NCERT Solutions Class 12- Mathematics Part I – Chapter 4 Determinants – Miscellaneous Exercises on Chapter 4

Question 1. Prove that the determinant is independent of θ.

Solution: 

A = 

A = x(x² – 1) – sinθ(-x sinθ – cosθ) + cosθ(-sinθ + x cosθ)

A = x³ – x + x sin2θ + sinθcosθ – sinθcosθ + x cos2θ

A = x³ – x + x(sin2θ + cos2θ)

A = x³ – x + x

A = x³(Independent of θ).

Hence, it is independent of θ

Question 2. Evaluate 

Solution: 

A = 

Expanding along C3 

A = -sinα(-sinα sin²β – cos²β sinα) + cosα(cosα cos²β + cosα sin²β)

A = sin²α(sin²β + cos²β) + cos²α(cos²β + sin²β)

A = sin²(1) + cos²(1)

A = 1

Question 3. If A^-1 =and B =. Find (AB)^-1

Solution:

|B| = 1(3 – 0) + 1(2 – 4) = 1

B₁₁ = 3 – 0 = 3

B₁₂ = 1

B₁₃ = 2 – 0 = 2

B₂₁ = -(2 – 4) = 2

B₂₂ = 1 – 0 = 1

B₂₃ = 2

B₃₁ = 0 + 6 = 6

B₃₂ = -(0 – 2) = 2

B₃₃ = 3 + 2 = 5

adj B = 

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

B^-1 = 

Now,

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

(AB)^-1 = 

= 

(AB)^-1 = 

Question 4. Let A = verify that

(i) [adj A]^-1 = adj(A^-1)

(ii) (A^-1)^-1 = A

Solution: 

A = 

|A| = 1(15 – 1) + 2(-10 – 1) + 1(-2 – 3) = 14 – 27 = -13

A₁₁ = 14

A₁₂ = 11

A₁₃ = -5

A₂₁ = 11

A₂₂ = 4

A₂₃ = -3

A₃₁ = -5

A₃₂ = -3

A₃₃ = -1

adj A = 

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

= 

= 

(i). |adj A| = 14(-4 – 9) – 11(-11 – 15) – 5(-33 + 20)

= 14(-13) – 11(-26) – 5(-13)

= -182 + 286 + 65 = 169

adj(adj A) = 

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

= 

= 

Now, A^-1 = 

= 

adj(A^-1) = 

= 

= 

Hence, [adj A]^-1 = adj(A^-1)

(ii). A^-1 = 

adj A^-1 = 

|A^-1| = (1/13)3[-14 × (-13) +11 × (-26) + 5 × (-13)]

= (1/13)3 × (-169)

= -1/13

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

= 

= 

= A

Hence, it is proved that (A^-1)^-1 = A

Question 5. Evaluate 

Solution: 

A = 

Applying R1 -> R1+R2+R3

A = 

= 2(x+y)

Applying C2-> C2 – C1 and C3-> C3 – C1

A = 2(x + y)

Expanding along R1

A = 2(x + y)[-x² + y(x – y)]

= -2(x + y)(x² + y² – yx)

A = -2(x³ + y³)

Question 6. Evaluate 

Solution:

A = 

Applying R2->R2 – R1 and R3->R3 – R1

A = 

Expanding along C1

A = 1(xy – 0)

A = xy

Question 7. Solve the system of the following questions:

2/x + 3/y + 10/z = 4

4/x – 6/y + 5/z = 1

6/x + 9/y – 20/z = 2

Solution:

Assume 1/x = p ; 1/y = q; 1/z = r

then. the above equations will be like

2p + 3Q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

This can be written in the form of AX=B

where,

A = 

X = 

B = 

We have,

|A| = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)

|A| = 150 + 330 + 720

|A| = 1200 ≠ 0

Hence A is invertible matrix.

A₁₁ = 75

A₁₂ = 110

A₁₃ = 72

A₂₁ = 150

A₂₂ = -100

A₂₃ = 0

A₃₁ = 75

A₃₂ = 30

A₃₃ = -24

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

A^-1 = 

Now,

X = A^-1B

= 

= 

= 

= 

From above p = 1/2; q = 1/3 ; r = 1/5

So, x = 2; y = 3; z = 5

Question 8. Choose the correct answer.

If x, y, z are non-zero real numbers, then the inverse of matrix A =  is 

(A) 

(B) xyz

(C) 

(D) 

Solution:

A = 

|A| = x(yz – 0) = xyz ≠ 0

Hence, the matrix is invertible

Now,

A₁₁ = yz

A₁₂ = 0

A₁₃ = 0

A₂₁ = 0

A₂₂ = xz

A₂₃ = 0

A₃₁ = 0

A₃₂ = 0

A₃₃ = xy

adj A = 

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

A^-1 = 

A^-1 = 

A^-1 = 

A^-1 = 

Hence, the correct answer is A.

Question 9. Choose the correct answer

Let A = , where 0 ≤ θ ≤ 2π, then

(A) Det(A) = 0                                      (B) Det(A) ∈ (2, ∞)

(C) Det(A) ∈ (2, 4)                              (D) Det(A) ∈ [2, 4]

Solution:

A = 

|A| = 1(1 + sin2θ) – sinθ(-sinθ + sinθ) + 1(sin2θ + 1)

|A| = 1 + sin2θ + sin2θ + 1

= 2 + 2 sin2θ

= 2(1 + sin2θ)

Now 0 ≤ θ ≤ 2π

So, 0 ≤ sinθ ≤ 1

0 ≤ sin2θ ≤ 1

0 + 1 ≤ 1 + sin2θ ≤ 1 + 1

2 ≤ 2(1 + sin2θ) ≤ 4

Det(A) ∈ [2, 4]

Hence, the correct answer is D.