NCERT Solutions Class 12 Maths  (Determinants)  Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-4 (Determinants) Miscellaneous Exercise 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.

Question 1. Prove that the determinantis

 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. 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 = -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 4. If a, b and c are real numbers, and Δ == 0 = 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.

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 = 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.

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

adj B = 

B-1 = (adj B)/|B|

B-1 = 

Now,

(AB)-1 = B-1A-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

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 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

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)(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.

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 + a2 – a2 + 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.

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 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,

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 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.