NCERT Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-4 (Determinants)Exercise 4.5 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of  NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 4.5

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) = 

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 . 

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 – 1A – 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.

Chapter-4 (Determinants)

• Exercise 4.1
• Exercise 4.2
• Exercise 4.3
• Exercise 4.4
• Exercise 4.6