Page(1)

Chap1:-[Adjoint of Matrix]

Cofactor of an element:-

Cofactor of an element is given byAij = (-1)i+j

.Mij

Adjoint of a Matrix:-

A transpose of a Matrix of cofactor is called Adjoint.

Example:-

Ex :-1 ] If A= 1 1 02 1 −11 2 3

Find the all Cofactor elements and Find adjA.

SolLet

A= 1 1 02 1 −11 2 3

We know that Aij=(-1)i+j

.Mij

...A11 = (-1)

i+j. M11

= (-1)2

1 −12 3

= 1(3+2)`

= 5

A12=(-1)1+2.

M12

=(-1)3

2 −11 3

= -1(6+1)

= -7

A13 =(-1)1+3 .

M13

=(-1)4

2 11 2

= (4-1)

= 3

A21 =(-1)2+1 .

M21

=(-1)3

1 02 3

= -(3-0)

= -3

A22 =(-1)2+2 .

M22

=(-1)4

1 01 3

Page(2)

= 3

A23 =(-1)2+3 .

M23

=(-1)5

1 11 2

= -1

A31 =(-1)3+1.

M31

=(-1)4

1 01 −1

= -1

A32 =(-1)3+2 .

M32

=(-1)5

1 02 −1

= 1

A33 =(-1)3+3 .

M33

=(-1)6

1 12 1

= (1-2)

= -1

Matrix of Cofactor = 5 −7 3

−3 3 −1−1 1 −1

adj A = 5 −3 −1

−7 3 13 −1 −1

Ex:-2] If A= 1 1 −11 1 01 2 −1

Find minor and cofactor of a11 ,a21 ,a33 .

SolLet A= 1 1 −11 1 01 2 −1

Minor of a11 = 1 02 −1

= -1-0

= -0

∴Cofactor of a11 = A11 = (-1)1+1.

M11

= (-1)2.

1 02 −1

= -1

Minor of a21 = 1 −12 −1

= (-1+2)

Page(3)

= 1

∴Cofactor of a21 = A21=(-1)2+1.

M21

= (-1)3 .

(1)

= -1

Minor of a33 = 1 11 1

= 0

∴Cofactor of a33 =A33 = (-1)3+3.

M33

= (-1)6.(0)

= 0

Ex:-3] If A= −3 1 0

2 −2 1−1 −1 1

Show that A(adjA) is a Null matrix.

SolLet A= −3 1 0

2 −2 1−1 −1 1

Matrix of minor = −1 3 −41 −3 41 −3 4

∴Matrix of Cofactor = −1 −3 −4−1 −3 −41 3 4

adj A = −1 −1 1−3 −3 3−4 −4 4

A.(adj A)=

−3 1 02 −2 1

−1 −1 1 .

−1 −1 1−3 −3 3−4 −4 4

= 3 − 3 + 0 3 − 3 + 0 −3 + 3 + 0−2 + 6 − 4 −2 + 6 − 4 2 − 6 + 41 + 3 − 4 1 + 3 − 4 −1 − 3 + 4

= 0 0 00 0 00 0 0

A.(adj A) is a Null matrix.

Ex:-4] If A= −1 −2 −22 1 −22 −2 1

Show that adjA=3A'

SolLet A= −1 −2 −22 1 −22 −2 1

Page(4)

matrix of minor = −3 6 −6−6 3 66 6 3

∴matrix of Cofactor= −3 −6 −66 3 −66 −6 3

adjA= −3 6 6−6 3 −6−6 −6 3

---------------------- 1

now,

3A' = 3 −1 2 2−2 1 −2−2 −2 1

= −3 6 6−6 3 −6−6 −6 3

---------------------- 2

From eqn (1) & (2), Hence proved that adj A= 3A'

Ex:-5] If A= 1 2 47 2 00 1 2

Show that A(adjA)= 𝐴 I

SolLet A= 1 2 47 2 00 1 2

Matrix of minor = 4 14 70 2 1

−8 −28 −12

Matrix of Cofactor= 4 −14 70 2 −1

−8 28 −12

adj A= 4 0 −8

−14 2 287 −1 −12

A (adj A)= 1 2 47 2 00 1 2

. 4 0 −8

−14 2 287 −1 −12

= 4 − 28 + 28 0 + 4 − 4 −8 + 56 − 4828 − 28 + 0 0 + 4 − 0 −56 + 56 − 00 − 14 + 14 0 + 2 − 2 0 + 28 − 24

= 4 0 00 4 00 0 4

---------------------- 1

Now,

Page(5)

𝐴 = 1 2 47 2 00 1 2

= 1 (4-0) -2 (14-0) +4 (7-0)

= 4 – 28 + 28

𝐴 = 4

∴ 𝐴 𝐼= 4. 1 0 00 1 00 0 1

= 4 0 00 4 00 0 4

---------------------- 2

From Eqn (1) and Eqn(2) ,

Hence proved A(adj A) = 𝐴 𝐼

Ex:-6]If A= 1 0 12 1 10 1 2

Show that adj(2A) = 22

adj A .

SolLet A= 1 0 12 1 10 1 2

∴2A = 2. 1 0 12 1 10 1 2

= 2 0 24 2 20 2 4

Matrix of Minor= 4 16 8

−4 8 4−4 −4 4

∴ Matrix of Cofactor= 4 −16 84 8 −4

−4 4 4

∴Adj 2A = 4 4 −4

−16 8 48 −4 4

---------------------- 1

Now,A = 1 0 12 1 10 1 2

Matrixof Minor = 1 4 2

−1 2 1−1 −1 1

Page(6)

∴Matrix of Cofactor = 1 −4 21 2 −1

−1 1 1

∴Adj A = 1 1 −1−4 2 1 2 −1 1

Now, 22 adj A = 4 adj A

= 4 1 1 −1−4 2 1 2 −1 1

= 4 4 −4−16 8 4 8 −4 4

---------------------- 2

From eqn (1) & (2),Hence proved that adj(2A) = 22

adj A

Ex:-7] If A = 2 −13 −2

, B = 5 01 2

Show that adj AB = adj B.adj A

Sol AB = 2 −13 −2

. 5 01 2

= 10 − 1 0 − 215 − 2 0 − 4

AB = 9 −2

13 −4

Matrix of minor of AB= −4 13−2 9

∴Matrix of Cofactorof AB = −4 −132 9

∴Adj AB = −4 2−13 9

---------------------- 1

Let B = 5 01 2

Matrix of minor of B = 5 01 2

∴Matrix of Cofactorof B = 2 −10 5

∴Adj B = 2 0

−1 5

Let A = 2 −13 −2

Matrix of minor of A = −2 3−1 2

∴Matrix of Cofactorof A = −2 −31 2

Page(7)

∴Adj A = −2 1−3 2

Now,

adj B .adj A =

2 0−1 5

× −2 1−3 2

= −4 2−13 9

---------------------- 2

From eqn (1) & (2),Hence proved that adj AB = adj B .adj A

Theorem:-1:-For any Square matrix A, show that A(adjA) = (adjA)A= 𝐴 𝐼

SolLet,

A=

𝑎11 𝑎12 𝑎13−−−−−− 𝑎1𝑛

𝑎21 𝑎22 𝑎23−−−−− 𝑎2𝑛

𝑎31 𝑎32 𝑎33−−−−−− 𝑎3𝑛

− − − − − − − −− − − − − − − −𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 𝑎𝑛𝑛

be a given matrix.

Let , Matrix of cofactor =

𝐴11 𝐴12 𝐴13−−−−−− 𝐴1𝑛

𝐴21 𝐴22 𝐴23−−−−− 𝐴2𝑛

𝐴31 𝐴32 𝐴33−−−−−− 𝐴3𝑛

− − − − − − − −− − − − − − − −𝐴𝑛1 𝐴𝑛2 𝐴𝑛3 𝐴𝑛𝑛

adj A =

𝐴11 𝐴21 𝐴31−−−−−− 𝐴𝑛1

𝐴12 𝐴22 𝐴32−−−−− 𝐴𝑛2

𝐴13 𝐴23 𝐴33−−−−−− 𝐴𝑛3

| | | || | | |

𝐴1𝑛 𝐴2𝑛 𝐴3𝑛 𝐴𝑛𝑛

A (adj A)=

𝑎11 𝑎12 𝑎13−−−−−− 𝑎1𝑛

𝑎21 𝑎22 𝑎23−−−−− 𝑎2𝑛

𝑎31 𝑎32 𝑎33−−−−−− 𝑎3𝑛

− − − − − − − −− − − − − − − −𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 𝑎𝑛𝑛

𝐴11 𝐴21 𝐴31−−−−−− 𝐴𝑛1

𝐴12 𝐴22 𝐴32−−−−− 𝐴𝑛2

𝐴13 𝐴23 𝐴33−−−−−− 𝐴𝑛3

| | | || | | |

𝐴1𝑛 𝐴2𝑛 𝐴3𝑛 𝐴𝑛𝑛

Page(8)

=

𝑎1𝑗

𝑛

𝑗 =1

𝐴1j 𝑎1𝑗

𝑛

𝑗 =1

𝐴2j 𝑎1𝑗

𝑛

𝑗 =1

𝐴3j

−−−−−−

𝑎1𝑗

𝑛

𝑗 =1

𝐴nj

𝑎2𝑗

𝑛

𝑗 =1

𝐴1j 𝑎2𝑗

𝑛

𝑗 =1

𝐴2j 𝑎2𝑗

𝑛

𝑗 =1

𝐴3j

−−−−−−

𝑎2𝑗

𝑛

𝑗 =1

𝐴nj

𝑎3𝑗

𝑛

𝑗 =1

𝐴1j 𝑎3𝑗

𝑛

𝑗 =1

𝐴2j 𝑎3𝑗

𝑛

𝑗 =1

𝐴3j

−−−−−−

𝑎3𝑗

𝑛

𝑗 =1

𝐴nj

− − − − − − − − − − − − − − − − − −

𝑎𝑛𝑗

𝑛

𝑗 =1

𝐴1j 𝑎𝑛𝑗

𝑛

𝑗 =1

𝐴2j 𝑎𝑛𝑗

𝑛

𝑗 =1

𝐴3j − − − − 𝑎𝑛𝑗

𝑛

𝑗 =1

𝐴nj

We Know that ,

1. 𝑎𝑛𝑗 𝐴𝑛𝑗 = 𝐴

2. 𝑎𝑛𝑗 𝐴𝑘𝑗 = 0

Above eqn becomes,

A (adj A) =

𝐴 0 0 − − − 00 𝐴 0 − − 00 0 𝐴 − − 0

− − − − − − − − −− − − − − − − − −

0 0 0 − − − 𝐴

= 𝐴

1 0 0 − − − 00 1 0 − − − 00 0 1 − − − 0

− − − − − − − − − −0 0 0 − − − − 1

A (adj A) = 𝐴 I---------------------- 1

Similarly We can show that

(adj A) A = 𝐴 I---------------------- 2

From eqn (1) & (2), Hence proved that A (adjA) = (adjA) A= 𝐴 𝐼

Inverse of a Matrix:-If A and Bare the two matrices such that

AB = BA = I

then B is called inverse of A and we write B = A-1 .

Theorem:-2] State and prove Necessary and sufficient condition for existence

of inverse.

Statement:- The Necessary and Sufficient condition for square matrix A to have

an inverse is that A is a non singular matrix.(i.e 𝐴 ≠ 0)

Page(9)

Proof:-Necessary part:-

Suppose that the matrix A has inverse say B

∴By Definition,

AB=BA=I

i.e. AB=I

taking determinant on both sides,

𝐴𝐵 = 𝐼

𝐴 B = 1 ≠ 0

⇨ 𝐴 ≠ 0and 𝐵 ≠ 0

∴A is a non Singular matrix.

Conversly,

Suppose that A is a non Singular matrix,

i.e. 𝐴 ≠ 0

We know that,

A(adj A)= (adj A)A = 𝐴 𝐼

Divide each term by 𝐴

We get,

A 1

𝐴 𝑎𝑑𝑗𝐴 =

1

𝐴 𝑎𝑑𝑗𝐴 𝐴 = 𝐼

∴ By Definition of Inverse ,

⇨1

𝐴 𝑎𝑑𝑗𝐴 = A−1

Ex:-1]Find the inverse of Following matrix

A = 1 2 −1

−1 1 22 −1 1

SolLet,

A = 1 2 −1

−1 1 22 −1 1

∴ 𝐴 = 1 2 −1

−1 1 22 −1 1

= 1(1+2)-2(-1-4)-1(1-2)

= 1(3)-2(-5)-1(-1)

= 3+10+1

= 14 ≠ 0

∴A−1exist.

Matrix of minor= 3 −5 −11 3 −55 1 3

∴Matrix of Cofactor = 3 5 −1

−1 3 55 −1 3

Page(10)

∴Adj A= 3 −1 55 3 −1

−1 5 3

∴A−1 =1

𝐴 𝑎𝑑𝑗𝐴

=1

14

3 −1 55 3 −1

−1 5 3

Ex:-2] If A = 1 24 −1

Find A-1

Sol Let A= 1 24 −1

∴ 𝐴 = 1 24 −1

= -1-8

= -9 ≠ 0

A-1

exist.

Matrix of minor = −1 42 1

∴Matrix of Cofactor = −1 −4−2 1

∴Adj A = −1 −2−4 1

We know that,

A-1

= 1

𝐴 adj A

A-1

= 1

−9 −1 −2−4 1

Ex:-3] Show that A= 4 −1

−3 2 Satisfy the eqn 𝐴2 − 6A + 5I = 0

Hence find A-1

Sol Let, A= 4 −1

−3 2

∴ A2= A

.A

= 4 −1

−3 2

4 −1−3 2

= 19 −6

−18 7

Consider, 𝐴2 − 6A + 5I= 19 −6

−18 7 − 6

4 −1−3 2

+ 5 1 00 1

= 19 −6

−18 7 −

24 −6−18 12

+ 5 00 5

Page(11)

∴𝐴2 − 6A + 5I= 0 00 0

∴𝐴2 − 6A + 5I= 0

Premultiplying by A-1

Then,

A-1

A2 - 6A

-1.A + 5 A−1 I = 0

A –6I+5A-1

= 0

5A-1

= 6 I– A

5A-1

= 6 1 00 1

− 4 −1

−3 2

= 6 00 6

− 4 −1

−3 2

5A-1

= 2 13 4

A-1

= 1

5 2 13 4

Theorem:-1]If A is a non-singular matrix then show that A−1 =1

𝐴

Proof:- We know that,

A A−1 = I

Taking the determinant on both sides,we get

A A−1 I

𝐴 A−1 = 1

A−1 =1

𝐴

Theorem:-2] If A & B are non –Singular matrices then show that AB is a non-

singular matrix 𝑎𝑛𝑑 (𝐴𝐵)−1 = 𝐵−1. 𝐴−1

Proof:-Since, A & B are non-singular matrix.

𝐴 0 𝑎𝑛𝑑 𝐵 ≠ 0

⇨ 𝐴 . 𝐵 ≠ 0

⇨ 𝐴𝐵 ≠ 0

AB is a non-singular matrix.

To show that (𝐴𝐵)−1 = 𝐵−1. 𝐴−1

Consider, 𝐴𝐵 (𝐵−1. 𝐴−1) = 𝐴(𝐵. 𝐵−1)𝐴−1

= 𝐴 𝐼𝐴−1

= 𝐴 𝐴−1

Page(12)

= 𝐼

Also,

𝐵−1. 𝐴−1 𝐴𝐵 = 𝐵−1 𝐴−1. 𝐴 𝐵

= 𝐵−1𝐼 𝐵

= 𝐵−1𝐵

= 𝐼

Thus; 𝐴𝐵 𝐵−1 . 𝐴−1 = 𝐵−1. 𝐴−1 𝐴𝐵 = 𝐼

By Definition of Inverse,

(𝐵−1. 𝐴−1) = (𝐴𝐵)−1

i.e.(𝐴𝐵)−1 = (𝐵−1. 𝐴−1)

Theorem:-3] If A is a non-singular matrix , then show that

1. 𝑎𝑑𝑗𝐴 = 𝐴 𝑛−1

2. 𝑎𝑑𝑗 𝐴is a non-singular matrix.

Proof:-

1. We know that,

A(𝑎𝑑𝑗𝐴) = 𝐴 I

Taking determinant on both sides ,

𝐴(𝑎𝑑𝑗𝐴) = 𝐴 𝐼

𝐴 𝑎𝑑𝑗𝐴 = 𝐴 𝑛 − − − 𝐹𝑜𝑟𝑚𝑢𝑙𝑎: −[ 𝐴 𝐼 =

𝐴 𝑛 ]

𝑎𝑑𝑗𝐴 = 𝐴 𝑛

𝐴

𝑎𝑑𝑗𝐴 = 𝐴 𝑛−1---------------------- 1

2. Since, A is a non-singular matrix.

A 0

𝐴 𝑛−1 0

𝑎𝑑𝑗𝐴 0 − − − − − (𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1 )

adj A is non singular matrix.

Page(13)

Principle of Mathematical Induction:-

Let, P(n) denote the given statement such that

1) P(1) is true

2) P(m) is true

P(m+1) is true

Then we say that the statement is true for every natural numbers.

Theorem:-1 ] Show that (𝐴𝑛)−1 = (𝐴−1)𝑛 , ∀ 𝑛 ∈ 𝑁

Proof:-We prove this result by the principle of Mathematical induction.

Step-1 Let n=1

L.H.S = (𝐴𝑛)−1

= (𝐴1)−1

= 𝐴−1

R.H.S = (𝐴−1)𝑛

= (𝐴−1)1

= 𝐴−1

L.H.S = R.H.S

The theorem is true for n=1.

Step-2 Assume that Theorem is true for 𝑛 = 𝑚

We get,

(𝐴𝑚 )−1 = (𝐴−1)𝑚 − − − − − 1

Now, we show that for 𝑛 = 𝑚 + 1

Consider,

(𝐴𝑚+1)−1 = (𝐴𝑚𝐴)−1

= 𝐴−1. (𝐴𝑚 )−1

= 𝐴−1. (𝐴−1)𝑚 − − − − − 𝑏𝑦 𝑒𝑞𝑛 1

Page(14)

= (𝐴−1)𝑚+1

Theorem is true for n = m+1

By the principle of Mathematical induction Theorem is true for every natural

number.

Page(15)

Theorem:-2]If A is a non-singular matrix and k is any scalar then show that

(𝐾𝐴)−1 =1

𝑘𝐴−1

Proof:- Consider

𝐾𝐴 1

𝑘𝐴−1 = 𝑘.

1

𝑘 𝐴. 𝐴−1

= 1 (I)

= I

Also,

1

𝑘𝐴−1 𝐾𝐴 =

1

𝑘. 𝑘 . (𝐴−1. 𝐴)

= 1 (I)

= I

Thus 𝐾𝐴 1

𝐾𝐴−1 =

1

𝐾𝐴−1 𝐾𝐴 = 𝐼

By definition of inverse,

1

𝐾𝐴−1 = 𝐾𝐴 −1

i.e. 𝐾𝐴 −1 =1

𝐾𝐴−1

Theorem:-3]If A is a non-singular matrix and k is any scalar then show that

𝑎𝑑𝑗 𝐾𝐴 = 𝑘𝑛−1(𝑎𝑑𝑗 𝐴)

Proof:- We have

𝑎𝑑𝑗 𝐾𝐴 = 𝐾𝐴 𝐾𝐴 −1

= 𝑘𝑛 𝐴 . 1

𝐾𝐴−1 − − − 𝐾𝐴 = 𝑘𝑛 𝐴 & (𝐾𝐴)−1 =

1

𝐾𝐴−1

= 𝑘𝑛−1 𝐴 𝐴−1

= 𝑘𝑛−1𝑎𝑑𝑗𝐴

Page(16)

Theorem:-4]If A and B are non-singular matrices then show that𝑎𝑑𝑗 𝐴𝐵 =𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴.

Proof:- We know that,

𝐴 𝑎𝑑𝑗𝐴 = 𝑎𝑑𝑗𝐴 𝐴 = 𝐴 𝐼 − − − − − − 1

𝐵 𝑎𝑑𝑗𝐵 = 𝑎𝑑𝑗𝐵 𝐵 = 𝐵 𝐼 − − − − − − 2

𝐴𝐵 𝑎𝑑𝑗𝐴𝐵 = 𝑎𝑑𝑗𝐴𝐵 𝐴𝐵 = 𝐴𝐵 𝐼 − − − − − − 3

Consider,

𝐴𝐵 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴 = 𝐴 𝐵. 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴

= 𝐴 𝐵 𝐼 𝑎𝑑𝑗𝐴 − − − −𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 2

= 𝐴 . 𝑎𝑑𝑗𝐴 ( 𝐵 𝐼)

= 𝐴 𝐼 𝐵 𝐼 − − − − − 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 1

= 𝐴 𝐵 𝐼

= 𝐴𝐵 𝐼

𝐴𝐵 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴 = 𝐴𝐵 𝑎𝑑𝑗𝐴𝐵

𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴 = 𝑎𝑑𝑗𝐴𝐵

𝑎𝑑𝑗𝐴𝐵 = 𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴

Theorem:-5]Inverse of matrix if exist.Show that it is unique

Proof:- Suppose that the matrix A has two inverses say B & C.

By Definition of inverse

𝐴𝐵 = 𝐵𝐴 = 𝐼 − − − − − − 1

𝐴𝐶 = 𝐶𝐴 = 𝐼 − − − − − − 2

Consider,

𝐵 = 𝐵. 𝐼

= 𝐵 𝐴𝐶 − − − − − 𝐹𝑟𝑜𝑚 𝑞𝑛 2

= 𝐵𝐴 𝐶

= 𝐼. 𝐶 − − − − − −𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1

= 𝐶

The matrix ‘A’ has unique inverse.