Adjoint And Inverse Of A Matrix

Here,

Now, we have to find adj A and for that we have to find co-factors:

a₁₁ (co – factor of 2) = (-1)¹⁺¹(9) = (-1)²(9) = 9

a₁₂ (co – factor of 3) = (-1)¹⁺²(5) = (-1)³(5) = -5

a₂₁ (co – factor of 5) = (-1)²⁺¹(3) = (-1)³(3) = -3

a₂₂ (co – factor of 9) = (-1)²⁺²(2) = (-1)⁴(2) = 2

Now, adj A = Transpose of co-factor Matrix

Calculating A (adj A)

= 3I

Calculating (adj A)A

= 3I

Calculating |A|.I

= (2 × 9 – 3 × 5)I

= (18 – 15)I

= 3I

Thus, A(adj A) = (adj A)A = |A|I = 3I

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A and for that we have to find co-factors:

a₁₁ (co – factor of 3) = (-1)¹⁺¹(2) = (-1)²(2) = 2

a₁₂ (co – factor of -5) = (-1)¹⁺²(-1) = (-1)³(-1) = 1

a₂₁ (co – factor of -1) = (-1)²⁺¹(-5) = (-1)³(-5) = 5

a₂₂ (co – factor of 2) = (-1)²⁺²(3) = (-1)⁴(3) = 3

Now, adj A = Transpose of co-factor Matrix

Calculating A (adj A)

= I

Calculating (adj A)A

= I

Calculating |A|.I

= [3 × 2 – (-1) × (-5)]I

= [6 – (5)] I

= (1)I

= I

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

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A and for that we have to find co-factors:

a₁₁ (co – factor of cos α) = (-1)¹⁺¹(cos α)= (-1)²(cos α) = cos α

a₁₂ (co – factor of sin α) = (-1)¹⁺²(sin α) = (-1)³(sin α) = -sin α

a₂₁ (co – factor of sin α) = (-1)²⁺¹(sin α) = (-1)³(sin α) = -sin α

a₂₂ (co – factor of cos α) = (-1)²⁺²(cos α)= (-1)⁴(cos α) = cos α

Now, adj A = Transpose of co-factor Matrix

Calculating A (adj A)

= (cos² α – sin² α) I

Calculating (adj A)A

= (cos² α – sin² α) I

Calculating |A|.I

= [cos α × cos α – (sin α) × (sin α)]I

= [cos² α – sin² α] I

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

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A, and for that, we have to find co-factors:

Calculating A (adj A)

= 12I

Calculating (adj A)A

= 12I

Calculating |A|.I

Expanding along C₁, we get

= [1(3 – 0) – (-1){9 – (-2)} + 2(0 – 1)]I

= [3 + 1(11) + 2(-1)] I

= (3 + 11 – 2 )I

= 12I

Thus, A(adj A) = (adj A)A = |A|I = 12I

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A, and for that, we have to find co-factors:

Calculating A (adj A)

= I

Calculating (adj A)A

= I

Calculating |A|.I

Expanding along C₁, we get

= [3(12 – 10) – (-15){-2 – (-2)} + 5(5 – 6)]I

= [3(2) + 15(0) + 5(-1)] I

= (6 – 5)I

= I

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

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A, and for that, we have to find co-factors:

Calculating A (adj A)

= -2I

Calculating (adj A)A

= -2I

Calculating |A|.I

Expanding along C₁, we get

= [0(2 – 3) – (1){1 – 2} + 3(3 – 4)]I

= [0 – 1(-1) + 3(-1)] I

= (1 – 3)I

= -2I

Thus, A(adj A) = (adj A)A = |A|I = -2I

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A and for that we have to find co-factors:

Calculating A (adj A)

=-70 I

Calculating (adj A)A

= -70 I

Calculating |A|.I

Expanding along C₁, we get

= [9(-2 – 32) – (5){14 – 24} + 6(28 – (-3))]I

= [9(-34) – 5(-10) + 6(31)] I

= (-306 + 50 + 186)I

=-70 I

Thus, A(adj A) = (adj A)A = |A|I = -70 I

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A and for that we have to find co-factors:

Calculating A (adj A)

= -132I

Calculating (adj A)A

= -132I

Calculating |A|.I

Expanding along C₁, we get

= [4(0 – 42) – (1){45 – 21} + 2(30 – 0 )]I

= [4(-42) – 1(24) + 2(30)] I

= (-168 – 24 + 60)I

= -132I

Thus, A(adj A) = (adj A)A = |A|I = -132I

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A, and for that, we have to find co-factors:

[∵cos² α + sin² α = 1]

Calculating A (adj A)

[∵cos² α + sin² α = 1]

= I

Calculating (adj A)A

[∵cos² α + sin² α = 1]

= I

Calculating |A|.I

Expanding along C₁, we get

= [0 – 0 + 1(cos² α – (-sin² α))]I

= [cos² α + sin² α] I

= (1)I [∵cos² α + sin² α = 1]

= I

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

⇒ A(adj A) = (adj A)A = |A|I

Hence Proved

Ans.

Here,

Now, we have to find adj A, and for that, we have to find co-factors:

Thus, adj A = A

Hence Proved

We have,

To show: adj A = 3A’

Firstly, we find the Transpose of A i.e. A’

So,

…(i)

Now, we have to find adj A, and for that, we have to find co-factors:

Now, taking Adj A i.e.

= 3A’ [from eq. (i)]

Hence Proved

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 3) = (-1)¹⁺¹(2) = (-1)²(2) = 2

a₁₂ (co – factor of -5) = (-1)¹⁺²(-1) = (-1)³(-1) = 1

a₂₁ (co – factor of -1) = (-1)²⁺¹(-5) = (-1)³(-5) = 5

a₂₂ (co – factor of 2) = (-1)²⁺²(3) = (-1)⁴(3) = 3

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [3 × 2 – (-1) × (-5)]

= (6 – 5)

= 1

Ans.

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 4) = (-1)¹⁺¹(3) = (-1)²(3) = 3

a₁₂ (co – factor of 1) = (-1)¹⁺²(2) = (-1)³(2) = -2

a₂₁ (co – factor of 2) = (-1)²⁺¹(1) = (-1)³(1) = -1

a₂₂ (co – factor of 3) = (-1)²⁺²(4) = (-1)⁴(4) = 4

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [4 × 3 – 1 × 2]

= (12 – 2)

= 10

Ans.

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 2) = (-1)¹⁺¹(6) = (-1)²(6) = 6

a₁₂ (co – factor of -3) = (-1)¹⁺²(4) = (-1)³(4) = -4

a₂₁ (co – factor of 4) = (-1)²⁺¹(-3) = (-1)³(-3) = 3

a₂₂ (co – factor of 6) = (-1)²⁺²(2) = (-1)⁴(2) = 2

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [2 × 6 – (-3) × 4]

= (12 + 12)

= 24

Ans.

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of a) = (-1)¹⁺¹(d) = (-1)²(d) = d

a₁₂ (co – factor of b) = (-1)¹⁺²(c) = (-1)³(c) = -c

a₂₁ (co – factor of c) = (-1)²⁺¹(b) = (-1)³(b) = -b

a₂₂ (co – factor of d) = (-1)²⁺²(a) = (-1)⁴(a) = a

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [a × d – c × b]

= ad – bc

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 1(1 – (-3)) – 1(-2 – 15) + 2(-2 – (-5))

= (1 + 3) – 1(-17) + 2(-2 + 5)

= 4 + 17 + 2(3)

= 21 + 6

= 27

Now, we have to find adj A, and for that, we have to find co-factors:

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 2(0 – (-6)) – 3(0 – 6) + 2(1 – 0)

= 2(6) – 3(-6) + 2(1)

= 12 + 18 + 2

= 32

Now, we have to find adj A, and for that, we have to find co-factors:

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 2(4 – (-6)) – 2(-6 – (-6)) + 3(-9 – 6)

= 2(4 + 6) – 2(-6 + 6) + 3(-15)

= 2(10) – 2(0) – 45

= 20 – 45

= -25

Now, we have to find adj A and for that we have to find co-factors:

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 0 – 3(0 – 4) – 2(0 – (-4))

= 12 – 2(4)

= 12 – 8

= 4

Now, we have to find adj A and for that we have to find co-factors:

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 2(0 – 1) + 3(-2 – 4) – 1(-1 – 0)

= 2(-1) + 3(-6) – 1(-1)

= -2 – 18 + 1

= -19

Now, we have to find adj A and for that we have to find co-factors:

Ans.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 8(0 – 6) – 10(-24 – 1) + 8(-24 – 0)

= 8(-6) – 10(-25) + 8(-24)

= -48 + 250 – 192

= 250 – 240

= 10

Now, we have to find adj A and for that we have to find co-factors:

Ans.

Here,

To show:

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 2) = (-1)¹⁺¹(-2) = (-1)²(-2) = -2

a₁₂ (co – factor of 3) = (-1)¹⁺²(5) = (-1)³(5) = -5

a₂₁ (co – factor of 5) = (-1)²⁺¹(3) = (-1)³(3) = -3

a₂₂ (co – factor of -2) = (-1)²⁺²(2) = (-1)⁴(2) = 2

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [2 × (-2) – 3 × 5]

= (-4 – 15)

= -19

[Taking (-1) common from the matrix]

Hence Proved

We have,

To show: A^-1 = A²

Firstly, we have to find A^-1 and

Calculating |A|

Expanding |A| along C₁, we get

= 1(0) – 2(0) + 1(0 – (-1))

= 1(1)

= 1

Now, we have to find adj A and for that we have to find co-factors:

…(i)

Calculating A²

= A^-1 [from eq. (i)]

Thus, A² = A^-1

Hence Proved

We have,

To show: A^-1 = A³

Firstly, we have to find A^-1 and

Calculating |A|

Expanding |A| along C₁, we get

= 3(-3 – (-4)) – 2(-3 – (-4)) + 0

= 3(- 3 + 4) – 2(-3 + 4)

= 3(1) – 2(1)

= 3 – 2

= 1

Now, we have to find adj A and for that we have to find co-factors:

…(i)

Calculating A³

= A^-1 [from eq. (i)]

Thus, A³ = A^-1

Hence Proved

We have,

To show: A^-1 = A’

Firstly, we find the Transpose of A, i.e. A.’

Here,

So, …(i)

Now, we have to find A^-1 and

Calculating |A|

Expanding |A| along C₁, we get

= -1

Now, we have to find adj A, and for that, we have to find co-factors:

= A’ [from eq. (i)]

Thus, A^-1 = A’

Hence Proved

Given: D = diag [d₁, d₂, d₃]

It is also given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0

A diagonal matrix D = diag(d₁, d₂, …d_n) is invertible iff all diagonal entries are non – zero, i.e. d_i ≠ 0 for 1 ≤ i ≤ n

If D is invertible then D^-1 = diag(d₁^-1, …d_n^-1)

By the Inverting Diagonal Matrices Theorem, which states that

Here, it is given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0

∴ D is invertible

⇒ D^-1 = diag [d₁^-1, d₂^-1, d₃^-1]

Hence Proved.

Given:

To Verify: (AB)^-1= B^-1A^-1

Firstly, we find the (AB)^-1

Calculating AB

We have to find (AB)^-1 and

Firstly, we find the adj AB and for that we have to find co-factors:

a₁₁ (co – factor of 34) = (-1)¹⁺¹(94) = (-1)²(94) = 94

a₁₂ (co – factor of 39) = (-1)¹⁺²(82) = (-1)³(82) = -82

a₂₁ (co – factor of 82) = (-1)²⁺¹(39) = (-1)³(39) = -39

a₂₂ (co – factor of 94) = (-1)²⁺²(34) = (-1)⁴(34) = 34

Now, adj AB = Transpose of co-factor Matrix

Calculating |AB|

= [34 × 94 – (82) × (39)]

= (3196 – 3198)

= -2

Now, we have to find B^-1A^-1

Calculating B^-1

Here,

We have to find A^-1 and

Firstly, we find the adj B and for that we have to find co-factors:

a₁₁ (co – factor of 6) = (-1)¹⁺¹(9) = (-1)²(9) = 9

a₁₂ (co – factor of 7) = (-1)¹⁺²(8) = (-1)³(8) = -8

a₂₁ (co – factor of 8) = (-1)²⁺¹(7) = (-1)³(7) = -7

a₂₂ (co – factor of 9) = (-1)²⁺²(6) = (-1)⁴(6) = 6

Now, adj B = Transpose of co-factor Matrix

Calculating |B|

= [6 × 9 – 7 × 8]

= (54 – 56)

= -2

Calculating A^-1

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 3) = (-1)¹⁺¹(5) = (-1)²(5) = 5

a₁₂ (co – factor of 2) = (-1)¹⁺²(7) = (-1)³(7) = -7

a₂₁ (co – factor of 7) = (-1)²⁺¹(2) = (-1)³(2) = -2

a₂₂ (co – factor of 5) = (-1)²⁺²(3) = (-1)⁴(3) = 3

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [3 × 5 – 2 × 7]

= (15 – 14)

= 1

Calculating B^-1A^-1

Here,

So,

So, we get

and

∴ (AB)^-1 = B^-1A^-1

Hence verified

Given:

To Verify: (AB)^-1= B^-1A^-1

Firstly, we find the (AB)^-1

Calculating AB

We have to find (AB)^-1 and

Firstly, we find the adj AB and for that we have to find co-factors:

a₁₁ (co – factor of -41) = (-1)¹⁺¹(26) = (-1)²(26) = 26

a₁₂ (co – factor of 31) = (-1)¹⁺²(-34) = (-1)³(-34) = 34

a₂₁ (co – factor of -34) = (-1)²⁺¹(31) = (-1)³(31) = -31

a₂₂ (co – factor of 26) = (-1)²⁺²(-41) = (-1)⁴(-41) = -41

Now, adj AB = Transpose of co-factor Matrix

Calculating |AB|

= [-41 × 26 – (-34) × (31)]

= (-1066 + 1054)

= -12

Now, we have to find B^-1A^-1

Calculating B^-1

Here,

We have to find A^-1 and

Firstly, we find the adj B and for that we have to find co-factors:

a₁₁ (co – factor of -4) = (-1)¹⁺¹(-4) = (-1)²(-4) = -4

a₁₂ (co – factor of 3) = (-1)¹⁺²(5) = (-1)³(5) = -5

a₂₁ (co – factor of 5) = (-1)²⁺¹(3) = (-1)³(3) = -3

a₂₂ (co – factor of -4) = (-1)²⁺²(-4) = (-1)⁴(-4) = -4

Now, adj B = Transpose of co-factor Matrix

Calculating |B|

= [(-4) × (-4) – 3 × 5]

= (16 – 15)

= 1

Calculating A^-1

Here,

We have to find A^-1 and

Firstly, we find the adj A and for that we have to find co-factors:

a₁₁ (co – factor of 9) = (-1)¹⁺¹(-2) = (-1)²(-2) = -2

a₁₂ (co – factor of -1) = (-1)¹⁺²(6) = (-1)³(6) = -6

a₂₁ (co – factor of 6) = (-1)²⁺¹(-1) = (-1)³(-1) = 1

a₂₂ (co – factor of -2) = (-1)²⁺²(9) = (-1)⁴(9) = 9

Now, adj A = Transpose of co-factor Matrix

Calculating |A|

= [9 × (-2) – (-1) × 6]

= (-18 + 6)

= -12

Calculating B^-1A^-1

Here,

So,

So, we get

and

∴ (AB)^-1 = B^-1A^-1

Hence verified

We have,

To find: (AB)^-1

We know that,

(AB)^-1 = B^-1A^-1

and here, B^-1 is given but we have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 1(8 – 6) – 0 + 3(-3 – 4)

= 1(2) + 3(-7)

= 2 – 21

= -19

Now, we have to find adj A and for that we have to find co-factors:

Now, we have

So,

Ans.

Let

To find: A^-1, B^-1 and C^-1

Calculating A^-1

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 1(1 – 0)

= 1

Now, we have to find adj A and for that we have to find co-factors:

…(i)

Calculating B^-1

We have,

We have to find B^-1 and

Firstly, we find |A|

Expanding |B| along C₁, we get

= 1(1 – 0) – q(0)

= 1

Now, we have to find adj B and for that we have to find co-factors:

…(ii)

Calculating C^-1

Here,

⇒ C = AB

⇒ C^-1 = (AB)^-1

We know that,

(AB)^-1 = B^-1A^-1

Substitute the values, we get

Ans. , and .

Given:

To verify: A² – 4A – I = 0

Firstly, we find the A²

Taking LHS of the given equation .i.e.

A² – 4A – I

= 0

= RHS

∴ LHS = RHS

Hence verified

Now, we have to find A^-1

Finding A^-1 using given equation

A² – 4A – I = O

Post multiplying by A^-1 both sides, we get

(A² – 4A – I)A^-1 = OA^-1

⇒ A².A^-1 – 4A.A^-1 – I.A^-1 = O [OA^-1 = O]

⇒ A.(AA^-1) – 4I – A^-1 = O [AA^-1 = I]

⇒ A(I) – 4I – A^-1 = O

⇒ A – 4I – A^-1 = O

⇒ A – 4I – O = A^-1

⇒ A – 4I = A^-1

Ans.

Given:

To show: Matrix A satisfies the equation x² + 4x – 42 = 0

If Matrix A satisfies the given equation then

A² + 4A – 42 = 0

Firstly, we find the A²

Taking LHS of the given equation .i.e.

A² + 4A – 42

= O

= RHS

∴ LHS = RHS

Hence matrix A satisfies the given equation x² + 4x – 42 = 0

Now, we have to find A^-1

Finding A^-1 using given equation

A² + 4A – 42 = O

Post multiplying by A^-1 both sides, we get

(A² + 4A – 42)A^-1 = OA^-1

⇒ A².A^-1 + 4A.A^-1 – 42.A^-1 = O [OA^-1 = O]

⇒ A.(AA^-1) + 4I – 42A^-1 = O [AA^-1 = I]

⇒ A(I) + 4I – 42A^-1 = O

⇒ A + 4I – 42A^-1 = O

⇒ A + 4I – O = 42A^-1

Ans. .

Given:

To verify: A² + 3A + 4I = 0

Firstly, we find the A²

Taking LHS of the given equation .i.e.

A² + 3A + 4I

= O

= RHS

∴ LHS = RHS

Hence verified

Now, we have to find A^-1

Finding A^-1 using given equation

A² + 3A + 4I = O

Post multiplying by A^-1 both sides, we get

(A² + 3A + 4I)A^-1 = OA^-1

⇒ A².A^-1 + 3A.A^-1 + 4I.A^-1 = O [OA^-1 = O]

⇒ A.(AA^-1) + 3I + 4A^-1 = O [AA^-1 = I]

⇒ A(I) + 3I + 4A^-1 = O

⇒ A + 3I + 4A^-1 = O

⇒ 4A^-1 = – A – 3I + O

Ans.

Given:

To find: value of x and y

Given equation: A² + xI = yA

Firstly, we find the A²

Putting the values in given equation

A² + xI = yA

On Comparing, we get

16 + x = 3y …(i)

y = 8 …(ii)

56 = 7y …(iii)

32 + x = 5y …(iv)

Putting the value of y = 8 in eq. (i), we get

16 + x = 3(8)

⇒ 16 + x = 24

⇒ x = 8

Hence, the value of x = 8 and y = 8

So, the given equation become A² + 8I = 8A

Now, we have to find A^-1

Finding A^-1 using given equation

A² + 8I = 8A

Post multiplying by A^-1 both sides, we get

(A² + 8I)A^-1 = 8AA^-1

⇒ A².A^-1 + 8I.A^-1 = 8AA^-1

⇒ A.(AA^-1) + 8A^-1 = 8I [AA^-1 = I]

⇒ A(I) + 8A^-1 = 8I

⇒ A + 8A^-1 = 8I

⇒ 8A^-1 = – A + 8I

Ans. 𝒳 = 8, 𝒴 =8 and A^-1 = . .

Given:

To find: value of λ

Given equation: A² = λA – 2I

Firstly, we find the A²

Putting the values in given equation

A² = λA – 2I

On Comparing, we get

3λ – 2 = 1 …(i)

-2λ = -2 …(ii)

4λ = 4 …(iii)

-2λ – 2 = -4 …(iv)

Solving eq. (iii), we get

4λ = 4

⇒ λ = 1

Hence, the value of λ = 1

So, the given equation become A² = A – 2I

Now, we have to find A^-1

Finding A^-1 using given equation

A² = A – 2I

Post multiplying by A^-1 both sides, we get

(A²)A^-1 = (A – 2I) A^-1

⇒ A².A^-1 = AA^-1 – 2IA^-1

⇒ A.(AA^-1) = I – 2A^-1 [AA^-1 = I]

⇒ A(I) = I – 2A^-1

⇒ A + 2A^-1 = I

⇒ 2A^-1 = – A + I

Ans. λ= 1, A^-1 = .

Given:

We have to show that matrix A satisfies the equation A³ – A² – 3A – I = O

Firstly, we find the A²

Now, we have to calculate A³

Taking LHS of the given equation .i.e.

A³ – A² – 3A – I

Putting the values, we get

= O

= RHS

∴ LHS = RHS

Hence, the given matrix A satisfies the equation A³ – A² – 3A – I

Now, we have to find A^-1

Finding A^-1 using given equation

A³ – A² – 3A – I

Post multiplying by A^-1 both sides, we get

(A³ – A² – 3A – I)A^-1 = OA^-1

⇒ A³.A^-1 – A².A^-1 – 3A.A^-1 – I.A^-1 = O [OA^-1 = O]

⇒ A².(AA^-1) – A.(AA^-1) – 3I – A^-1 = O

⇒ A²(I) – A(I) – 3I – A^-1 = O [AA^-1 = I]

⇒ A² – A – 3I – A^-1 = O

⇒ O + A^-1 = A² – A – 3I

⇒ A^-1 = A² – A – 3I

Ans. .

(i) To Prove: adj I = I

We know that, I means the Identity matrix

Let I is a 2 × 2 matrix

Now, we have to find adj I and for that we have to find co-factors:

a₁₁ (co – factor of 1) = (-1)¹⁺¹(1) = (-1)²(1) = 1

a₁₂ (co – factor of 0) = (-1)¹⁺²(0) = (-1)³(0) = 0

a₂₁ (co – factor of 0) = (-1)²⁺¹(0) = (-1)³(0) = 0

a₂₂ (co – factor of 1) = (-1)²⁺²(1) = (-1)⁴(1) = 1

Now, adj I = Transpose of co-factor Matrix

Thus, adj I = I

Hence Proved

(ii) To Prove: adj O = O

We know that, O means Zero matrix where all the elements of matrix are 0

Let O is a 2 × 2 matrix

Calculating adj O

Now, we have to find adj O and for that we have to find co-factors:

a₁₁ (co – factor of 0) = (-1)¹⁺¹(0) = 0

a₁₂ (co – factor of 0) = (-1)¹⁺²(0) = 0

a₂₁ (co – factor of 0) = (-1)²⁺¹(0) = 0

a₂₂ (co – factor of 0) = (-1)²⁺²(0) = 0

Now, adj O = Transpose of co-factor Matrix

Thus, adj O = O

Hence Proved

(iii) To Prove: I^-1 = I

We know that,

From the part(i), we get adj I

So, we have to find |I|

Calculating |I|

= [1 × 1 – 0]

= 1

Thus, I^-1 = I

Hence Proved