# Class 12 RD Sharma Solutions – Chapter 6 Determinants Exercise Ex. 6.6 | Set 2

### Question 19. Let A = [a_{ij}] be a square matrix of order 3 × 3 and C_{ij} denote the cofactor of a_{ij} in A. If |A| = 5, find the value of a_{11} C_{21} + a_{12} C_{22} + a_{13} C_{23}.

**Solution:**

As we know that, if a matrix is square matrix of order n, then the sum of the products of elements of a row or a column with the cofactors of the corresponding elements of some other row or column is zero.

So,

A = [a

_{ij}] is a square matrix of order n.Also we have,

And

=> a

_{11}C_{21}+ a_{12}C_{22}+ a_{13}C_{23}= 0

Therefore, the required value is 0.

### Question 20. Find the value of .

**Solution:**

Given that,

A =

=> |A| =

= sin 20° cos 70° + cos 20° sin 70°

= sin (20° + 70°)

= sin 90

= 1

### Question 21. If A is a square matrix satisfying A^{T} A = I, write the value of |A|.

**Solution:**

Let us assume that A = [a

_{ij}] be a square matrix of order n.So, by using the property of determinants, we get

=> |A| = |A

^{T}|Here, we have

=> A

^{T}A = I=> |A

^{T}A| = 1So, the determinants are of same order, we get

=> |A

^{T}A| = |A^{T}| |A|=> |A

^{T}| |A| = 1=>

=>

=> |A|

^{2}= 1=> |A| = ±1

Therefore, the value of |A| is ±1.

### Question 22. If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.

**Solution:**

According to the question, A and B are square matrices of the same order.

So, by using the property of determinants we get,

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

Here, |A| = 3, AB = I.

=> |AB| = 1

=> |A| |B| = 1

=> 3 |B| = 1

=> |B| = 1/3

Therefore, the value of |B| is 1/3.

### Question 23. A is a skew-symmetric of order 3, write the value of |A|.

**Solution:**

Here, |A| = 4.

So we have,

Order of the matrix (n) = 3

Using the properties of matrices, we get

For a square matrix of order n and constant k, we know,

=> |k A| = k

^{n}|A|=> |- A| = (-1)

^{3}|A|= (-1) (4)

= -4

Therefore, the value of |A| is -4.

### Question 24. If A is a square matrix of order 3 with determinant 4, then write the value of |−A|.

**Solution:**

Given that, |A| = 4.

Order of the matrix (n) = 3

So, by using the properties of matrices, we get

=> |k A| = k

^{n }|A|=> |- A| = (-1)

^{3}|A|= (-1) (4)

= -4

Therefore, the value of |A| is -4.

### Question 25. If A is a square matrix such that |A| = 2, write the value of |A A^{T}|.

**Solution:**

Given that, |A| = 2

As we know that in a square matrix, |A| = A

^{T}So, they are of sane order

Hence, |A A

^{T}| = |A| |A^{T}|=> |A A

^{T}| = 2 (2)= 4

Therefore, the value of |A A^{T}| is 4.

### Question 26. Find the value of the determinant .

**Solution:**

Given that,

A =

|A| =

On applying R

_{1}-> R_{1}– 3R_{2}we have,=

=

= 0

Therefore, the value of the determinant is 0.

### Question 27. Find the value of the determinant .

**Solution:**

Given that,

A =

|A| =

On applying R

_{2}-> R_{2}– 2R_{1}we get,=

=

= 0

Therefore, the value of the determinant is 0.

### Question 28. If the matrix is singular, find the value of x.

**Solution:**

As we know that a matrix is singular only when its determinant is zero.

According to the question,

is a singular matrix

So,

=> |A| = = 0

On expanding the determinant we get,

=> 5x + 20 = 0

=> x = -20/5

=> x = -4

Therefore, the value of x is -4.

### Question 29. If A is a square matrix of order n × n such that |A| = λ, then write the value of |−A|.

**Solution:**

Given that,

A is a square matrix of order n × n

So, |A| = λ

=> |- A| = (-1)

^{n }A=> |-A| = (-1)

^{n}λ

Therefore, the value of |-A| is (-1)^{n}λ.

### Question 30. Find the value of the determinant .

**Solution:**

Given that,

A =

|A| =

On taking out common factors from R

_{1}, R_{2}and R_{3}we get,=

Here, the two rows are identical, so we get

=

= 0

Therefore, the value of the determinant is 0.

### Question 31. If A and B are non-singular matrices of the same order, prove whether AB is singular or non-singular.

**Solution:**

According to the question, A and B be non-singular matrices of order n.

Here, |A| ≠ 0 and |B| ≠ 0.

So, the order of these matrix are same, we get

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

=> |AB| = 0 if |A| = 0 or |B| = 0

But as it is not the case here, so |AB| is non- zero matrix and AB is non-singular matrix.

Hence proved.

### Question 32. A matrix of order 3 × 3 has determinant 2. What is the value of |A (3I)|, where I is the identity matrix of order 3 × 3.

**Solution:**

Given that a matrix of order 3 x 3 has determinant 2.

So let us assume B be the matrix. so the order of the matrix is 3

and |B| = 2

Let us consider I be the identity matrix, so we get

=> |I| = 1

=> 3 |I| = 3

=> |A (3I)| = |3 A|

= 3

^{3}|A|= 27 (2)

= 54

=> |A (3I)| = 54

Therefore, the value of |A (3I)| is 54.

### Question 33. If A and B are square matrices of order 3 such that |A| = −1, |B| = 3, then find the value of |3 AB|.

**Solution:**

We have,

A and B are square matrices of order 3.

Also |A| = −1, |B| = 3.

We know,

As n is the order of A, we get

=> |K A| = K

^{n}|A|=> |3 AB| = 3

^{3}|AB|If the order of A and B matrix are same and they are square matrix then |AB| = |A| |B|.

So, we have,

=> |3 AB| = 3

^{3}|A| |B|= 27 (-1) (3)

= -81

Therefore, the value of |3 AB| is -81.

### Question 34. Write the value of .

**Solution:**

We have,

A =

|A| =

= a

^{2}– iab + iab – i^{2}b^{2}– (-c^{2}– icd + icd + i^{2}d^{2})= a

^{2}– i^{2}b^{2}+ c^{2}– i^{2}d^{2}Here, we have, i

^{2}= – 1.So we get,

|A| = a

^{2}– (-1) b^{2}+ c^{2}– (-1) d^{2}= a

^{2}+ b^{2}+ c^{2}+ d^{2}

Therefore, the value is a^{2}+ b^{2}+ c^{2}+ d^{2}.

### Question 35. Write the cofactor of a_{12} in the following matrix .

**Solution:**

We have

So,

=> a

_{12}= -3Now we find the cofactor of a

_{12}a

_{12}= (-1)^{1+2}= – (- 42 – 4)

= 46

Therefore, the value of the required cofactor is 46.

### Question 36. If , find x.

**Solution:**

Here we have,

A =

=> |A| = = 0

=> 9(2x + 5) – 3(5x + 2) = 0

=> 18x + 45 – 15x – 6 = 0

=> 3x + 39 = 0

=> 3x = – 39

=> x = -39/3

=> x = -13

Therefore, the value of x is -13.

### Question 37. Find the value of x from the following :

**Solution:**

We have,

A =

|A| =

=> = 0

=> 2 x

^{2}– 8 = 0=> 2 x

^{2 }= 8=> x

^{2}= 8/2=> x

^{2}= 4=> x = √4

=> x = ±2

Therefore, the value of x is ±2.