEBRA
3T6
prove that the following set of
28. Usin$ Swccp out mcthod
vectors is linearlY indePendent :
(-3' -l' -4}}'
{1, g,21, (1,-7, -8),
zg.F]ndtherarrkarrdthebasisofagivensetofvectors
AnsEGr: Rank is 3 -, -
q
1)l'
Basrs = (1, 1,5.P, l, -r\(0, o'
3O. Shour ttrat the set S = {(l' O'
O}' (1' I' O)'
of IRs and hence find the coordinates of
(l' I' l}} is a basis
the vector (a' b' c) with
respect to the above b:asls'
.AoErtr : (a-b'b - e' c)'
of IRs ' t}ren
31. Shonr that if {u' v, w} is a basts
32. IfWl ls a subspace of IRa
' Sr
= (1, 1' O, -l)'
generated by a set of veltors
(1, 2' 3' O)' A' g'g'-lD andWz !s a subsPace
of IRa generated by the set of vectors
(l' 3' 4' -3))'
Sz = ((1,2, 2,-21.@'g'2'-3)'
(Wr nW2)'
flnd (i) dtrn (Wr +Wd (it) dim
(Wr oW2) = 1'
(ii)
Anglrlre : (f) dim (Wr +Wd = 3' dtm
(1' 2' Ol' vz = (O' 5' 7)' and
33. Prove that the vectors v1 =
find the coordinates of
v3 = (-1, 1,3) form abasis of IR3and
['' U'Ia 1990' 9U
the vector v = (2' 3, 1)'
Aosrtr: M = [O. l,-21.
!
of the vector syf W of
94. LetS be the following basis
2x2ralsYmmetric matrlces : -
ili';l Ii ll l-; -?D
Find the coordinate vector of ttre
the above o""o
matrix AeW relative to
*r'.rJr;='1 ; -: I '"0 (b)A = lL '^l
Ansrlr: (a) [2. -r, U O) [3' r' -21'