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# 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

### =

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

### Question 3. 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

### 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.

### Question 5. 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

### Question 6. 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 = 2ab2

Applying C2 ⇢ C2 – C1

A = 2ab2

Expanding along R3

A = 2ab2c[a(c – a) + a(a + c)]

= 2ab2c[ac – a2 + a2 + ac]

= 2ab2c(2ac)

= 4a2b2c2

Hence, it is proved.

### Question 7. 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

B-1

Now,

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

(AB)-1

(AB)-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

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

= 14(-13) – 11(-26) – 5(-13)

= -182 + 286 + 65 = 169

Now, A-1

(ii). A-1

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

= (1/13)3 × (-169)

= -1/13

= A

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

### Question 9. 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 10. Evaluate

Solution:

A =

Applying R2->R2 – R1 and R3->R3 – R1

A =

Expanding along C1

A = 1(xy – 0)

A = xy

### = (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.

### =(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.

### = 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.

### = 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.

### = 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.

### 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

Now,

X = A-1B

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

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

### (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.

### (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

A-1

A-1

A-1

A-1

Hence, the correct answer is A.

### (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.

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