Matrix
21.2 EIGEN VALUES
«11
^5
«ln
011
Ojj
■■■
?i'
*1
*2
7s
*3
°31
■1
A
+
1 r T
Am
attl
flrt3
TBP
1 P ■■
A
°rtrt. _v
3?
TBP
________ 1
P 1 H
AV = 7
.„(!)
Where J is the matrix, Xis the column vector and Xis also column vector.
Here column vector X is transformed into the column vector Y by means of the square
matrix A.
Let A’be a such vector which transforms into XX by means of the transformation (1). Suppose
the linear transformation F = AY transforms X into a scalar multiple of itself i.e. XV.
AY = Y = XX
AX-<
kLY = O
„(2)
Thus the unknown scalar X is known as an eigen value or the matrix A and the corresponding
non zero vector X aS eigen vector.
(A-1J)X = 0
The cigen values arc also called characteristic values or proper values or latent values,
'2 2 1
Let
A= 1
3
1
1
2
2
'2
2
1'
x-w = 1
3
1
-A
1
0
o’
0
1
0 = 1
2-X.
2
1
3-X
1
characteristic matrix
12 2
0 0 1_
1
2
2-X
(ft) Characteristic Polynomial: The determinant | .4 - XJ | vs hen expanded will give a
polynomial, which we call as characteristic polynomial of matrix J.
—
1
1 1
—
—
■
For example \
2-1
2
1
I
3-1
1
I
2
2-1
= (2 - X) (6 - 51 + la - 2) - 2 p -1 - 1) 4-1( 2 - 3 +1}
= - J? + 7 X1 - 11 X + 5
(c)
Characteristic Equation: Tlx? equation | A - rd | = 0 is called die characteristic equation
of the matrix J e.g.
V-7Xa+ 111-5 = 0
(rf) Characteristic Roots or Eigen Values: The roots of characteristic equation | A -1/1 = 0
arc called characteristic roots of matrix 4. e.g,
lJ-7la+ 111-5 = 0
=>
(1-1) (X - I) (X-5) = 0
Characteristic roots are 1, 1, 5
.-.1=1,1, 5
Some Important Properties of Eigen Values
{/LVf/FTE
2009)
(1) Any square matrix>4 and its transpose ?lr have the same eigenvalues.
Nott The sum of the elements on the principal diagonal of a matrix is called the trace of the
matrix.
(2) The sum of the eigen values of a matrix is equal lo the trace of the matrix.
(3) The produel of the eigen values of a matrix /I is equal to lhe determinant of J.
(4) If X
are the eigen values of .4, then the eigenvalues of
(0 Aj4 are
..... f
Example 1> bind the characteristic mats of the matrix
6-2
2
-2
3
-1
2
-I
3
Solution. The characteristic equation of tltc given matrix is
(6 - A) (9 - 61 + I2 -1)+ 2 (-6+ 21 +2)+ 2(2-6 + 21) = 0
-V + 12 I2 - 36A + 32 = 0
=>
By trial, 1 = 2 is a root of this equation.
(1-2) (A2 - 101 + 16> - 0 => (1 - 2) (1 - 2) (X -8) - 0
=>
A = 2S 2.8 are Lite characteristic roots or Eigen values.
Ans.
3
2
0 0
Find die eigen values of 3 A' + 5 J2 — 6.-1 + 2/,
Solution. |X-17( = 0
-2
Example 3. The matrix A is defined as A = 0
l-l
2
-3
0
3-JI
2
0
0
-2-X.
=0
=>
(1-X) (3-1) (-2 - A) = 0 or K = 1, 3,-2
Eigen values of/T = L 27, -8;
Eigen values of4: = 1, 9, 4
Eigen values or A = l, 3, -2;
Eigen values of / = 1.1, 1
Eigen values of 3j43 + 5.-12 - 6.-1 + 21
First eigen value
=3 (if + 5 (if - 6 (1) + 2(1) =4
Second eigen value
= 3 (27) + 5 (9) - 6 (3) + 2(1) = 110
Third cigen value
= 3 (-8) + 5 (4) - 6 (-2) + 2 (1) = 10
Required eigen values arc 4, 110, 10
characteristic wfr o/ a fria?rgw/ar mafr/x areyujtf rfie diagoHa/
Example 12- Show
qf tite jnalrix.
Solution. Let us consider the triangular matrix.
xc
-4 =
Characteristic equation is
0
®31
"22sxn
a3]
£T32
£J4I
^42
0
0
0
^43
1-?JI = 0
c7-
■
a2i
JI
A.
0
0
0
«22 “*■
0
0
0
rtj]
fl4i
=0
^2
Oil expansion it gives
(", i - M 0 J2 "
<*© -
~ fl|l’
which are diagonal elements of matrix A.
^12’
<«*♦ “ M “ 0
°3J’
344
Proved-
Example 14.
(he eigen values of the orthogonal mairix.
1
Solution. The characteristic equation of
1-1
2
1
2
2'
=>
=>
=>
2
I
-2
2
-2
I
is
2
2
/
-2
-2
/
2
2
1 -1 -2
2
-2
1-1
(l-Z)[(l-X)(l-l)-4]-2[2(l-Z) + 4'| + 2 -4-2(l-k)]=0
(1-1) (1 -21 + y} -4) -2(2-21 + 4) + 2 (-4 - 2 + 21) = 0
I3-31s -91+27=0
(l-3)2(l+3) = O
The cigen values of A are T 3* -3, so the eigen values of B
3
are lh 1,-1-
Nate, If X - l is an eigen value of B then its reciprocal — = - = 1 is also an eigen value of B. Ant,
l
21.3 CAYLEY-HAMILTON THEOREM
Satcnmit. Every square matrix satisfies its own characteristic equation.
if |>1-U =(-l)J,(x* i V"1 i tip/"1 + -■ i
be the characteristic polyitoniialof n x «
matrix A = (a. then the matrix equation
+flLAr” 1+a2XJT 2+... + <<7 = 0 is satisfied by
+£j]?1n_L
j4 f.e.,
-...+«nZ =0
4 L-
Exgnpie 15+ J erify Cayley-Hamilton theorem for the matrix
p
2
1
□4 =
J
and hence find A
b
Solution. The characteristic equation of the matrix is |^ - X J| = 0
1-1
2
2
-l-l = 0
(!-?.)(- I- X) - 4 = 0 =>- I + V -4 = 0=>??-5=0
By Cayley-Hamilton Theorem,
v4z-5Z =0
(U.P..I Sem., Dec 2008)
Now;
A = A.A =
A2 -5/ =
‘ 1
2‘
2
-1
5
0
0
5
-5
1
21
5
O'
.0
5.
1
0
0
0
I
5
+
0
0
0 -5
Io
-5
0
0
From (I) and (2), Cayley-Hamilton iboorcm is verified
Again from (J), we have
,7
A1 - 5/ = 0
Multiplying by jT;, we get
A-5A-{, = f>
i = -A
1
j-i_
-0
...(2)
J.
2
1 ]
2
5
5
5 2
-I
2
]
5
5
Ans
Example 16. Find the characteristic equation of the matrix A,
■4
3
j4 = 2
l
Fence find A
]’
-2
2
]
I
1
(R.<1P. I; Bhopal. Feb. 2006)
J
Solution Characteristic cq nation is
4-A
3
I
2
l-Z
-2
I
2
1--A
=0
(4-A)[l + J?-2X + 41 -3(2—23. +2) + b(4-1 + A) = 0
(4-;.)()? -23L+s)-3(-2A+4)+(3+k) = 0
4>?-8X+20-A3+2Az-5A+6A-12+3+A = 0
-A3+6A2-6X+11 = 0
cr
V -6V +6)1-11 =0
By Ciiyley-Hamilton Theorem
^3_6J2+6X-111=0
Multiplying (J) tty J-1, we get
/i2- 6/1+6/ -i vr1 = 0 or 1N-' = J 2-6J+6Z
4 3
T 4 3
r
■4 3
r
"1 0 o'
IN ' = 2 1 -2 2 1 -2 -6 2 1 -2 +6 0 1 0
1 2
1 1 2
1
_1 2
1
_0 0 1
23 17
+
'-24
-l’
K
3
-2 + -12
9
7
-2_
5
^=1II -4
3
-18 -6_
-6
1,2 + 0
-+> -12 -6
-1 -7’
3
10
-5
-2
"6 0 0"
5
-1
’71
6 0 = -4
3
10
0 0 6
3
“5 A
Ans.
21.5 CHARACTERISTIC VECTORS OR EIGEN VECTORS
As we have discussed in Art 21.2,
A column vector A-is transformed into column vector 7 by means of a square matrix A.
Now we want to multiply the column vector X by a scalar quantity 1 so that we can find the
same uansfomed column vector F.
Le.,
AF is known as eigen vector.
Example 19.Sftow that rhe vector (1, /, 2) is an eigen vector ofthe matrix
1
2 -I ctJrrespaWmg tn the eigen value 2.
2
0
Solution. Let A - (h 1,2).
"3
Now;
I
AA = 2 2
2
2
-]' I
-1
1
0_ 2
3 + 1-2
—
2+2-2
2+2+0
2
—
]
2 =2 1
4
2
Corresponding to each characteristic root 1. we have a corresponding non-zero vector A
which satisfies the equation [-4-JJJA' =0. The non-zero vector JT is called characteristic
vector or Eigen vector.
21.6 PROPERTIES OF EIGEN VECTORS
1. The eigen vector A'of a matrix J is not unique.
2. If
....... X* be dislinci eigen values of an n * n matrix then corresponding eigen
vectors A, X
JT form a linearly independent sei.
3. If two or more eigen values arc equal it may or may not be possible to get linearly
mdepctidenl eigen vectors corresponding to tlic equal roots.
4, Two eigen vectors X and X ate called orthogonal vectors if X[ A\ = (),
5. Eigen vectors of a symmetric matrix corresponding to different eigenvalues arc orthogonal.
a
Normalised form of vectors. To find normalised form of
1
For eraropte, normalised form of
2
2
&
we divide each element by
1/3
IS
2/3
2/3
l2+22 + 21
Example 20. Determine whether the eigen vectors of the matrix
0
-1
2
]
2
3
tw orthogonal.
Solution. Characteristic equation is
l-X 0
-I
=>
(]-Z>|(2-X)(3-X)-2]-0-l[2-2(2-X)] = 0
=>
(1-X)(6-5X+X2-2)-(2-4+2k) = 0 =*
=>
(l-l)(k2-51 + 4)-2(k-l)=0
=> (k-l)(k2-5k + 4 + 2|=C
=>
(l-l)(k2-5k+6} = 0
=> (X-I)(X-2)C.-3) = 0
SoT 3k. = 1,2, 3 arc three distinct eigen values of?L
(k - I) {X? - 51 + 4) + 2 (X - 1) = 0
For 1 = 1
j
'o’
F k
_
_
0
—
4
3-1
2
«
-1
1
2-1
0
"J
L
_
L
0
_1
0
-]
V
’O’
0
0
2-1
]
*2
— 0
J
1
2
3-1
_*s_
0
2
2
x3 = 0
x <■ x, ■* x = 0
Let x = fr then x_, = 0 - k = - fc’
i
_______
k
a; = k -i
A-
=>
0
i
1
o
*1 =
For A = 2
1-2
0
-1
’<
1
2-2
1
*2
_2
2
3-2
.*3.
—
'o'
0
0
0
_0_
2
jt 4- Or + jr = 0
2r[ + 2x? + / = Q
Ji
0-2
X2 =k
_ -r2 _ J'1 _ {2-1
2-0
For k = 3
1-3
0
-l
rL
1
2-3
1
*2
2
2
3-3
*3
=>
*, = k,
-1
—
—7
X
0
-T
XL
0
1
-]
I
-V2
0
2
2
0
_*3_
0
_
-k
J2
X]-x2+Xj=0 J
-1+2
X2 = -*, X3 =
^1
A3 = Xj = -k
_*d -2k
X^-[1,-1, 0]
-2
-2*t +0*i -*j = 0]
o'
-
*3
0
2-0
-2k
l'
k
2
-
0
=
k -I
-2
r r
= 3, X2X3 = [2,-1.-21 -1
-2.
xyr, =[).-], -2]
0
Since X{X2 = 3*0, XfX, = 7 * 0, XfXt = 2 # 0
Thus, there are three distinct eigen vectors. Sa A r
A^ are not orthogonal eigen vectors.
Example 23. Find the eigen value and corresponding eigeit vectors of the matrix
Solution, pl- M| = 0
-5-1
2
= 0 => (- 5- 1) (- 2 -1) - 4 = 0
2
-2-1
=> I2 + 71 + 10 - + = 0 => I2 + Th + 6 = 0
(1+ 1) (1 + 6) = 0=> 1=-1 ,-6
The cigen values of the given matrix arc - 1 and - 6.
(i) When 'K = - 1, the corresponding eigen vectors are given by
'-5 + 1
2
2
-2 + 1
__
_*2_
’O’
-4
2'
*i'
_0_
2
-1
-X2-
=H
h
I
k
Let x. = Jt, then x. = 2k. Hence, eigen vector X( =
2k
(i0 When < = - 6 . the corresponding eigen vectors arc given by
'-5 + 6
2
[<
21 L .
-2 + fiJ
tj
Xj + 2x, = 0
Let x. = A., then x, =
lienee eigen vector X2 =
o'
1
2'
0
2
4. -X2,
*i
=> *i = - 2r2
"O'
_0_
3
1
4
Example 24, bind the eigen values and eigen vectors of matrix A = 0
2
6
5
(AM/ETE. June 20/0. 2009)
= (3-1) (2-X) (5-1)
0
3-1
1
4
0
2-1 6
0
0
5-1
Hence the characteristic equation of matrix J is given by
Solution, |J-U| =
0
X-1/I=O
(3-1) (2-l)(5-l)=0
1 = 2,3,5.
Thus the eigenvalues of matrix A are 2, 3, 5.
Tlie eigen vectors of I he matrix A corresponding to the ci gem value l is given by Ute non­
zero solution of the equation (/I - ?J)A’ = 0
5-1 _*3.
0
1
1
o o
—
-------------------
_______
----------
0
1
i
o
i
_0_
3-2
0
0
1
4
2-2 6
0
5-2
*2
1
’
When X = 2, (he corresponding eigen vector is given by
o o
_______
or
rr
'3-X
=
0
I 1
0 0
0 0
xL + +4*3 = 0
0X| + Oxj +
=0
*i
*2
6-0
0-6'0-0'
1 " -I ’ 0
=>
X] = £, Xn =
*3=0
r
Ilcncc.V. - -k = k -1
0
_ oj
can be taken as an eigen vector of A corresponding to (he eigen
value Z = 2
2-3
6
_0
0
5-3
1
"0
1
4"
0
-1
6
_0_
0
0
2 .*3.
o
0
*i
*2
0
_____
4
—
Ox, * x2 + 4xt = 0
Ox, - Xj + fix3 = 0
x, _ x, _ Xj
6+4_0-0~0-0
= Ar, X-, = 0, x, = 0
x, _ Xt _ Xj _ £
)0=0~0~10
*2
—
1
1
0 0
_____
■3-3
1
’
When X = 3, substituting in (J), tlw corresponding eigen vector is given by
0
fr
1
Hence, A\ = 0
= A- 0
0
1
eigenvalue X — i
0
can be taken as an eigen vector of A corresponding to the
When X = 5.
Again, when X = 5, substituting in (1), the corresponding eigenvector is given by
3-5
1
4
0
0
2-5
6
= 0
0
0
5-5
_r3_
-2
1
4"
0
-3
6
*2
0
_ 0
0
0_ .*3.
0_
=>
0
0
-2xj + Xj + 4 a\ = 0
—3 Xj. +
=0
By cross-multiplication method, we have
_
r( =
_
6 + 12 0+12
6-0
r2 = 2th
= k
3£
Hence,
X2
value X - 5.
12
6
_3"
= 2k = k 2
k
18
can be taken as an eigen vector of A corresponding to the eigen
1
Ans.
21.9 NON-SYM METRIC MATRIX WITH REPEATED EIGENVALUES
Example 25. JW the eigen values and eigen vectors of the matrix:
2
I
l
1
2
]
: Bhopal, June 2004)
0 0 ]
"2 I
I
Solution. We have, /I = 1
2
1
0
0
1
Characteristic equation of A is | A - ?J | = 0
2-X
l
]
l
2-X
0
1
0
=0
1-X
On expanding llic determinant by (he third row. we get
=>
(1-Z){(2-a.)(2-a.)-1} = 0
=>
(i - ?-) {(i - ?.)2 -i}=o
=>
(l-k}(2-?-+1)(2-X-1) = 0
=>
(1-5L)(3-1) ()-X) =0
X. = IU
m titn X = 1
r
2-1
1
1
1
2 -1
1
0
0
1-1
1
1
"o
= 0
z
Q
]' X
'o'
I
I
0
x I y f z = 0
Lctx- kf and m = A.
W*=°
*i
r
= k
^2
A] =
z = -(fc,+^)
o
i
11 f k} = A, = A |
-2
J
0
Again 1 = 1, -V2 -
|Again if A, = 1, ^=0, -(A + *,) = - l|
-1
1
1
2-3
I
0
0
1-3
K
1
"o"
"-1
I
1
X
"0"
V
"2-3
________ j
1
when 1=3
= 0
1
-1
]
y
__ 0
z
Q
_ 0
0
-2
z
0
"-1
1
I
x
0
0
2
>' = 0
0
0
-2
z
0
-x + 7 + z = 0
2z = 0 => z = 0
-.v-fj/ + 0 = 0 => x - y ■ i(say J
T
A
*3 = A = A 1
0
Ans.
0
Example 26. Find a// fAe £f£en ra/eres mJ £j#en vectors o/JAe /nufrix
-2
2
-31
(dAfffTE Dec 2009)
Solution. Characteristic equation of j4 is
=>
-2-1
2
2
]-k
-6
-]
-2
U-A
=0
(-2 - a) L-X+V -12] - 2(-2l -6) - 3 (-4 +1 - a) = 0
V+V-211-45 = 0
By trial: If A. = -3, then -27 + 9 + 63 - 45 = 0, so (X. + 3) is one factor of (1).
The remaining factors are obta i tied o n div i di tig (1') by X + 3.
]
1
-21
-45
-3
45
-15
0
1
“2
(X-5)(X + 3) = 0
Xl-2X-I5-O
=> (X+3)(X i 3)(X 5)-0 =>■ 1 = 5, 3, J
■v "
~J
XT.""
V “
X
“ X
- p
To find die eigenvectors for corresponding eigen values, we will consider die matrix equation
-2-X
2
-3
X
2
1-K
-6
y = 0
0-1
z
-3’ Jf
’O’
G4-l/)A'=0
-2
-1
'-7
2
2
-4
Oil pulling Z - 5 in eq. (2), it becomes
-1
Wc have
x
-12-12
-6
-2 -5
y
...(2)
0
= 0
0
- 7x + 2> - 3z = 0,
2x - 4y - 6z = 0
k
— ----—
-6-42
x = t,
28-4
y = 2fc
z
X
?
-- —
or
24
z = -it
f
' k'
Hence, die eigen vector jf. - 2k
-k
=k
2
-1
-48
24
or
]
2
-1
Pul x = -3 in eq. (2). it becomes
1
2
-3
2
4
-2
-C,
sl
-I
We have
x 4 2y - 3z ™ 0,
2x + 4y - 62 = 0,
- x - 2_v + 3z = 0
Here first, second and third equations arc the same.
Hence, the eigen vector is
Let Jt| =0.^2 = X Hence A\ = 3
2
0
0
Since the matrix is non-symnielric, the corresponding eigen vectors Ar, and X must be
linearly independent. This can be done by choosing
U
O
vs =
2
-1
■
j]
O
*2 =
--------------- 1
2 ,
CjJ
I fence, .Y, =
r
_________
■
--------------- II
0, and Hence T3 = 0
1
Enample 2 Ji. bind the eigen values, eigen vectors the modal matrix given below.
0 0"
0 3 -I
0 -] 3_
]
(R.G.P.V. Bhopal, I Sent,, 2003)
Solution. The characteristic equation of the given matrix is
1-A
0
0
0
3-X
-I
0
-I
3-X
(1-X)(3-X,4-1)C3-3L-1) = O
(l-X>(4-X)(2-X) = 0
*-=1,2,4
When 1 = 1,
-
'0
0
0
2
0
-1
ooo
O’ \i roi
jJ = 0 ~ 0 2-1
-1
-*J H
0 o 2
2
L
2
=>
2x2-x5 = 0
3
-x}
= 0
■ (!)
x3 == 0
(2)
Pulling i3= 0 from (2) in (1)? we get 2x2 - 0
Eigcti Vector = 0
0
0=>x2 = 0
When k = 2,
—> W- + R-.
Eigenvector = 1
1
When 31 = 4,
-3
0
0
-1
0
-1
-3jTj = 0
-x2 -
= 0
S " *3
0
Eigen Vfccior =
I
-I
]
0
0
Modal matrix = 0
1
1
Ans.
21.19 DIAGONALISATION OF A MATRIX
Diagonalisation of a matrix X is the process of reduction of X to a diagonal form
IfX is
related to D by a similarity transformation such that D = P • XP then X is reduced to the diagonal
matrix D through modal matrix P. Z> is also called .specfro/ matrix of A
Notes L The square matrix P, which diagonalises X, is found by grouping the eigen vectors ofX
into square-matrix and the resulting diagonal matrix has the eigen values of 4 as its
diagonal elements.
2» The transformation of a matrix X to P 1 XP is known as a
3r The reduction of 4 to a diagonal matrix is, obviously. a particular case of similarity
transformation.
4. The matrix P which diagonalises X is called die mods/ JNdhx of X and die resulting
diagonal matrix D is known as the rpeefra nrafrix of X.
Example 35. Ftrtda matrix P which diagonalizes the matrix
4
2
I
3 L verify Pr' AP = D where D is the diagonal matrix. (UP, I Semester, Dec. 2008}
Solution. Tlic characteristic equal ion of matrix A is
4-k
(4 - k) (3 - k) - 2 = 0
2
=> V - 71 + 12 - 2 = 0
=>
k1 - 7k + 10 = 0
=> (k - 2) (k -5) = 0
=>
k = 2, k = 5
Eigen values arc 2 and 5,
(f) When X = 2 , eigen vectors are given by the matrix equation
[4:2 A xi
[
2
3-2
*2
0
2
1
J>i
0
2
1
*2
2X, + x, = 0 => x2 = -2x,
X| = k , x^ = - 2 k
■ k
Hence, die eigen vector
rn
xi = -2k or M
L°.
(ii) When X = 5 , eigen vectors are given by the matrix equation
4-5
2
Let< = A . then
u
=4
J.
T
or
1
Hence, the eigen vector
—
Modal tnairix
“J
1
=>?■’=!
1
-1
]
3
2
I
1 n
-ip
n
i
L2
lJL2
3J
-2 ]
6
o'
For diagonalization
.
D = F' AP= t
]
-r
’ 2 !
3 2
i
-4
]
d 3
:
0 15
r
'2 O'
.0
5
Verified♦
6
-2
Example 3(i, Let A = -2
2
3-1
Find matrix P such that P 1 AP is diagonal matrix.
2-13
Solution. The characteristic equation of the matrix 4 is
6-X
-2
2
-2
3-1
-1
2
-I
3-1
=0
(6-X)[9+l -6X-l] + 2[-6+2X + 2]+2[2-6+2X] = 0
(6 - XXX3 -6X + 8) - K + 4X - 8 +4X = 0
6X' - 36X + 48 - X3 + 6X2 - SX - J6 +81 = 0
-X3 +12?? -36?. +32=0
(X-2):(X-8)-0
??-J2X2+36X-32 = 0
X “2,2, 8
Eigen vector for X = 2
2
-]
or -2
0_
= 2
]
4
-2
2'
0"
-2
]
-I
= 0
2
-1
1
"2
-J
i"
’o’
0
0
0
— 0
0
0
0
0
This equation is satisfied by
-1
OL 2x1
= 0, r2 = l, x3 = l
0
*1- ]
]
arid again
je
x = 3 *3
x
*] = 1’ ’ *1
r
-^2 “ 3
i
I.
jR-i —> J?| 4-
/l; —> J?; + A?
Eigen vector for X = 8
'-2
"o'
2’
-2
-2-5-1
*2
— 0
_ 2 -1 “5_
0
- 2 xL -
=0
-2xj -5x2 -r3 =0
*1
2+10
*3 _ _ *J
-4-2'10-
A=
12 -6
=
6
2
-1
1
2
?3 = -I
1
"o
P=
Now
p
1
2
1 3 -1 .
1 1
]_
4
F-'=-l -2
6
-2
1 -7
2 2
I -1
4
I -7ir 6 -2 2 [0 ]
2" '2 0 O'
I
=-2 -2
2 -2
3 -1 1 3 -1 — 0 2 0
6
0 0 8
-2
1 -I
2 -1
3 1 1
1
Ans.