Adjoint And Inverse Of A Matrix

A =

Cofactors of A are

C₁₁ = 4

C₁₂ = – 2

C₂₁ = – 5

C₂₂ = – 3

Since, adj A =

(adj A) =

=

Now, (adj A)A =

(adj A)A =

And, |A|.I =

Also, A(adj A) =

A(adj A) =

Hence, (adj A)A = |A|.I = A.(adj A)

A =

Cofactors of A are

C₁₁ = d

C₁₂ = – c

C₂₁ = – b

C₂₂ = a

Since, adj A =

(adj A) =

=

Now, (adj A)A =

(adj A)A =

And, |A|.I =

Also, A(adj A) =

Hence, (adj A)A = |A|.I = A.(adj A)

A =

Cofactors of A are

C₁₁ =

C₁₂ =

C₂₁ =

C₂₂ =

Since, adj A =

(adj A) =

=

Now, (adj A)A =

(adj A)A =

And, |A|.I =

=

=

=

Also, A(adj A) =

=

Hence, (adj A)A = |A|.I = A.(adj A)

A =

Cofactors of A are

C₁₁ = 1

C₁₂ =

C₂₁ =

C₂₂ = 1

Since, adj A =

(adj A) =

=

Now, (adj A)A =

=

(adj A)A =

And, |A|.I =

=

Also, A(adj A) =

=

=

Hence, (adj A)A = |A|.I = A.(adj A)

A =

Cofactors of A are:

C₁₁ = – 3 C₂₁ = 2 C₃₁ = 2

C₁₂ = 2 C₂₂ = – 3 C₂₃ = 2

C₁₃ = 2 C₂₃ = 2 C₃₃ = – 3

adj A =

=

Now, (adj A).A =

=

=

Also, |A|.I = = ( – 3 + 4 + 4)

=

Then, A.(adj A) =

=

=

Since, (adj A).A = |A|.I = A(adj A)

A =

Cofactors of A

C₁₁ = 2 C₂₁ = 3 C₃₁ = – 13

C₁₂ = – 3 C₂₂ = 6 C₃₂ = 9

C₁₃ = 5 C₂₃ = – 3 C₃₃ = – 1

adj A =

=

adj A =

Now, (adj A).A =

=

=

Also, |A|.I = = [1(3 – 1) – 2(2 + 1) + 5(2 + 3)]

= (21)

=

Then, A.(adj A) =

=

=

Hence, (adj A).A = |A|.I = A(adj A)

A =

Cofactors of A

C₁₁ = – 22 C₂₁ = 11 C₃₁ = – 11

C₁₂ = 4 C₂₂ = – 2 C₃₂ = 2

C₁₃ = 16 C₂₃ = – 8 C₃₃ = 8

adj A =

=

adj A =

Now, (adj A).A =

=

=

Also, |A|.I =

= [2( – 2 – 20) + 1( – 4 – 0) + 3(16 – 0)]

= ( – 44 – 4 + 48)

=

Then, A.(adj A) =

=

=

Hence, (adj A).A = |A|.I = A(adj A)

A =

Cofactors of A

C₁₁ = 3 C₂₁ = – 1 C₃₁ = – 1

C₁₂ = – 15 C₂₂ = 7 C₃₂ = – 5

C₁₃ = 4 C₂₃ = – 2 C₃₃ = 2

adj A =

=

adj A =

Now, (adj A).A =

=

=

Also, |A|.I =

= [2(3 – 0) + 0(15 – 0) – 1(5 – 1)]

= (6 – 4)

=

Then, A.(adj A) =

=

=

Hence, (adj A).A = |A|.I = A(adj A)

A =

Cofactors of A

C₁₁ = 30 C₂₁ = 12 C₃₁ = – 3

C₁₂ = – 20 C₂₂ = – 8 C₃₂ = 2

C₁₃ = – 50 C₂₃ = – 2 0 C₃₃ = 5

adj A =

=

So, adj(A) =

Now, A.(adj A) =

=

=

Hence, A(adj A) = 0

A =

Cofactors of A

C₁₁ = – 4 C₂₁ = – 3 C₃₁ = – 3

C₁₂ = 1 C₂₂ = 0 C₃₂ = 1

C₁₃ = 4 C₂₃ = 4 C₃₃ = 3

adj A =

=

So, adj A =

Hence, adj A = A

A =

Cofactors of A are:

C₁₁ = – 3 C₂₁ = 6 C₃₁ = 6

C₁₂ = – 6 C₂₂ = 3 C₃₂ = – 6

C₁₃ = – 6 C₂₃ = – 6 C₃₃ = 3

adj A =

=

So, adj A =

Now, 3A^T = 3

Hence, adj A = 3.A^T

A =

Cofactors of A are:

C₁₁ = 9 C₂₁ = 19 C₃₁ = – 4

C₁₂ = 4 C₂₂ = 14 C₃₂ = 1

C₁₃ = 8 C₂₃ = 3 C₃₃ = 2

adj A =

=

So, adj A =

Now, A. adj A =

=

=

Hence, A. adj A = 25.I₃

Now, |A| = cos (cos ) + sin (sin )

= 1

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ =

C₁₂ =

C₂₁ =

C₂₂ =

Since, adj A =

(adj A) =

=

Now, A ^{– 1} = .adj A

A ^{– 1} = .

A ^{– 1} =

Now, |A| = – 10

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ =

C₁₂ = – 1

C₂₁ = – 1

C₂₂ = 0

Since, adj A =

(adj A) =

=

Now, A ^{– 1} = .adj A

A ^{– 1} = .

A ^{– 1} =

Now, |A| = bc =

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ =

C₁₂ = – c

C₂₁ = – b

C₂₂ = a

Since, adj A =

(adj A) =

=

Now, A ^{– 1} = .adj A

A ^{– 1} = .

A ^{– 1} =

Now, |A| = 2 + 15 = 17

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ = 1

C₁₂ = 3

C₂₁ = – 5

C₂₂ = 2

Since, adj A =

(adj A) =

=

Now, A ^{– 1} = .adj A

A ^{– 1} = .

A ^{– 1} = .

|A| =

= 1(6 – 1) – 2(4 – 3) + 3(2 – 9)

= 5 – 2 – 21

= – 18

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 5 C₂₁ = – 1 C₃₁ = – 7

C₁₂ = – 1 C₂₂ = – 7 C₃₂ = 5

C₁₃ = – 7 C₂₃ = 5 C₃₃ = – 1

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 1(1 + 3) – 2( – 1 + 2) + 5(3 + 2)

= 4 – 2 + 25

= 27

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 4 C₂₁ = 17 C₃₁ = 3

C₁₂ = – 1 C₂₂ = – 11 C₃₂ = 6

C₁₃ = 5 C₂₃ = 1 C₃₃ = – 3

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 2(4 – 1) + 1( – 2 + 1) + 1(1 – 2)

= 6 – 2

= – 4

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 3 C₂₁ = 1 C₃₁ = – 1

C₁₂ = + 1 C₂₂ = 3 C₃₂ = 1

C₁₃ = – 1 C₂₃ = 1 C₃₃ = 3

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 2(3 – 0) – 0 – 1(5)

= 6 – 5

= 1

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 3 C₂₁ = – 1 C₃₁ = 1

C₁₂ = – 15 C₂₂ = 6 C₃₂ = – 5

C₁₃ = – 5 C₂₃ = – 2 C₃₃ = 2

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 0 – 1(16 – 12) – 1( – 12 + 9)

= – 4 + 3

= – 1

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 0 C₂₁ = – 1 C₃₁ = 1

C₁₂ = – 4 C₂₂ = 3 C₃₂ = – 4

C₁₃ = – 3 C₂₃ = 3 C₃₃ = – 4

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 0 – 0 – 1( – 12 + 8)

= 4

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = – 8 C₂₁ = 4 C₃₁ = 4

C₁₂ = 11 C₂₂ = – 2 C₃₂ = – 3

C₁₃ = – 4 C₂₃ = 0 C₃₃ = 0

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| = – 0 + 0

= ()

= – 1

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = – 1 C₂₁ = 0 C₃₁ = 0

C₁₂ = 0 C₂₂ = C₃₂ =

C₁₃ = 0 C₂₃ = C₃₃ =

adj A =

=

So, adj A =

Now, A ^{– 1} = .adj A

So, A ^{– 1} = .

Hence, A ^{– 1} =

|A| =

= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3

= 1

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 7 C₂₁ = – 3 C₃₁ = – 3

C₁₂ = – 1 C₂₂ = – 1 C₃₂ = 0

C₁₃ = – 1 C₂₃ = 0 C₃₃ = 1

adj A =

=

So, adj A =

Now, A ^{– 1} =

Also, A ^{– 1}.A =

=

=

Hence, A ^{– 1}.A = I

|A| =

= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

= 2 – 9 + 9

= 2

Hence, A ^{– 1} exists

Cofactors of A are:

C₁₁ = 1 C₂₁ = 1 C₃₁ = – 1

C₁₂ = – 3 C₂₂ = 1 C₃₂ = 1

C₁₃ = 9 C₂₃ = – 5 C₃₃ = – 1

adj A =

=

So, adj A =

Now, A ^{– 1} =

Also, A ^{– 1}.A =

=

=

Hence, A ^{– 1}.A = I

A = , |A| = 1

Then, adj A =

A ^{– 1} =

B = , |B| = – 10

Then, adj B =

B ^{– 1} =

Also, A.B =

AB =

|AB| = 936 – 946 = – 10

Adj(AB) =

(AB) ^{– 1} =

Now B ^{– 1}A ^{– 1} =

=

=

Hence, (AB) ^{– 1 =} B ^{– 1} A ^{– 1}

|A| = 1

Adj A =

A ^{– 1} =

B =

|B| = – 1

B ^{– 1} =

Also, AB =

=

|AB| = 407 – 406 = 1

And, adj(AB) =

(AB) ^{– 1} =

=

Now, B ^{– 1}A ^{– 1 =}

=

Hence, (AB) ^{– 1 =} B ^{– 1}A ^{– 1}

A =

|A| = 15 – 14 = 1

adj A =

A ^{– 1} =

B =

|B| = 54 – 56 = – 2 adj B =

B ^{– 1} =

Now, (AB) ^{– 1 =} B ^{– 1}A ^{– 1}

=

=

=

(AB) ^{– 1 =}

A =

|A| = 14 – 12 = 2 adj A =

A ^{– 1 =}

To Show: 2A ^{– 1} = 9I – A

L.H.S 2A ^{– 1} = 2.

R.H.S 9I – A =

=

Hence, 2A ^{– 1} = 9I – A

A =

|A| = 4 – 10 = – 6 adj A =

A ^{– 1 =}

To Show: A – 3I = 2 (I + 3A ^{– 1})

LHS A – 3I =

=

R.H.S 2 (I + 3A ^{– 1}) = 2I + 6A ^{– 1 =}

=

=

Hence, A – 3I = 2 (I + 3A ^{– 1})

A =

Now, |A| = bc =

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ = C₁₂ = – c

C₂₁ = – b C₂₂ = a

Since, adj A =

(adj A) =

=

Now, A ^{– 1} = .adj A

A ^{– 1} = .

A ^{– 1} =

To show. aA ^{– 1} = (a² + bc + 1) I – aA.

LHS aA ^{– 1} = a

=

RHS (a² + bc + 1) I – aA =

=

Hence, LHS = RHS

A = and B ^{– 1} =

Here , (AB) ^{– 1 =} B ^{– 1} A ^{– 1}

|A| = – 5 + 4 = – 1

Cofactors of A are:

C₁₁ = – 1 C₂₁ = 8 C₃₁ = – 12

C₁₂ = 0 C₂₂ = 1 C₃₂ = – 2

C₁₃ = 1 C₂₃ = – 10 C₃₃ = 15

adj A =

=

So, adj A =

Now, A ^{– 1} =

(AB) ^{– 1 =} B ^{– 1} A ^{– 1}

=

Hence, =

F(α) =

| F(α)| = = 1

Cofactors of A are:

C₁₁ = cos α C₂₁ = sin α C₃₁ = 0

C₁₂ = – sin α C₂₂ = cos α C₃₂ = 0

C₁₃ = 0 C₂₃ = – 10 C₃₃ = 1

adj F(α) =

=

So, adj F(α) = …… (i)

Now, [F(α)] ^{– 1} =

And, F( – α) = …… (ii)

=

Hence, [F (α)] ^{– 1} = F( – α)

|G(β)| = = 1

Cofactors of A are:

C₁₁ = cos β C₂₁ = sin α C₃₁ = sin β

C₁₂ = 0 C₂₂ = 1 C₃₂ = 0

C₁₃ = sin β C₂₃ = 0 C₃₃ = cos β

Adj G(β) =

=

So, adj G(β) = …… (i)

Now, [G(β)] ^{– 1} =

And, G( – β) =

=

Hence, [G (β)] ^{– 1} = G( – β)

We have to show that

[F(α)G(β)] ^{– 1} = G( – β) F( – α)

We have already shown that

[G (β)] ^{– 1} = G( – β)

[F (α)] ^{– 1} = F( – α)

And LHS = [F(α)G(β)] ^{– 1}

= [G (β)] ^{– 1} [F (α)] ^{– 1}

= G( – β) F( – α)

Hence = RHS

A² =

=

4A =

I =

Now, A² – 4 A + I =

=

Hence, =

Now, A² – 4 A + I = 0

A.A – 4A = – I

Multiply by A ^{– 1} both sides

A.A(A ^{– 1}) – 4A A ^{– 1} = – IA ^{– 1}

AI – 4I = – A ^{– 1}

A ^{– 1 =} 4I – A =

A ^{– 1} =

A =

A² = =

=

4A = 4

42I = 42

Now,

A² + 4A – 42I =

=

Hence, =

Now, A² + 4A – 42I = 0

= A ^{– 1}. A . A + 4 A ^{– 1}.A – 42 A ^{– 1}.I = 0

= IA + 4I – 42A ^{– 1} = 0

= 42A ^{– 1 =} A + 4I

= A ^{– 1 =}

=

A ^{– 1 =}

A =

A² =

=

Now, A² – 5A + 7I =

=

=

So, A² – 5A + 7I = 0

Multiply by A ^{– 1} both sides

= A.A. A ^{– 1} – 5A. A ^{– 1} + 7I. A ^{– 1} = 0

= A – 5I + 7 A ^{– 1} = 0

= A ^{– 1} =

= A ^{– 1} =

= A ^{– 1} =

A =

A² =

=

Now, A² – xA + yI =

= 22 – 4x + y = 0 or 4x – y = 22

= 18 – 2x = 0 or X = 9

= Y = 14

So, A² – 5A + 7I = 0

Multiply by A ^{– 1} both sides

= A.A. A ^{– 1} – 9A. A ^{– 1} + 14I. A ^{– 1} = 0

= A – 9I + 14 A ^{– 1} = 0

= A ^{– 1} =

= A ^{– 1} =

= A ^{– 1} =

A =

A² =

=

Now, A² = λA – 2I

= λA = A² + 2I

=

= λ

=

= = 3 or λ = 1

So, A² = A – 2I

Multiply by A ^{– 1} both sides

= A.A. A ^{– 1} = A. A ^{– 1} – 2I. A ^{– 1} = 0

= 2A ^{– 1} =

Hence, A ^{– 1} =

A =

A² =

=

Now, A² – 3A – 7 = 0

=

=

=

So, A² – 3A – 7I = 0

Multiply by A ^{– 1} both sides

= A.A. A ^{– 1} – 3A. A ^{– 1} – 7I. A ^{– 1} = 0

= A – 3I – 7A ^{– 1} = 0

= 7A ^{– 1} = A – 3I

= A ^{– 1} =

Hence, A ^{– 1} =

A =

We have A² – 12A + I = 0

A^{2 =}

=

Now, A² – 12A + 1 = 0

=

=

Hence, =

Also, A² – 12A + 1 = 0

= A – 12I + A ^{– 1} = 0

= A ^{– 1} = 12I – A

= 12

=

Hence, A ^{– 1} =

A =

A³ = A².A

A² =

=

A².A =

=

=

Now, A³ – 6A² + 5A + 11I

+ 11

=

=

=

Thus, A³ – 6A² + 5A + 11I

Now, (AAA) A ^{– 1}. – 6(A.A) A ^{– 1} + 5.A A ^{– 1} + 11I.A ^{– 1} = 0

AA(A ^{– 1}A) – 6A(A ^{– 1}A) + 5(A ^{– 1}A) = – 1(A ^{– 1}I)

A² – 6A + 5I = 11 A ^{– 1}

= A ^{– 1} =

Now,

A² – 6A + 5I

=

=

=

=

Hence, A ^{– 1} =

A =

A³ = A².A

A² =

=

A².A =

=

=

Now, A³ – A² – 3A – I

=

=

=

Thus, A³ – A² – 3A – I

Now, (AAA) A ^{– 1}. – (A.A) A ^{– 1} – 3.A A ^{– 1} – I.A ^{– 1} = 0

AA(A ^{– 1}A) – A(A ^{– 1}A) – 3(A ^{– 1}A) = – 1(A ^{– 1}I)

A² – A – 3A – I = 0

= A ^{– 1} =

Now,

=

=

=

Hence, A ^{– 1} =

A =

A³ = A².A

A² =

=

A².A =

=

=

Now, A³ – 6A² + 9A – 4I

=

=

Thus, A³ – 6A² + 9A – 4I

Now, (AAA) A ^{– 1}. – 6(A.A) A ^{– 1} + 9.A A ^{– 1} – 4I.A ^{– 1} = 0

A² – 6A + 9I = 4A ^{– 1}

= A ^{– 1} =

Now,

=

=

=

Hence, A ^{– 1} =

A = A^T =

|A| = [ – 8(16 + 56) – 1(16 – 7) + 4( – 32 – 4)]

= – 81

Cofactors of A are:

C₁₁ = 72 C₂₁ = – 36 C₃₁ = – 9

C₁₂ = – 9 C₂₂ = – 36 C₃₂ = 72

C₁₃ = – 36 C₂₃ = – 63 C₃₃ = – 36

adj A =

=

So, adj A =

Now, A ^{– 1} =

Hence, A ^{– 1} = = A^T

A =

|A| = 3 + 6 – 8 = 1

Cofactors of A are:

C₁₁ = 1 C₂₁ = – 1 C₃₁ = 0

C₁₂ = – 2 C₂₂ = 3 C₃₂ = – 4

C₁₃ = – 2 C₂₃ = 3 C₃₃ = – 3

adj A =

=

So, adj A =

Now, A ^{– 1} =

Also, A² =

=

=

A³ = A².A =

=

Hence, A ^{– 1} = A³

|A| =

|A| = – 1(0 – 1) – 2(0) + 0

= 1 – 0 + 0

|A| = 1

A =

A² = A.A =

=

=

Cofactors of A are:

C₁₁ = – 1 C₂₁ = 0 C₃₁ = 2

C₁₂ = 0 C₂₂ = 0 C₃₂ = 1

C₁₃ = – 1 C₂₃ = 1 C₃₃ = 1

adj A =

=

So, adj A =

Now, A ^{– 1} =

Hence, A ^{– 1} = = A²

Let A = B =

So, AX = B

Or, X = A ^{– 1}B

|A| = 1

Cofactors of A are

C₁₁ = 1 C₁₂ = – 1

C₂₁ = – 4 C₂₂ = 5

Since, adj A =

(adj A) =

=

Now, A ^{– 1} =

A ^{– 1} =

So, X =

Hence, X =

Let A = B =

So, AX = B

Or, X = A ^{– 1}B

|A| = – 7

Cofactors of A are

C₁₁ = – 2 C₁₂ = 1

C₂₁ = – 3 C₂₂ = 5

Since, adj A =

(adj A) =

=

Now, A ^{– 1} =

A ^{– 1} =

So, X =

Hence, X =

X =

Let A = B = C =

Then The given equations becomes as

AXB = C

= X = A ^{– 1}CB ^{– 1}

|A| = 35 – 14 = 21

|B| = – 1 + 2 = 1

A ^{– 1} =

B ^{– 1} =

= X = A ^{– 1}CB ^{– 1} =

=

=

=

Hence, X =

Let A = B = C =

Then The given equations becomes as

AXB = I

= X = A ^{– 1}B ^{– 1}

|A| = 6 – 5 = 1

|B| = 10 – 9 = 1

A ^{– 1} =

B ^{– 1} =

= X = A ^{– 1}B ^{– 1} =

=

=

Hence, X =

A =

A² =

=

=

A² – 4A + 5I = 0

=

=

=

Also, A² – 4A – 5I = 0

Now, 6(A.A) A ^{– 1} – 4.A A ^– 1 – 5I.A ^{– 1} = 0

= A – 4I – 5A ^{– 1} = 0

= A ^{– 1} =

=

=

Hence, A ^{– 1} =

|A adj A| = |A|ⁿ

LHS |A adj A|

|A|.|adj A|

|A|.|A|^{n – 1}

|A|^{n – 1 + 1}

|A|ⁿ = RHS

Hence, LHS = RHS

A ^{– 1} = B =

|B| = 1(3 – 0) – 2( – 1 – 0) – 2(2 – 0)

= 3 + 2 – 4

|B| = 1

Now, B ^{– 1} =

Cofactors of B are:

C₁₁ = – 3 C₂₁ = 2 C₃₁ = 6

C₁₂ = 1 C₂₂ = 1 C₃₂ = 2

C₁₃ = 2 C₂₃ = 2 C₃₃ = 5

adj B =

=

So, adj B =

Now, B ^{– 1} =

(AB) ^{– 1 =} B ^{– 1} A ^{– 1}

=

Hence, =

A =

Let B = A^T =

|B| =

= ( – 1 – 8) – 0 – 2( – 8 + 3) = – 9 + 10 = 1

Cofactors of B are:

C₁₁ = – 9 C₂₁ = 8 C₃₁ = – 5

C₁₂ = – 8 C₂₂ = 7 C₃₂ = – 4

C₁₃ = – 2 C₂₃ = 2 C₃₃ = – 1

adj B =

=

So, adj B =

Now, B ^{– 1} =

Hence, (A^T) ^{– 1}

A =

|A| =

= – 1(1 – 4) + 2(2 + 4) – 2( – 4 – 2)

= 3 + 12 + 12

|A| = 27

Cofactors of A

C₁₁ = – 3 C₂₁ = – 6 C₃₁ = 6

C₁₂ = – 6 C₂₂ = 3 C₃₂ = – 6

C₁₃ = – 6 C₂₃ = – 6 C₃₃ = 3

adj A =

=

So, adj A =

A(adj A) =

=

A(adj A) = 27

Hence, A(adj A) = |A|I

A = |A| = 0 – 1(0 – 1) + 1(1 – 0) = 0 + 1 + 1 = 2

Cofactors of A are:

C₁₁ = – 1 C₂₁ = 1 C₃₁ = 1

C₁₂ = 1 C₂₂ = – 1 C₃₂ = 1

C₁₃ = 1 C₂₃ = 1 C₃₃ = – 1

adj A =

=

So, adj A =

Now, A ^{– 1} =

A² – 3I =

=

=

=

Hence, A ^{– 1} = (A² – 3I)

Given:- 2 x 2 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₂A

Where I₂ is 2 x 2 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 2 x 2 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₂A

Where I₂ is 2 x 2 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

⇒

Applying

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 2 x 2 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₂A

Where I₂ is 2 x 2 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 2 x 2 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₂A

Where I₂ is 2 x 2 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 2 x 2 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₂A

Where I₂ is 2 x 2 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying and

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Hence, it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying and

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying and

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying and

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A

Given:- 3 x 3 square matrix

Tip:- Algorithm to find Inverse of a square matrix of ‘n’ order by elementary row transformation

(i) Obtain the square matrix, say A

(ii) Write A = I_nA

(iii) Perform a sequence of elementary row operation successively on A on the LHS and pre-factor I_n on the RHS till we obtain the result

I_n = BA

(iv) Write A^-1 = B

Now,

We have,

A = I₃A

Where I₃ is 3 x 3 elementary matrix

⇒

Applying

⇒

Applying

⇒

Applying

⇒

Applying and

⇒

Applying

⇒

Applying

⇒

Hence , it is of the form

I = BA

So, as we know that

I = A^-1A

Therefore

A^-1 = B

⇒ inverse of A