Mathematics

Session

Matrices and Determinants - 3

1. Singular and Non-singular Matrix

2. Adjoint of a Square Matrix and its Properties

3. Inverse of a Matrix and its Properties

4. Solution of Simultaneous Linear Equations (Matrix Method)

5. Class Exercise

Session Objectives

Singular Matrix

A square matrix A is said to be singular if lAl = 0 .

A is non-singular if A 0.

For Example:

1 1 31 3 35 3 3

Let A=

A is a singular matrix .

A=1(9+9)+1(3+15)+3(3-15) = 18+18-36 = 0

Non-Singular Matrix

Let B=1 1 12 1 11 2 3

B is a non-singular matrix.

B= 1(-3+2)-1(6-1)+1(-4+1)

= -1 – 5 – 3

= -9 0

Example -1

Find the value of x for which the matrix

is singular.

x 1 0

A = 2 -1 1

3 4 -2

For matrix A to be singular

A = 0

x 1 0

2 -1 1 = 0

3 4 -2

- 1 -4 - 3 = 0

7-2x +7 = 0 x =

2

x 2 4

Solution:

Adjoint of a Square Matrix

The transpose of the matrix of cofactors of elements of a square matrix A is called the adjoint of A and is denoted by adjA.

Tij jiadjA = [C ] adjA = C

Adjoint of a Square Matrix

11 12 13 11 21 31

21 22 23 12 22 32

31 32 33 13 23 33

C C C C C C

adjA = C C C C C C

C C C C C C

T

where Cij denotes the cofactor of aij in A.

11 12 13

21 22 23

31 32 33

a a a

Let A = a a a ,

a a a

then

Example - 2

Find the adjoint of matrix

a bA

c d

Solution :

11 12

21 22

T

a bA= c d

C =d , C =-c

C =-b , C =a

d -cadjA= -b a

d -badjA =

-c a

Example - 3

Find the adjoint of matrix

1 2 1

A 3 1 0

0 1 1

Solution:

11 12 13

21 22 23

31 32 33

1 2 1

We have A 3 1 0

0 1 1

C (1 0) 1 , C ( 3 0) 3, C (3 0) 3

C ( 2 1) 3, C (1 0) 1 , C ( 1 0) 1

C (0 1) 1 , C (0 3) 3 , C (1 6) 5

Solution cont.

T1 3 3

adjA 3 1 1

1 3 5

1 3 1

adj A = 3 1 3

3 1 -5

Properties

A (adjA) = |A| In = (adjA) A

Proof: Let A = [aij] be a square matrix and let Cij be cofactor of aij in A, then (adjA) = [Cji] for all i, j = 1, 2, ..., n

n

ir rjr 1

A , if i jwe know a C

0, if i j

Properties (Con.)

Therefore, each diagonal element of A (adjA) is equal to |A| and all non-diagonal elements are equal to zero.

n

| A | 0 0 .. 0

0 | A | 0 ..0i.e. A (adjA) = = A I

: : : :

0 0 0 ..| A |

n

ri rjr=1

A , ifi = jSimilarly, (adjA)A = C a =

O, ifi j

Hence, A (adjA) = |A| In = (adjA)A

Properties (Con.)

2. If A is a non-singular square matrixof order n, then |adjA| = |A|n – 1

3. If A and B are non-singular square matrices of same order, then adj AB = (adjB) (adjA)

4. If A is a non-singular square matrix, then adj (adjA) = |A|n–2 A.

Example-4

Compute the adjoint of matrix A= 1 23 -5

and verify that A(adj A)=|A|I.

Solution:

1 2We have A = 3 -5

11 12 21 22

T

C =-5 , C =-3 , C =-2 , C =1

-5 -3adjA= -2 1

-5 -2= -3 1

Solution (Con.)

Hence verified.

1 2 -5 -2L.H.S. =A adj.A = 3 -5 -3 1

-5- 6 -2+2 -11 0= =-15+15 -6-5 0 -11

A = -5- 6=-11

1 0 -11 0R.H.S.= A I=(-11) =0 1 0 -11

L.H.S =R.H.S

Example-5

If a matrix p qA= r s , find det. {A(adjA)}.

Solution:

Ts -r s -qadjA= = -r p-q p

p q s -q ps- qr -pq+pqA(adjA)= =r s -r p rs- rs -rq+sp

1 0=(ps- qr) 0 1

Now, det.{A(adj.A)}=(ps- qr)×1

=ps- qr

p qWe have A= r s

Inverse of a Matrix

Steps to find inverse of a matrix:

-1 -1 1then A is given by A = .(adjA)

Aexists and

I f matrix A is non - singular i.e. A 0,a

(i) Find out |A| and if , then the matrix is invertible.

(ii) Find out (adjA).

Then -1 1

A = .(adjA)A

A 0

Example-6

Find the inverse of the Matrix

-1 4 2

A = 2 -1 4

1 2 3

-1 4 2

A = 2 -1 4 = -1 -3 - 8 - 4 6 - 4 +2 4+1 =13 0

1 2 3

Solution:

-1 4 2

We have A = 2 -1 4

1 2 3

Solution cont.

T T11 12 13

21 22 23

31 32 33

C C C -11 -2 5 -11 -8 18

adjA = C C C = -8 -5 6 = -2 -5 8

C C C 18 8 -7 5 6 -7

-1-11 -8 18

1 1Hence, A = .(adjA) = -2 -5 8

13A5 6 -7

Properties

(i) A square matrix is invertible if it is non-singular.

(ii) Every invertible matrix possesses a unique inverse.

Hence, an invertible matrix possesses a unique inverse.

Proof: Let A be an invertible matrix of order n x n.

Let B and C be two inverses of A.

Then AB = BA = In

and AC = CA = In

Now AB = In

Multiplying by C C(AB) = CIn

(CA)B = C In

In B = C In

B = C

Properties (Con.)

(iii) (AB)–1 = B–1 A–1

or (ABC)–1 = C–1 B–1 A–1

(iv) (AT)–1 = (A–1)T

Example-7

Show that 2 -3A = 3 4

satisfies the equation x2 - 6x + 17 = 0.

Hence, find A-1.

Solution:

2 2 -3 2 -3 4- 9 -6-12 -5 -18A =A.A= = =3 4 3 4 6+12 -9+16 18 7

22

-5 -18 2 -3 1 0A - 6A+17I = - 6 +1718 7 3 4 0 1

2 -3We have A= 3 4

-5 -18 12 -18 17 0= - +18 7 18 24 0 17

Solution (Cont.)

A2 - 6A + 17I = 0

Multiplying each side by A-1, we getA-1A2 - 6(A-1A) + 17(A-1I) = A-1.0

1 1 1 1( A A I, A I A , A .0 0) (A-1A)A - 6I + 17A-1 = 0

IA - 6I + 17A-1 = 0, 17A-1 = 6I - A

-1

4 317 171 1 16 0 2 -3 4 3A = (6I - A)= - = =0 6 3 4 -3 217 17 17 3 2-17 17

-1

4 317 17

Hence, A =3 2

-17 17

0 0 O0 0

Hence, A satisfies the equation x2 - 6x + 17 = 0.

Example-8

Find A-1 , if0 1 1

A= 1 0 11 1 0

. Also show that A-1=2A - 3I

.2

Solution:

11 12 13

21 22 23

31 32 33

Now,C =-1,C =1,C =1

C =1,C =-1,C =1

C =1,C =1,C =-1

0 1 1A = 1 0 1 =0(0-1)-1(0-1)+1(1- 0) =1+1= 2 0

1 1 0

0 1 1We have A= 1 0 1

1 1 0

T-1 1 1 -1 1 1

adj.A= 1 -1 1 = 1 -1 11 1 -1 1 1 -1

Solution cont.

-1

11 1- 22 2

1-1 1 11 1 1 1Hence, A = adj A = 1 -1 1 = - 2A 2 2 21 1 -1

1 1 1-

2 2 2

2 0 1 1 0 1 1 1 0 0A - 3I 1Also = 1 0 1 1 0 1 - 3 0 1 0

2 2 1 1 0 1 1 0 0 0 1

2 1 1 3 0 01= 1 2 1 - 0 3 0

2 1 1 2 0 0 3

-1-1 1 11

= 1 -1 1 = A2 1 1 -1

Example-9

Solution:

2 1 4 5(AB) = 5 3 3 4

11 14= 29 37

-1 -1 -12 1 4 5If A = and B= , verify that AB = B A .

5 3 3 4

T

-1 37 -29 37 -14LHS = AB = =-14 11 -29 11

Solution (Cont.)

T-1 4 -3 4 -5B = =-5 4 -3 4

T-1 3 -5 3 -1A = =-1 2 -5 2

-1 -1 4 -5 3 -1 37 -14RHS =B A = = =LHS-3 4 -5 2 -29 11

Solution of Simultaneous Linear Equations (Matrix Method)

Let the system of 3 linear equations be

1 1 1 1

2 2 2 2

3 3 3 3

a x +b y +c z = d

a x +b y +c z = d

a x +b y +c z = d

This system of linear equation can be written in matrix form as

11 1 1

2 2 2 2

3 3 3 3

da b c x

a b c y = d

a b c z d

AX = B ... i

Solution of Simultaneous Linear Equations (Matrix Method)

1X = (adjA)B

A

The matrix A is called the coefficient matrix of the system of linear equations.

Multiplying (i) by A–1, we get

-1I f A 0 i.e. A is non - singular, then A exists.

1 1A AX A B

1 1A A X A B

1IX A B

Important Results

(i) If A is a non-singular matrix, then the system of equations given by AX = B has a unique solution given by X = A–1B

(ii) If A is a singular matrix and (adjA)B = 0, then the system of equations given by AX = B is consistent with infinitely many solutions.

(iii) If A is a singular matrix and (adjA)B 0, then the system of equations given by AX = B is inconsistent.

Example-10Using matrix method, solve the following system of linear equationsx + 2y -3z = -42x + 3y + 2z = 23x - 3y - 4z = 11

Solution:

The given system of equations is

x + 2y - 3z = -4 ...(i)

2x + 3y + 2z = 2 …(ii)

3x -3y - 4z = 11 …(iii)

1 2 -3 x -4or 2 3 2 y = 2

z3 -3 -4 11

AX = B

Solution (Cont.)

1 2 -3 x -4where A= 2 3 2 , X= y , B= 2

z3 -3 -4 11

2 2 1 3 3 1

-1

1 2 -3A = 2 3 2

3 -3 -4

1 0 0= 2 -1 8 Applying C C - 2C and C C +3C

3 -9 5

=1(-5+72)=67 0

A exists.

ij ij ijLet C be the cofactor a in A = a , then

Solution Cont.

11 12 13c =(-12+6) c =- -8- 6 c =(-6- 9)=- 6 =14 =-15

21 22 23c =- (-8- 9) c =(-4+9) c =-(-3- 6)=17 =5 = 9

31 32 33c =(4+9) c =-(2+6) c =(3- 4)=13 = - 8 =-1

31 32 33c =(4+9) c =-(2+6) c =(3- 4)=13 = - 8 =-1

-6 17 1314 5 -8-15 9 -1

T-6 14 -15

adjA = 17 5 9 =13 -8 -1

Solution (Con.)

-1Now, X=A B

-6 17 13 -41X= 14 5 -8 2

67 -15 9 -1 11

x 201 31y = -134 = -2

67z 67 1

x=3 , y=-2 , z=1

-1-6 17 131 1

A = .adj A = 14 5 -8A 67 -15 9 -1

Example-11

Using matrices, solve the following system of equationsx + y + z = 6x + 2y + 3z = 14x + 4y + 7z = 30

Solution:

x + y + z = 6 …(i)

x + 2y + 3z = 14 …(ii)

x + 4y + 7z = 30 …(iii)

1 1 1 x 6or 1 2 3 y = 14

z1 4 7 30

AX = B

The given system of linear equations is

Solution (Cont.)

1 1 1Now, A = 1 2 3 =1(14-12)-1(7- 3)+1(4- 2)

1 4 7

=2- 4+2=0

1 1 1 x 6where A= 1 2 3 ; X= y ; B= 14

z1 4 7 30

11 12 13c =(14-12) c =-(7- 3) c = 4- 2=2 =- 4 =2

ij ij ijLet C be the cofactor a in A = a , then

Solution cont.

The given system of equations is consistent with infinitely many solutions.

21 22 23c =-(7- 4) c =(7-1) c =-(4-1)=- 3 = 6 =- 3

31 32 33c =(3- 2) c =-(3-1) c =(2-1)= 1 =- 2 =1

T2 -4 2 2 -3 1

adjA = -3 6 -3 = -4 6 -21 -2 1 2 -3 1

2 -3 1 6 0and (adjA)B= -4 6 -2 14 = 0 =0

2 -3 1 30 0

Solution (Con.)

Putting z = k in first two equations, we get

x + y = 6 - k

x + 2y = 14 - 3k

1 1 x 6 - k= AX = B

1 2 y 14 - 3kor

-1

1 1A = = 2 - 1 = 1 0

1 2

A exists.

T2 - 1 2 - 1adjA = =

-1 1 -1 1

Solution (Con.)

-1 2 - 11A = adjA =

| A | -1 1

-1 x 2 - 1 6 - kNow, X = A B =

y -1 1 14 - 3k

x 12 - 2k - 14 + 3k -2 + k= =

y -6 + k +14 - 3k 8 - 2k

x = -2 + k and y = 8 - 2k

These values of x, y and z = k also satisfy (iii) equation.

Hence, x = -2 + k, y = 8 - 2k and z = k, where k R.

Thank you