NCERT Solutions Class 12 – Mathematics Part I – Chapter 4 Determinants – Exercise 4.4

Find adjoint of each of the matrices in Exercises 1 and 2.

Question 1. 

Solution: 

A = 

A₁₁ = 4; A₁₂ = -3; A₂₁ = -2; A₂₂ = 1

adj A = 

adj A = 

Question 2. 

Solution: 

A = 

A₁₁ = 

A₁₁ = 3 – 0 = 3

A₁₂ = 

A₁₂ = -(2 + 10) = -12

A₁₃ = 

A₁₃ = 0 + 6 = 6

A₂₁ = 

A₂₁ = -(-1 – 0) = 1

A₂₂ = 

A₂₂ = 1 + 4 = 5

A₂₃ = 

A₂₃ = -(0 – 2) = 2

A₃₁ = 

A₃₁ = -5 – 6 = -11

A₃₂ = 

A₃₂ = -(5 – 4) = -1

A₃₃ = 

A₃₃ = 3 + 2 = 5

adj A = 

adj A = 

Verify A(adj A) = (adj A)A = |A| I in exercises 3 and 4.

Question 3. 

Solution: 

|A| = -12 -(-12) 

|A| = -12 + 12 = 0

so, |A|*I = 0 * 

|A|*I = ……… (1)

Now, for adjoint of A

A₁₁ = -6

A₁₂ = 4

A₂₁ = -3

A₂₂ = 2

adj A = 

adj A = 

Now, A(adj A) = 

A(adj A) = [Tex]\begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]

A(adj A) = …………(2)

Now, (adj A)A = 

(adj A)A = 

(adj A)A =  …………….(3)

From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I

Question 4. 

Solution:

A = 

|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)

|A| = 11

|A| * I = 

|A| * I = 

Now, for adjoint of A

A₁₁ = 0

A₁₂ = -(9 + 2) = -11

A₁₃ = 0

A₂₁ = -(-3 – 0) = 3

A₂₂ = 3 – 2 = 1

A₂₃ = -(0 + 1) = -1

A₃₁ = 2 – 0 = 2

A₃₂ = -(-2 – 6) = 8

A₃₃ = 0 + 3 = 3

adj A = 

Now,

A(adjA) = 

A(adj A) = 

A(adj A) = 

Also,

(adj A).A = 

(adj A).A = 

(adj A).A = 

From above, you can see,

A(adj A) = (adj A)A = I

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Question 5. 

Solution:

|A| = 6 – (-8) = 14

|A| ≠ 0, So inverse exists.

A₁₁ = 3

A₁₂ = 2

A₂₁ = -4

A₂₂ = 2

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

Question 6. 

Solution: 

A = 

|A| = -2 + 15 = 13 ≠ 0

Hence, inverse exists.

A₁₁ = 2

A₁₂ = 3

A₂₁ = -5

A₂₂ = -1

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

Question 7. 

Solution:

A = 

|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10

For adj A

A₁₁ = 10 – 0 = 0

A₁₂ = -(0 – 0) = 0

A₁₃ = 0 – 0 = 0

A₂₁ = -(10 – 0) = -10

A₂₂ = 5 – 0 = 5

A₂₃ = -(0 – 0) = 0

A₃₁ = 8 – 6 = 2

A₃₂ = -(4 – 0) = -4

A₃₃ = 2 – 0 = 2

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

Question 8. 

Solution:

A = 

|A| = 1(-3 – 0) – 0 + 0 = -3 ≠ 0

Hence, inverse exists.

For adj A

A₁₁ = -3 – 0 = -3

A₁₂ = -(-3 – 0) = 3

A₁₃ = 6 – 15 = -9

A₂₁ = -(0 – 0) = 0

A₂₂ = -1 – 0 = -1

A₂₃ = -(2 – 0) = -2

A₃₁ = 0 – 0 = 0 

A₃₂ = -(0 – 0) = 0

A₃₃ = 3 – 0 = 3

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

Question 9. 

Solution:

A = 

|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 ≠ 0

Hence, inverse exists.

For adj A

A₁₁ = -1 – 0 = -1 

A₁₂ = -(4 – 0) = -4

A₁₃ = 8 – 7 = 1

A₂₁ = -(1 – 6) = 5

A₂₂ = 2 + 21 = 23

A₂₃ = -(4 + 7) = -11

A₃₁ = 0 + 3 = 3

A₃₂ = -(0 – 12) = 12

A₃₃ = -2 – 4 = -6

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

Question 10. 

Solution:

A = 

|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1

Now for adj A

A₁₁ = 8 – 6 =2 

A₁₂ = -(0 + 9) = -9

A₁₃ = 0 – 6 = -6

A₂₁ = -(-4 + 4) =0

A₂₂ = 4 – 6 = -2

A₂₃ = -(-2 + 3) = -1

A₃₁ = 3 – 4 = -1

A₃₂ = -(-3 – 0) = 3

A₃₃ = 2 – 0 = 2

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

A^-1 = 

Question 11. 

Solution: 

A = 

|A| = 1(-cos2α – sin2α) = -1

Now,

A₁₁ = -cos2α – sin2α = -1

A₁₂ = 0

A₁₃ = 0

A₂₁ = 0

A₂₂ = -cosα

A₂₃ = -sinα

A₃₁ = 0

A₃₂ = -sinα

A₃₃ = cosα

adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

A^-1 = 

Question 12. Let A =  and B = , verify that (AB) ^{– 1} = B ^{– 1}A ^{– 1} 

Solution:

A = 

|A| = 15 – 14 = 1

A₁₁ = 5

A₁₂ = -2

A₂₁ = -7

A₂₂ = 3

A^-1 = (adj A)/|A|

A^-1 = 

B = 

|B| = 54 – 56 = -2

adj B = 

B^-1 = (adj B)/|B|

B^-1 = 

B^-1 = 

Now,

B^-1A^-1 = 

B^-1A^-1 = 

B^-1A^-1 = 

Now, AB = 

AB = 

AB = 

|AB| = 67 * 61 – 87 * 47 = -2

adj (AB) = 

(AB)^-1 = (adj AB)/|AB|

(AB)^-1 = 

(AB)^-1= 

From above, you can see that (AB)^-1 = B^-1A^-1.

Hence, it is proved.

Question 13. A = , show that A² – 5A + 7I = O. Hence find A^-1.

Solution: 

A = 

A² = 

A² = 

A² = 

So, A² – 5A + 7I

= – 5+ 7

=  – + 

= 

= O

Hence, A² – 5A + 7I = O

It can be written as 

A.A – 5A = -7I

Multiplying by A^-1 in both sides

A.A(A^-1) – 5AA^-1 = 7IA^-1

A(AA^-1) – 5I = -7A^-1

AI – 5I = -7A^-1

A^-1 = -(A – 5I)/7

A^-1 =1/7( – )

A^-1 = 

Question 14. For the matrix A =, find the numbers a and b such that A2 + aA + bI = O.

Solution: 

A = 

A² = 

A² = 

A2 = 

Now,

A² – aA + bI = O

Multiplying by A^-1 in both sides

(AA)A^-1 + aAA^-1 + bIA^-1 = O

A(AA^-1) + aI + b(IA^-1) = O

AI + aI + bA^-1 = O

A + aI = -bA^-1

A^-1 = -(A + aI)/b

Now,

A^-1 = (adj A)/|A|

A^-1 = 

Now, 

= -1/b

= 

On comparing elements you will get

-1/b = -1

b = 1

(-3 – a)/b = 1

-3 – a = 1

a = -4

Hence, a = -4 and b = 1

Question 15. A = , show that A³ – 6A² + 5A + 11I = O. Hence find A^-1

Solution:

A = 

A² = 

A² = 

A² = 

A³ = A².A

A³ = 

A³ = 

A³ = 

A³ – 6A² + 5A + 11I

– 6+ 5 + 11

= –  +  + 

= 

= O

Hence, A³ – 6A² + 5A + 11I = O

Now,

A³ – 6A² + 5A + 11I = O

(AAA)A^-1 – 6(AA)A^-1 + 5(AA^-1) + 11IA^-1 = O

AA(AA^-1) – 6A(AA^-1) + 5(AA^-1) = -11(IA^-1)

A² – 6A + 5I = -11 A^-1

A^-1 = -1/11(A2 – 6A + 5I) ………….(1)

Now, A² – 6A + 5I

= – 6 + 5

= –  + 

= 

From eq(1) you have 

A^-1 = -1/11 

Question 16. A = , verify that A³ – 6A² + 9A – 4I = O and hence fin A^-1.

Solution:

A = 

A² = 

A² = 

A²= 

A³ = A².A

A³ = 

A³ = 

A³ = 

Now,

A³ – 6A² + 9A – 4I

 – 6 + 9– 4

= –  + – 

= – 

= 

= O

So, A³ – 6A² + 9A – 4I = O

Now,

A³ – 6A² + 9A – 4I = O

Multiplying by A^-1 in both sides

(AAA)A^-1 – 6(AA)A^-1 + 9AA^-1 – 4IA^-1 = O

AA(AA^-1) – 6A(AA^-1) + 9(AA^-1) = 4(IA^-1)

AAI – 6AI +9I = 4A^-1

A² – 6A + 9I = 4A^-1

A^-1 = 1/4(A2 – 6A + 9I) ……….(1)

A² – 6A + 9I

=  – 6 + 9

= 

From eq(1), you have 

A^-1 = 

Question 17. Let A be a non-singular matrix of order 3 * 3. Then |adj A| is equal to 

(A) |A|      (B) |A|²      (C) |A|³      (D) 3|A|

Solution: 

You know,

(adj A)A = |A|I

(adj A)A = 

|(adj A)A| = 

|(adj A)A| = |A|³

|(adj A)A| = |A|³ I

|adj A| = |A|³

Hence, option B is correct.

Question 18. If A is an invertible matrix of order 2, then det(A^-1) is equal to 

(A) det(A)      (B) 1/(det A)      (C) 1      (D) 0

Solution:

Since A is an invertible matrix then A^-1 exists.

And A^-1 = (adj A)/|A|

Suppose a 2 order matrix is A = 

Then |A| = ad – bc

and adj A = 

A^-1 = (adj A)/|A|

A^-1 = 

A^-1 = 

|A^-1| = 

|A^-1| = 

|A^-1| = (ad – bc)/|A|²

|A^-1| = |A|/|A|²

|A^-1| = 1/|A|

det(A^-1) = 1/(det A)

Hence, option B is correct.