Algebra
ear
Dong
Chuol
Exercise
1
False:we know
unique
-
However, there
be
solution
No
-
Infinetly
solutions
subspace
solution
distinct
of
a
combination
solution to
solutions
Therefore,
we
to
can
have
and
a
solution
exactly
two
system of
solution, because
Linear
equation
the vector space.
If there
and it to the
system, then
must
are
two
any
linear
be
also
ant by,
U and V, such
the
system. This means there are infinitelymany
of
a
system
True:Since,
a
must
as
two solutions.
2.
I
system ofequation
many
be
cannot
set of
Ruei
solution
A
-
the
a
that
Set2
Problem
linear equations
of
zero matix
always
have
cannot
yield
zero determinant
itis not
always invertible. Because for a matrixto
be invertible, determinant
must
always not equal to zero.
-
3.
with
matrix
False:A
zeros onethe
or
is
since,
not
C
invertible. Hence,
CR,
c
can
be
C
any
same
column or
F0,2
number
thiscountoneit
sine
letc=oF
invertible
is
row
Here,
every
4. False:Not
diagonalis
a
must
not be
diagonal
EX:
is
This
invertible
only
entries
True:We know
only if
determinantis
diagonal
the
enties
from definition that,
A
that
not
A.
Now,
same
definition
can
exists
A
Ar (A)
=
a
are
none
matrix
matrix
we
is invertible
=
(AY2A
(i3yB
one
zero
A
AAATAT AIA'
Exercise
A
ofthe
can
is also invertible with inverse
also show that
:
=
diagonal
none
2 and
AA=A.A=I,
Assuch that
we
a
is invertible because itas
is invertible if there
also
(c F0,2)
door the
Tmisisnotinvertiseas
5.0
of
5.
if
and
2
arezeros.
[680]
However,
0 and
to
matrices with
diagonal
invertible.
is because
are
matrices
equal
=
by the
AA
I
=
I
=
II
(y-yi)c te
=
=
1.
Solve
the problem
AF
3,
=
=
(x, x2, 43)
I
where
=
(1,
-
[23- [x]=
↳
x1
into
row
3x 4x3 1
+
=
+
2
2x1 5x 6x3
=>
+
-
Reduce this
2,4)
-
2x1
-
=
+
742
-
9x3 4
=
form
echoleon
L i j(] (*)
1
i
-
7
=
-
9
Lxc'js17cm :3 SJCR+,
e R2
[OsclR]Rs,moS Ear
Re
+
Therefore:
x1 3x
+
x2
4x3 1
=
+
2x3 4
=
+
x2
=
4
=
-
4
=
X 3 18
x2
=
x(
1
-
=
1
x=
x1
3( 16)
48
+
q
=
-
-
-
4(10)
40
9
=
-
=
240)
20
-
16
16
=
Therefore,
the solution
Hence, we
2)
We
can
have
set
(9, 16, 10)
-
=
(x) (2)
uniquesolution
a
=
inverse ofA
multiplicative
find the
using the
identity
of343
a
(x1, Xu, (3)
is
matrix
(i3x)
=
=
70
=
!33s0O+RcanNOR
↑
[03/I
↓
2P2 23
+
- >
R3
~
Fo'12I -8)
:32 1 100m
car
an
(888(3, 72]
I
-
-
z
re
I
ooolic)
=
-
I
A
1
(iii)
=
A
·
Now, we
is correct
A.A=
check
if
can
I
to
-
A
our
that
confirm
answer
[Ec] i ) 5
=
3
1
-
18 16 1
+
3
+
2
-
-
=
0
4
6 +4
-
30 24 0
0
=
=
+
2 +5
=
Hence:AA
3.
6
=
4
-
10
-
1
6
=
second
can be
if
=
the product
-
the
of
by verying that
is
first matrix
equal the
to
6
2
42
+
-
-
36
7 a
+
4 14
+
-
0
=
0
=
9 1
=
matrices
the
or not
exist
number
of
number rows
of
columns
the
of
If this condition is satisfied, then the matrices
multiplied and the product exist.
matrix.
3x3
i)
-
6 0
+
we need to check
the
-
I
To determine
of
=
=
BC
2xx2
2x3
->
We can
columns
2x2
see
B
of
the number of
that
and C are different
Therefore,
BC does
the productofmatrix
not exists
ii)
B
2458
(B
=
A
since, the
+3
number
the
that
column
ofB
number
the same, we
are
of
CB
product
ofcard
can
row
say
exist.
[ii[,,5i] 9 10i]
-
=
=
(iii)
CB
[9 103]
-
=
2x3
3x3
iii)
ACB
A and
Since, the columns of
3x32x3
CB do
We
can
have
not
say
that
rows
of
the same size (i.e3 #2)
product:A'CB
the
does
not exist.
Exercise
A
III
a a aare
a
=
i) Find the upper triangular
method.
aosof
=
using elimination
matrix
Camaaat-nta
↑ ac 1
an
obGasaaca
1
[
a
&
IaObaz.aBa-RInr
a
en
aa
ob-ab-ab
0bac
-
a
-
R3
R2
+
a
a
I
a
-
b-ac-ad-a
0
R3
+
R2
P4
+
1·aa a 1
!·9.acacasade
O
b. ab.ab
-
a
re
a
iv
=
O
-
aaa
0b.ab ab.a
-
00
2
-
bc
-
j
b
00C -bd -5
2) This
a
[oaaasasaI
is the upper
matrix
-
re
->
R4
23 R4 Ry
+
+
triangular
ii)
find
To
factorization,
A's LU
!
a
o
u.
=
O
-
we
know
that
scai
c
osni
n
.
"
ee
Y
a
①
0
all
L
i.
L
=
we
can
1
that
determine
a a a re I
a
=
LV
=
a
I
↓
Pu
->
↓
a
a
a
b
b
b
bcc
triangular:Lower
Theyare
accaand
0
a
-
C
:A LU
=
triangula:Upper is
Diagonal:All enteries are
upper
-
L
the
u
lower
-
=
L
abcd
1
A
↑2000
A
by-1
and we take
:.
Now,
rows
is
zeros
zero
except
the
zeros
①
diagonal
Exercise. Find
factorization
11 2 4 -I
-
1
A
-
-
-
-
I
2
126
4
-
4
Ry
q
1
80
1
5R3 14 84
+
-
j
-
u
1
.
=
+
+
-
0
2
-
-
3 Ri R4
+
4
-
0
L
O
0
I
-
I
0
D
O
2
L
I
o
O
I
4
-
4
2
-
-
4
o
->
D
O
2
00
-10
R4
+
-
4
I
↓
!!
1
3
->
2
3
2
4
-
I
O
-
DU
-
-
-31
o
-
2
=
D
many e
o
20
R2
2R1 Rc
7
I
L
A
1 ·- o
1.3
1
&
1
2
20
I
-
where
q
1
2.1
is
1
I
I
LDU
the
1
=
↓
oo
o
⑧
0
7 43
-
->
↓
-
5
We take the
of
rows
0
I
1/
opposite
multiples.
1
0
I
D
00
0
0
0
O
2
O
o
O
=
-
D
20
3
1
.6
I
u
-
2
-
o
D
20
Now,
4
as
1
show
the
itis with zeros
diagonal
↓
that
A
=
from
each
their
diagonal
of
upper
parts.
divide
We
u
diagonal
take
and lower
osI
00
we can
We
I/
ils
=
I
of
rows
and
U to 1.
by
change
the
LDU
1 ils
[2- z-p 2,200000
=
I
-
4
A
DV
=
0
3
0
-
D
O
2
o
O
LDU
-
I
-
2
d
4
:A
4
LDU
=
-
3
0
Yz
I/
0
Yz
[0:"if
I
=
/280l0-..of
7
-
D
-
=
2
0
120.
ils
+;1,
o
O
-
2
L
I
I
0
d
7
q
O
-
2
-
2
=
31
2 e)
1
--
6
A
4
Exercise 1.
1
Factor
*
flf system
the
into
A
use PAL and
We need to
avoid any division
2x1
-
by
-
-
the
LU
rows
or
PA=LU
exchange
to
zero
x 2Xz 1
=
+
6x1 +0xi
8x1
do
=
x2
-
2xz
0
=
5x3 4
=
+
[ ]3](x]] ]
7
en
i
z]
:07
I =
5/
O
A
p
Now,
pA
y
=
we can one
=
y
n
z
-
PA
Gaussian elimination
+
-
R2 Rj
+
u
-132)3m
10
imo
R5127
4R1 + R2 R2
182-23)m
to find
R3 1R3
re
+
ener
(0 j, 3]. noTaking
u
.z
=
te
multiples ofrows
=
exchanging
sign
their
Now,
We
we can
obtain
(5 3
=
they
=b
(i
solve
and
then
us
=
value
[4007 for me
=
i.y) 14y
yz
+
=
yz
=
-
0
-
=
4
-
3y1
3(1)
-
3
-
yz
y3 4
=
+
( 4) y3 4
-
-
4
+
13
+
=
+
4
=
43 3
=
i.[zy)=fy)
ux
y
=
when
5
[0j, 3(x)=te
2xy X272 1
x
-
8x3
3
=
43
=
24.
1
(x1
( z3) x(71
3x2
124
23 1
+
1
=
+
24
48x1 23 18
=
+
+
48x1
48x1
41
+
24
=
24
-
=
48x1
=
x
-
4/
17
=
-
3(8)
=
=
+
-
-
342
=
342
3
x
-
=
4 +
-
4
1
32 + 9
-
-
=
32 +9
-
-
23 =
3 13
24
-
=
xz
=
=
2
-
E
48
:(Ey)=8s
-1 1
23
&