# Class 12 RD Sharma Solutions – Chapter 6 Determinants Exercise Ex. 6.6 | Set 1

• Last Updated : 07 Sep, 2021

### Question 1. If A is a singular matrix, then find the value of |A|.

Solution:

Given that A is a singular matrix.

So, as we know if A is a n×n matrix and it is singular, the value of its determinant is always 0.

Thus, |A| = 0.

### Solution:

Given that As we know if A is a n×n matrix and it is singular, so, the value of its determinant is always 0.

=> |A| = 0

=> => 4(5 – x) – 2(x + 1) = 0

=> 20 – 4x – 2x – 2 = 0

=> 18 – 6x = 0

=> 18 = 6x

=> x = 3

Therefore, the value of x is 3.

### Question 3. Find the value of the determinant .

Solution:

Given that

A = |A| = So, on taking out x common from R2 we get,

|A| = As R1 = R2, we get

|A| = 0

Therefore, the value of the determinant is 0.

### Question 4. State whether the matrix is singular or non-singular.

Solution:

Given that

A = |A| = |A| = 2 (4) – 6 (3)

= 8 – 18

= -10

As we know if A is a n×n matrix and it is singular, so the value of its determinant is always 0.

As |A| = -10 here, the given matrix is non-singular.

### Question 5. Find the value of the determinant .

Solution:

Given that

A = |A| = On applying C2 -> C2 – C1, we get

|A| = |A| = |A| = 4200 – 4202

|A| = -2

Therefore, the value of determinant is -2.

### Question 6. Find the value of the determinant .

Solution:

Given that

A = |A| = On applying C2 -> C2 – C1 and C3 -> C3 – C1, we get

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

|A| = As R2 = R3, we get

|A| = 0

Therefore, the value of the determinant is zero.

### Question 7. Find the value of the determinant .

Solution:

Given that

A = |A| = On applying C1 -> C1 + C3 we get,  = (a + b + c) (0)

= 0

Therefore, the value of determinant is 0.

### Question 8. If A = and B = , find the value of |A| + |B|.

Solution:

Given that

A = |A| = = 0 – i2

= – (-1)

= 1

Also, we have

B = |B| = = 0 – 1

= -1

So,

|A| + |B| = 1 + (-1)

= 1 – 1

= 0

Therefore, the value of |A| + |B| is 0.

### Question 9. If A = and B = , find |AB|.

Solution:

We have,

A = and B = So, we get

AB =   Now we have,

|AB| = = -1 (0) – 0 (4)

= 0 – 0

= 0

Therefore, the value of |AB| is 0.

### Question 10. Evaluate .

Solution:

Given that

A = |A| = On applying C2 -> C2 – C1 we get,

|A| =  On taking out 2 common from R2 we get, = 2 (4785 – 4789)

= 2 (-4)

= -8

Therefore, the value of the determinant is 0.

### Question 11. If w is an imaginary cube root of unity, find the value of .

Solution:

Given that,

A = |A| = On applying C1 -> C1 + C_2 + C_3 we get,  = 0

### Question 12. If A = and B = , find |AB|.

Solution:

Given that

A = |A| = -1 – 6

= -7

B = |B| = – 2 + 12

= 10

We know if A and B are square matrices of the same order, then we have,

=> |AB| = |A|. |B|

= (-7) (10)

= -70

Therefore, the value of |AB| is -70.

### Question 13. If A = [aij]  is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.

Solution:

Given that  a11 = 1, a22 = 2 and a33 = 3.

If A is a diagonal matrix of order n x n, then we have

=> So, we get

|A| = 1 (2) (3)

= 6

Therefore, the value of |A| is 6.

### Question 14. If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then find the value of |A|.

Solution:

Given that A = [aij] which is a 3 × 3 scalar matrix and a11 = 2,

As we know that a scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number.

=> A =  On expanding along C1, we get = 2 (2) (2)

= 8

Therefore, the value of |A| is 8.

### Question 15. If I3 denotes an identity matrix of order 3 × 3, find the value of its determinant.

Solution:

As we know that in an identity matrix, all the diagonal elements are 1 and the remaining elements are 0.

Here,

I3  On expanding along C1, we get = 1

Therefore, the value of the determinant is 1.

### Question 16. A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?

Solution:

Given that matrix A is of order 3 x 3 and the determinant = 5.

If A is a square matrix of order n and k is a constant, then we have

=> |kA| = kn |A|

Here,

Number of rows = n

Also, k is a common factor from each row of k.

Hence, we get

3A = 33 |A|

= 27 (5)

= 135

Therefore, the value of |3A| is 135.

### Question 17. On expanding by the first row, the value of the determinant of 3 × 3 square matrix A = [aij] is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by the second column.

Solution:

As we know that if a square matrix(let say  A) is of order n, then the sum of the products of elements of a row or a column with their cofactors is always equal to det (A).

So, Also, On expanding along R1 we get,

|A| = a11 C11 + a12 C12 + a13 C13

Now,

On expanding along C2 we get,

|A| = a12 C12 + a22 C22 + a32 C32

### Question 18. On expanding by the first row, the value of the determinant of 3 × 3 square matrix A = [aij ] is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by the second column.

Solution:

As we know that if a square matrix(let say  A) is of order n, then the sum of the products of elements of a row or a column with their cofactors is always equal to det (A).

So, Also, On expanding along R1 we get,

|A| = a11 C11 + a12 C12 + a13 C13

Now,

On expanding along C2 we get,

|A| = a12 C12 + a22 C22 + a32 C32 = 5

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