# 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(x

^{2 }– 1) – sinθ(-x sinθ – cosθ) + cosθ(-sinθ + x cosθ)A = x

^{3 }– x + x sin2θ + sinθcosθ – sinθcosθ + x cos2θA = x

^{3 }– x + x(sin2θ + cos2θ)A = x

^{3 }– x + xA = x

^{3}(Independent of θ).Hence, it is independent of θ

### Question 2. 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 =

### Question 3. Evaluate

**Solution: **

A =

Expanding along C3

A = -sin

α(-sinαsin^{2}β – cos^{2}β sinα) + cosα(cosαcos^{2}β + cosαsin^{2}β)A = sin

^{2}α(sin^{2}β + cos^{2}β) + cos^{2}α(cos^{2}β + sin^{2}β)A = sin

^{2}(1) + cos^{2}(1)A = 1

### Question 4. 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)[-b

^{2}– c^{2}+ 2bc – bc + ba + ac – a^{2}]= 2(a + b + c)[ab + bc + ca – a

^{2}– b^{2}– c^{2}]According to the question Δ = 0

2(a + b + c)[ab + bc + ca – a

^{2 }– b^{2 }– c^{2}] = 0From above, you can see that either a + b + c =0 or ab + bc + ca – a

^{2 }– b^{2 }– c^{2 }= 0Now,

ab + bc + ca – a

^{2 }– b^{2 }– c^{2}= 0-2ab – 2bc – 2ac + 2a

^{2 }+ 2b^{2 }+ 2c^{2 }= 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)[a

^{2}] = 0a

^{2}(3x + a) = 0But a ≠ 0

Therefore,

3x + a = 0

x = a/3

### Question 6. Prove that = 4a^{2}b^{2}c^{2}

**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 = 2ab

^{2}cApplying C2 ⇢ C2 – C1

A = 2ab

^{2}cExpanding along R3

A = 2ab

^{2}c[a(c – a) + a(a + c)]= 2ab

^{2}c[ac – a^{2 }+ a^{2 }+ ac]= 2ab

^{2}c(2ac)= 4a

^{2}b^{2}c^{2}Hence, it is proved.

### Question 7. If A^{-1} =and B =. Find (AB)^{-1}

**Solution:**

|B| = 1(3 – 0) + 1(2 – 4) = 1

B

_{11 }= 3 – 0 = 3B

_{12}= 1B

_{13 }= 2 – 0 = 2B

_{21}= -(2 – 4) = 2B

_{22 }= 1 – 0 = 1B

_{23 }= 2B

_{31 }= 0 + 6 = 6B

_{32 }= -(0 – 2) = 2B

_{33 }= 3 + 2 = 5adj B =

B

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

^{-1}=Now,

(AB)

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

^{-1 }==

(AB)

^{-1}=

### Question 8. 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

A

_{11 }= 14A

_{12 }= 11A

_{13 }= -5A

_{21 }= 11A

_{22 }= 4A

_{23 }= -3A

_{31 }= -5A

_{32 }= -3A

_{33 }= -1adj 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 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)[-x

^{2 }+ y(x – y)]= -2(x + y)(x

^{2 }+ y^{2 }– yx)A = -2(x

^{3 }+ y^{3})

### Question 10. Evaluate

**Solution:**

A =

Applying R2->R2 – R1 and R3->R3 – R1

A =

Expanding along C1

A = 1(xy – 0)

A = xy

### Question 11. 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.

### Question 12. 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)(xy

^{2}+ x^{3}+ x^{2}y) + 1 + px^{3 }+ p(x + y + z)(xy)]= (x – y)(y – z)(z – x)[-pxy

^{2 }– px^{3 }– px^{2}y + 1 + px^{3 }+ px^{2}y + pxy^{2 }+ pxyz]= (x – y)(y – z)(z – x)(1 + pxyz)

Hence it is proved.

### Question 13. 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 + a

^{2 }– a^{2 }+ ac + ba – bc]=(a + b + c)(3ab + 3bc + 3ac)

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

Hence, it is proved.

### Question 14. 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.

### Question 15. 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.

### Question 16. 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.

A

_{11 }= 75A

_{12 }= 110A

_{13 }= 72A

_{21 }= 150A

_{22 }= -100A

_{23 }= 0A

_{31 }= 75A

_{32 }= 30A

_{33 }= -24A

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

^{-1}=Now,

X = A

^{-1}B=

=

=

=

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

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

### Question 17. 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.

### Question 18. 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,

A

_{11 }= yzA

_{12}= 0A

_{13}= 0A

_{21 }= 0A

_{22}= xzA

_{23}= 0A

_{31 }= 0A

_{32}= 0A

_{33 }= xyadj A =

A

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

^{-1}=A

^{-1 }=A

^{-1}=A

^{-1 }=Hence, the correct answer is A.

### Question 19. 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.