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

### Question 38. Write the value of the determinant .

**Solution:**

We have,

A =

|A| =

On taking 2x common from R

_{3}we get,|A| =

As R

_{1 }and R_{3 }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 ?

**Solution:**

We have,

A =

|A| =

= 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 singular?

**Solution:**

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

Given that,

A =

So,

|A| =

=>

=> (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| = K

^{n}|A|So,

|2 A| = 2

^{3}(4)= 8 (4)

= 32

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

### Question 43. Evaluate: .

**Solution:**

We have,

A =

|A| =

= 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 = , write the cofactor of the element a_{32}.

**Solution:**

We have,

A =

So, the minor of a

_{32}is,M

_{32}== 5 – 16

= -11

Now, the cofactor of a

_{32}is,A

_{32}= (−1)^{3+2}M_{32}= 11

Therefore, the cofactor of element a_{32}is 11.

### Question 45. If , then write the value of x.

**Solution:**

We have,

On expanding the determinants of both sides, we get

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

=> x

^{2}+ 3x + 2 – x^{2}+ 4x – 3 = 13=> 7x – 1 = 13

=> 7x = 14

=> x = 2

Therefore, the value of x is 2.

### Question 46. If , then write the value of x.

**Solution:**

We are given,

=>

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 , find the value of x.

**Solution:**

We have,

=>

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 , write the value of x.

**Solution:**

Here, we have,

=>

On expanding the determinants of both sides, we get

=> 2x

^{2}– 40 = 18 + 14=> 2x

^{2}– 40 = 32=> 2x

^{2}= 72=> x

^{2}= 72/2=> x

^{2}= 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 =

3A =

|3A| =

On taking 3 common from each row we get,

=

= 27 |A|

Therefore, the value of k is 27.

### Question 50. Write the value of the determinant .

**Solution:**

We have,

A =

|A| =

On expanding the determinant we have,

= p

^{2}– (p + 1) (p – 1)= p

^{2}– (p^{2}– 1)= p

^{2}– p^{2}+ 1= 1

Therefore, the value of the determinant is 1.

### Question 51. Write the value of the determinant .

**Solution:**

We have,

A =

|A| =

On applying R

_{1}-> R_{1}+ R_{2}we get,=

On taking x + y + z common from R

_{1}we have,=

On applying R

_{3}-> R_{3}+ 3 R_{1}we get,=

=

On expanding along the last row we get,

= 0

Therefore, the value of the determinant is 0.

### Question 52. If A = , then for any natural number, find the value of Det(A^{n}).

**Solution:**

Given that, A =

=> A

^{2}==

=

Similarly, A

^{n}=So,

|A

^{n}| == (cos nθ) (cos nθ) + (sin nθ) (sin nθ)

= cos

^{2}(nθ) + sin^{2}(nθ)= 1

Therefore, Det(A^{n}) = 1.

### Question 53. Find the maximum value of .

**Solution:**

We have,

A =

|A| =

On applying R

_{2}-> R_{2}– R_{1}and R_{3}-> R_{3}– R_{1}, we get|A| =

= 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 = 8, then find the value of x.

**Solution:**

Here we have,

A =

|A| = 8

On expansion we get,

=> (x + 3) (2x) – (-2) (-3x) = 8

=> 2 x

^{2}+ 6x – 6x = 8=> 2 x

^{2}= 8=> 2x

^{2}– 8 = 0=> x

^{2}– 4 = 0=> x

^{2}= 4As x ∈ N, we get

=> x = 2

Therefore, the value of x is 2.

### Question 55. If , write the value of x.

**Solution:**

We have,

A =

|A| =

=>

On expanding along R

_{1}, we get=> x (-x

^{2}– 1) – sin θ (-x sin θ – cos θ) + cos θ (-sin θ + x cos θ) = 8=> -x

^{3}– x + x sin^{2}θ + sin θ cos θ – sin θ cos θ + x cos^{2}θ = 8=> -x

^{3}– x + x (sin^{2}θ + cos^{2}θ) = 8=> -x

^{3}– x + x = 8=> x

^{3}+ 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

Therefore we get,

=>

As we know that,

=

=

= |A|

|A

^{-1}| = |A^{k}|, we get=> k = 1

Therefore, the value of k is 1.

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