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Class 12 RD Sharma Solutions – Chapter 6 Determinants Exercise Ex. 6.6 | Set 3

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Question 38. Write the value of the determinant \begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix} .

Solution:

 We have,

A = \begin{bmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{bmatrix}

|A| = \begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}

On taking 2x common from R3 we get,

|A| = 2x\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4\end{vmatrix}

As R1 and R3 are identical we get

|A| = 0

Therefore, the value of the determinant is 0.

Question 39. If |A| = 2, where A is 2 × 2 matrix, find |adj A|.

Solution:

Given that |A| = 2 and the order of A matrix is 2 x 2

As we know that |adj A| = |A|n-1

Here the value of n is 2

So, 

|adj A| = |2|2-1

= 2

Therefore, the value of the |adj A| is 2.

Question 40. What is the value of the determinant \begin{vmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{vmatrix} ?

Solution:

We have,

A = \begin{bmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{bmatrix}

|A| = \begin{vmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{vmatrix}

= 0(18 – 20) – 2(12 – 16) + 0(10 – 12) 

= 0 + 8 + 0

= 8

Therefore, the value of the determinant is 8.

Question 41. For what value of x is the matrix \begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}  singular?

Solution:

As we know, a matrix is singular matrix, when the value of its determinant is 0.

Given that,

A = \begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}

So, 

|A| = \begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix} = 0

=> \begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix} = 0

=> (6 – x) – 4(3 – x) = 0

=> 6 – x – 12 + 4x = 0

=> 3x – 6 = 0  

=> 3x = 6

=> x = 6/3

=> x = 2

Therefore, the value of x is 2.

Question 42. A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2 A|.

Solution:

We have,

A matrix A is of order 3 × 3 so the value of n is 3.

And |A| = 4.

As we know,

=> |K A| = Kn |A|

So, 

|2 A| = 23 (4)

= 8 (4)

= 32

Therefore, the value of |2 A| is 32.

Question 43. Evaluate: \begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix} .

Solution:

We have,

A = \begin{bmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{bmatrix}

|A| = \begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}

= cos 15° cos 75° – sin 15° sin 75°

As cos A cos B – sin A sin B = cos (A + B), we get 

= cos (15° + 75°) 

= cos 90°

= 0

Question 44. If A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix} , write the cofactor of the element a32.

Solution:

We have, 

A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}

So, the minor of a32 is,

M32 = \begin{vmatrix}5 & 8 \\ 2 & 1\end{vmatrix}

= 5 – 16 

= -11  

Now, the cofactor of a32 is,

A32 = (−1)3+2 M32 

= 11

Therefore, the cofactor of element a32 is 11.

Question 45. If \begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix} , then write the value of x.

Solution:

We have,

\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}

On expanding the determinants of both sides, we get

=> (x + 1) (x + 2) – (x – 1) (x – 3) = 12 + 1

=> x2 + 3x + 2 – x2 + 4x – 3 = 13

=> 7x – 1 = 13

=> 7x = 14

=> x = 2 

Therefore, the value of x is 2.

Question 46. If \begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix} , then write the value of x.

Solution:

We are given,

=> \begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}

On expanding the determinants of both sides, we get

=> (2x) (x + 1) – 2 (x + 1) (x + 3) = 3 – 15  

=> (x + 1) (2x – 2x – 6) = -12 

=> -6x – 6 = – 12  

=> -6x = -6

=> x = 1

Therefore, the value of x is 1. 

Question 47. If \begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix} , find the value of x.

Solution:

We have,

=> \begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}

On expanding the determinants of both sides, we get

=> 12x + 14 = 32 – 42  

=> 12x + 14 = -10

=> 12x = -24

=> x = -24/12  

=> x = -2

Therefore, the value of x is -2. 

Question 48. If \begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix} , write the value of x.

Solution:

Here, we have,

=> \begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}  

On expanding the determinants of both sides, we get

=> 2x2 – 40 = 18 + 14  

=> 2x2 – 40 = 32

=> 2x2 = 72  

=> x2 = 72/2

=> x2 = 36

=> x = ±6

Therefore, the value of x is ±6.

Question 49. If A is a 3 × 3 matrix, |A| ≠ 0 and |3A| = k |A| then write the value of k.

Solution:

We are given,

A is a 3 × 3 matrix.

Also |A| ≠ 0 and |3A| = k |A|.

Let us considered A = \begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}   

3A = \begin{bmatrix}3 a_1 & 3 a_2 & 3 a_3 \\ 3 b_1 & 3 b_2 & 3 b_3 \\ 3 c_1 & 3 c_2 & 3 c_3\end{bmatrix}

|3A| = \begin{vmatrix}3 a_1 & 3 a_2 & 3 a_3 \\ 3 b_1 & 3 b_2 & 3 b_3 \\ 3 c_1 & 3 c_2 & 3 c_3\end{vmatrix}

On taking 3 common from each row we get,

= 3^3 \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}

= 27 |A|

Therefore, the value of k is 27. 

Question 50. Write the value of the determinant \begin{vmatrix}p & p + 1 \\ p - 1 & p\end{vmatrix} .

Solution:

We have,

A = \begin{bmatrix}p & p + 1 \\ p - 1 & p\end{bmatrix}

|A| = \begin{vmatrix}p & p + 1 \\ p - 1 & p\end{vmatrix}

On expanding the determinant we have,

= p2 – (p + 1) (p – 1)

= p2 – (p2 – 1)

= p2 – p2 + 1

= 1

Therefore, the value of the determinant is 1.

Question 51. Write the value of the determinant \begin{vmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix} .

Solution:

We have,

A = \begin{bmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{bmatrix}

|A| = \begin{vmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}

On applying R1 -> R1 + R2 we get,

= \begin{vmatrix}x + y + z & x + y + z & z + x + y \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}

On taking x + y + z common from R1 we have,

= \left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}  

On applying R3 -> R3 + 3 R1 we get,

= \left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}  

= \left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ 0 & 0 & 0\end{vmatrix}

On expanding along the last row we get,

= 0

Therefore, the value of the determinant is 0.

Question 52. If A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix} , then for any natural number, find the value of Det(An).

Solution:

Given that, A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}

=> A2 = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}

= \begin{bmatrix}\cos^2 \theta - \sin^2 \theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ - \sin\theta\cos\theta - \cos\theta\sin\theta & - \sin^2 \theta + \cos^2 \theta\end{bmatrix}

= \begin{bmatrix}\cos2\theta & \sin2\theta \\ - \sin2\theta & \cos2\theta\end{bmatrix}

Similarly, An = \begin{bmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{bmatrix}

So,  

|An| = \begin{vmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{vmatrix}

= (cos nθ) (cos nθ) + (sin nθ) (sin nθ)

= cos2 (nθ) + sin2 (nθ)

= 1

Therefore, Det(An) = 1.

Question 53. Find the maximum value of \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix} .

Solution:

We have,

A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{bmatrix}

|A| = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}

On applying R2 -> R2 – R1 and R3 -> R3 – R1, we get

|A| = \begin{vmatrix}1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ 0 & 0 & \cos\theta\end{vmatrix}   

= sin θ cos θ

= (sin 2θ)/2

We know that −1 ≤ sin2θ ≤ 1.

So, the maximum value of |A| = (1/2) (1) 

= 1/2

Therefore, the maximum value is 1/2.

Question 54. If x ∈ N and \begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}  = 8, then find the value of x.

Solution:

Here we have,

A = \begin{bmatrix}x + 3 & - 2 \\ - 3x & 2x\end{bmatrix}

|A| = 8

On expansion we get,

=> (x + 3) (2x) – (-2) (-3x) = 8

=> 2 x2 + 6x – 6x = 8

=> 2 x2 = 8

=> 2x2 – 8 = 0

=> x2 – 4 = 0

=> x2 = 4

As x ∈ N, we get

=> x = 2 

Therefore, the value of x is 2.

Question 55. If \begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8 , write the value of x.

Solution:

We have,

A = \begin{bmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{bmatrix}

|A| = \begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8

=> \begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8

On expanding along R1, we get

=> x (-x2 – 1) – sin θ (-x sin θ – cos θ) + cos θ (-sin θ + x cos θ) = 8

=> -x3 – x + x sin2 θ + sin θ cos θ – sin θ cos θ + x cos2 θ = 8

=> -x3 – x + x (sin2 θ + cos2 θ) = 8

=> -x3 – x + x = 8

=> x3 + 8 = 0

=> x = -2

Therefore, the value of x is -2.

Question 56. If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k.

Solution:

Given that A is a 3 × 3 invertible matrix.

So, we know that

A^{- 1} = \frac{Adj A}{\left| A \right|}

Therefore we get,

=> \left| A^{- 1} \right| = \frac{\left| Adj A \right|}{\left| A \right|}

As we know that, \left|adj\left( A \right) \right| = \left| A \right|^{n - 1}

= \frac{\left| A \right|^{3 - 1}}{\left| A \right|}

= \frac{\left| A \right|^2}{\left| A \right|}

= |A|

|A-1| = |Ak|, we get

=> k = 1

Therefore, the value of k is 1.


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Last Updated : 21 Jul, 2021
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