Class 12 NCERT Solutions- 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.

Deleted Questions

Without expanding the determinant, prove that

= 

Solution: 

L.H.S. = 

= 

= 

(Taking abc out from C3)

= 

= 

(Applying column transformation between C1 and C3 and between C2 and C3)

= R.H.S.

Hence, it is proved that = 

If a, b and c are real numbers, and Δ = = 0

Show that either a + b + c = 0 or a = b = c

Solution:

 Δ = 

Applying R1 ⇢ R1 + R2 + R3

 Δ = 

= 2(a + b + c) 

Applying C2 ⇢ C2-C1 and C3 ⇢ C3 – C1

Δ = 2(a + b + c) 

Expanding along R1

Δ = 2(a + b + c)(1)[(b – c)(c – b) – (b – a)(c – a)]

= 2(a + b + c)[-b² – c² + 2bc – bc + ba + ac – a²]

= 2(a + b + c)[ab + bc + ca – a² – b² – c²]

According to the question Δ = 0

2(a + b + c)[ab + bc + ca – a² – b² – c²] = 0

From above, you can see that either a + b + c =0 or ab + bc + ca – a² – b² – c² = 0

Now,

ab + bc + ca – a² – b² – c² = 0

-2ab – 2bc – 2ac + 2a² + 2b² + 2c² = 0

(a – b)² + (b – c)² + (c – a)² = 0

(a – b)² = (b – c)² = (c – a)² = 0  (because (a – b)², (b – c)², (c – a)² are non negative)

(a – b) = (b – c) = (c – a) = 0

a = b = c

Hence, it is proved that if Δ = 0 then either a + b + c = 0 or a = b = c.

Solve the equations = 0, a ≠ 0

Solution:

= 0

Applying R1 ⇢ R1 + R2 + R3

= 0

(3x + a)= 0

Applying C2 ⇢ C2-C1 and C3 ⇢ C3 – C1

(3x + a)= 0

Expanding along R1

(3x + a)[a²] = 0

a²(3x + a) = 0

But a ≠ 0

Therefore,

3x + a = 0

x = a/3

Prove that = 4a²b²c²

Solution:

A = 

Taking out common factors a, b and c from C1, C2 and C3

A = abc 

Applying R2 ⇢ R2 – R1 and R3 ⇢ R3 – R1

A = abc 

Applying R2 ⇢ R2 + R1

A = abc 

A = 2ab²c 

Applying C2 ⇢ C2 – C1

A = 2ab²c 

Expanding along R3

A = 2ab²c[a(c – a) + a(a + c)]

= 2ab²c[ac – a² + a² + ac]

= 2ab²c(2ac)

= 4a²b²c²

Hence, it is proved.

Using properties of determinants, prove that:

= (β – γ)(γ – α)(α – β)(α + β + γ)

Solution:

A = 

Applying R2->R2 – R1 and R3->R3 – R1

A = 

A = (γ – α)(β – α)

Applying R3->R3 – R2

A = (γ – α)(β – α)

Expanding along R3

A = (γ – α)(β – α)[-(γ – β)(-α – β – γ)]

A = (γ – α)(β – α)(γ – β)(α + β + γ)

A = (β – γ)(γ – α)(α – β)(α + β + γ)

Hence, it is proved.

Using the properties of determinants, prove that:

=(1 + pxyz)(x – y)(y – z)(z – x)

Solution:

 A = 

Applying R2->R2 – R1 and R3-> R3 – R1

A = 

A = (y – x)(z – x)

Applying R3->R3 – R2

A = (y – x)(z – x)

A = (y – x)(z – x)(z – y)

Expanding along R3

A = (x – y)(y – z)(z – x)[(-1)(p)(xy² + x³ + x²y) + 1 + px³ + p(x + y + z)(xy)]

= (x – y)(y – z)(z – x)[-pxy² – px³ – px²y + 1 + px³ + px²y + pxy² + pxyz]

= (x – y)(y – z)(z – x)(1 + pxyz)

Hence it is proved.

Using properties of determinants, prove that

= 3(a + b + c)(ab + bc + ca)

Solution:

A = 

Applying C1->C1 + C2 + C3

A = 

A = (a + b + c)

Applying R2->R2 – R1 and R3 ->R3 – R1

A = (a + b + c)

Expanding along C1

A = (a + b + c)[(2b + a)(2c + a) – (a – b)(a – c)]

= (a + b + c)[4bc + 2ab + 2ac + a² – a² + ac + ba – bc]

=(a + b + c)(3ab + 3bc + 3ac)

A = 3(a + b + c)(ab + bc + ca)

Hence, it is proved.

Using properties of determinants, prove that:

= 1

Solution:

A = 

Applying R2->R2 – 2R1 and R3->R3 – 3R1

A = 

Applying R3->R3 – 3R2

Expanding along C1

A = 1(1 – 0)

A = 1

Hence, it is proved.

Using properties of determinants, prove that

= 0

Solution:

A = 

A = 

Applying C1->C1 + C3

A = 

from above, you can see that two columns C1 and C2 are identical.

Hence A = 0

Hence, it is proved.

Choose the correct answer.

If a, b, c are in A.P. then the determinant

(A) 0                                    (B) 1

(C) x                                     (D) 2x

Solution:

A = 

a, b and c are in A.P So, 2b = a + c

A = 

Applying R1->R1 – R2 and R3->R3 – R2

A = 

Applying R1->R1 + R3

A = 

All the elements in the first row are 0.

Hence A = 0

So, the correct answer is A.