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

### 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.

### Question 2. For what value of x, the following matrix is singular?

**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 R

_{2}we get,|A| =

As R

_{1}= R_{2}, 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 C

_{2}-> C_{2}– C_{1}, 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 C

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

|A| =

On taking out 2 common from R

_{3}we get,|A| =

As R

_{2}= R_{3}, 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 C

_{1}-> C_{1 }+ C_{3}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 – i

^{2 }= – (-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 C

_{2}-> C_{2}– C_{1}we get,|A| =

=

On taking out 2 common from R

_{2}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 C

_{1}-> C_{1}+ 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 = [a_{ij}] is a 3 × 3 diagonal matrix such that a_{11} = 1, a_{22} = 2 a_{33} = 3, then find |A|.

**Solution:**

Given that a

_{11}= 1, a_{22}= 2 and a_{33}= 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 = [a_{ij}] is a 3 × 3 scalar matrix such that a_{11} = 2, then find the value of |A|.

**Solution:**

Given that A = [a

_{ij}] which is a 3 × 3 scalar matrix and a_{11}= 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 C

_{1}, we get=

= 2 (2) (2)

= 8

Therefore, the value of |A| is 8.

### Question 15. If I_{3} 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,

I

_{3}==

On expanding along C

_{1}, 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| = k

^{n}|A|Here,

Number of rows = n

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

Hence, we get

3A = 3

^{3}|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 = [a_{ij}] is a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}, where [C_{ij}] is the cofactor of a_{ij} 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 R

_{1}we get,|A| = a

_{11}C_{11}+ a_{12 }C_{12}+ a_{13}C_{13}Now,

On expanding along C

_{2}we get,|A| = a

_{12}C_{12}+ a_{22}C_{22 }+ a_{32}C_{32}

### Question 18. On expanding by the first row, the value of the determinant of 3 × 3 square matrix A = [a_{ij }] is a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}, where [C_{ij}] is the cofactor of a_{ij} 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 R

_{1}we get,|A| = a

_{11}C_{11}+ a_{12}C_{12}+ a_{13}C_{13}Now,

On expanding along C

_{2}we get,|A| = a

_{12}C_{12}+ a_{22}C_{22}+ a_{32}C_{32}= 5