## NCERT Solutions for Class 12 Maths Chapter 3 – Matrices Ex 3.1

#### Page No 64:

#### Question 1:

In the matrix, write:

(i) The order of the matrix (ii) The number of elements,

(iii) Write the elements

*a*_{13},*a*_{21},*a*_{33},*a*_{24},*a*_{23}#### Answer:

**(i)**In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4.

**(ii)**Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.

**(iii)**

*a*

_{13}= 19,

*a*

_{21}= 35,

*a*

_{33}= −5,

*a*

_{24}= 12,

*a*

_{23}=

#### Question 2:

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?

#### Answer:

We know that if a matrix is of the order

*m*×*n*, it has*mn*elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4)

Hence, the possible orders of a matrix having 24 elements are:

1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

#### Question 3:

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

#### Answer:

We know that if a matrix is of the order

*m*×*n*, it has*mn*elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)

Hence, the possible orders of a matrix having 18 elements are:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

#### Question 4:

Construct a 2 × 2 matrix,, whose elements are given by:

(i)

(ii)

(iii)

#### Answer:

In general, a 2 × 2 matrix is given by

**(i)**

Therefore, the required matrix is

**(ii)**

Therefore, the required matrix is

**(iii)**

Therefore, the required matrix is

#### Question 5:

Construct a 3 × 4 matrix, whose elements are given by

(i) (ii)

#### Answer:

In general, a 3 × 4 matrix is given by

**(i)**

Therefore, the required matrix is

**(ii)**

Therefore, the required matrix is

#### Question 6:

Find the value of

*x*,*y*, and*z*from the following equation:
(i) (ii)

(iii)

#### Answer:

**(i)**

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x*= 1,

*y*= 4, and

*z*= 3

**(ii)**

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x*+

*y*= 6,

*xy*= 8, 5 +

*z*= 5

Now, 5 +

*z*= 5 ⇒*z*= 0
We know that:

(

*x*−*y*)^{2}= (*x*+*y*)^{2}− 4*xy*
⇒ (

*x*−*y*)^{2}= 36 − 32 = 4
⇒

*x*−*y*= ±2
Now, when

*x*−*y*= 2 and*x*+*y*= 6, we get*x*= 4 and*y*= 2
When

*x*−*y*= − 2 and*x*+*y*= 6, we get*x*= 2 and*y*= 4
∴

*x*= 4,*y*= 2, and*z*= 0 or*x*= 2,*y*= 4, and*z*= 0**(iii)**

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x*+

*y*+

*z*= 9 … (1)

*x*+

*z*= 5 … (2)

*y*+

*z*= 7 … (3)

From (1) and (2), we have:

*y*+ 5 = 9

⇒

*y*= 4
Then, from (3), we have:

4 +

*z*= 7
⇒

*z*= 3
∴

*x*+*z*= 5
⇒

*x*= 2
∴

*x*= 2,*y*= 4, and*z*= 3#### Question 7:

Find the value of

*a*,*b*,*c*, and*d*from the equation:#### Answer:

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*a*−

*b*= −1 … (1)

2

*a*−*b*= 0 … (2)
2

*a*+*c*= 5 … (3)
3

*c*+*d*= 13 … (4)
From (2), we have:

*b*= 2

*a*

Then, from (1), we have:

*a*− 2

*a*= −1

⇒

*a*= 1
⇒

*b*= 2
Now, from (3), we have:

2 ×1 +

*c*= 5
⇒

*c*= 3
From (4) we have:

3 ×3 +

*d*= 13
⇒ 9 +

*d*= 13 ⇒*d*= 4
∴

*a*= 1,*b*= 2,*c*= 3, and*d*= 4#### Page No 65:

#### Question 8:

is a square matrix, if

**(A)**

*m*<

*n*

**(B)**

*m*>

*n*

**(C)**

*m*=

*n*

**(D)**None of these

#### Answer:

The correct answer is C.

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, is a square matrix, if

*m*=*n*.#### Question 9:

Which of the given values of

*x*and*y*make the following pair of matrices equal**(A)**

**(B)**Not possible to find

**(C)**

**(D)**

#### Answer:

The correct answer is B.

It is given that

Equating the corresponding elements, we get:

We find that on comparing the corresponding elements of the two matrices, we get two different values of

*x*, which is not possible.
Hence, it is not possible to find the values of

*x*and*y*for which the given matrices are equal.#### Question 10:

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

**(A)**27

**(B)**18

**(C)**81

**(D)**512

#### Answer:

The correct answer is D.

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 2

^{9}= 512_{}

^{}