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

• Last Updated : 21 Jul, 2021

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

Solution:

We have,

A = |A| = On taking 2x common from R3 we get,

|A| = 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,

= 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| = Kn |A|

So,

|2 A| = 23 (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 a32.

Solution:

We have,

A = So, the minor of a32 is,

M32 = 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 , 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

=> x2 + 3x + 2 – x2 + 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

=> 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 = 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,

= 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 .

Solution:

We have,

A = |A| = On applying R1 -> R1 + R2 we get, On taking x + y + z common from R1 we have, On applying R3 -> R3 + 3 R1 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(An).

Solution:

Given that, A = => A2   Similarly, An So,

|An| = = (cos nθ) (cos nθ) + (sin nθ) (sin nθ)

= cos2 (nθ) + sin2 (nθ)

= 1

Therefore, Det(An) = 1.

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

Solution:

We have,

A = |A| = On applying R2 -> R2 – R1 and R3 -> R3 – R1, 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 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 , write the value of x.

Solution:

We have,

A = |A| = => 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 Therefore we get,

=> As we know that,   = |A|

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

=> k = 1

Therefore, the value of k is 1.

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