In this article, we share MP Board Class 12th Maths Book Solutions Chapter 3 Matrices Ex 3.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

## MP Board Class 12th Maths Solutions Chapter 3 Matrices Ex 3.1

Question 1.

In the matrix A = \(\left[\begin{array}{cccc} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{array}\right]\), 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}.

Solution:

i. The matrix has 3 rows and 4 columns. Hence the order of A is 3 x 4

ii. Number of elements = 3(4) = 12

iii. a_{13} = 19,

a_{21} = 35,

a_{33} = – 5,

a_{24} = 12,

a_{23} = \(\frac { 5 }{ 2 }\)

Question 2.

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

Solution:

The ordered pairs of natural numbers whose product is 24 are (1, 24), (2, 12), (3, 8), (4, 6), (6,4), (8,3), (12,2) and (24,1). Hence the possible orders are 1 X 24, 2 x 12, 3 x 3, 8, 4 x 6, 6 x 4, 8 x 3, 12 x 2, 24 x 1.

The possible orders with 13 elements are 1 x 13, 13 x 1

Question 3.

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

Solution:

The ordered pairs of natural numbers whose product is 18 are (1, 18), (2, 9), (3, 6), (6, 3), (9, 2) and (18, 1).

Hence the possible orders are 1 x 18, 2 x 9, 3 x 6, 6 x 3, 9 x 2, 1.8 x 1

The possible orders with 5 elements are 1 x 5, 5 x 1.

Question 4.

Construct a 2 x 2 matrix, A = [a_{ij}], whose elements are given by

i. a_{ij} = \(\frac{(i+j)^{2}}{2}\)

ii. a_{ij} = \(\frac{i}{j}\)

iii. a_{ij} = \(\frac{(i+2j)^{2}}{2}\)

Solution:

Question 5.

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

i. a_{ij} = \(\frac{1}{2}\)|-3i + j|

i. a_{ij} = 2i – j

Solution:

The general form of a 3 x 4 matrix is

ii. The general form of a 3 x 4 matrix is

Question 6.

Find the values of x, y and z from the following equations:

i. \(\left[\begin{array}{ll}

4 & 3 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

y & z \\

1 & 5

\end{array}\right]\)

ii. \(\left[\begin{array}{cc}

x+y & 2 \\

5+z & x y

\end{array}\right]=\left[\begin{array}{cc}

6 & 2 \\

5 & 8

\end{array}\right]\)

iii. \(\left[\begin{array}{c}

x+y+z \\

x+z \\

y+z

\end{array}\right]=\left[\begin{array}{l}

9 \\

5 \\

7

\end{array}\right]\)

Solution:

i. \(\left[\begin{array}{ll}

4 & 3 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

y & z \\

1 & 5

\end{array}\right]\)

Comparing the corresponding elements of the two matrices, we get x = 1, y = 4, z = 3.

ii. \(\left[\begin{array}{cc}

x+y & 2 \\

5+z & x y

\end{array}\right]=\left[\begin{array}{cc}

6 & 2 \\

5 & 8

\end{array}\right]\)

Equating the corresponding elements, we get

x + y = 6, 5 + z = 5, xy = 8

Solving we get x = 2, y = 4, z = 0

or x = 4, y = 2, z = 0

iii. \(\left[\begin{array}{c}

x+y+z \\

x+z \\

y+z

\end{array}\right]=\left[\begin{array}{l}

9 \\

5 \\

7

\end{array}\right]\)

⇒ x + y + z = 9 … (1)

x + z = 5 … (2)

y + z = 7 … (3)

(2) + (3) ⇒ x + y + 2z = 12

(1) ⇒ x + y + z = 9

Subtracting we get z = 3

∴ x = 2,y = 4, z = 3

Question 7.

Find the value of a, b, c and d from the equation:

\(\left[\begin{array}{cc}

a-b & 2 a+c \\

2 a-b & 3 c+d

\end{array}\right]=\left[\begin{array}{cc}

-1 & 5 \\

0 & 13

\end{array}\right]\)

Solution:

\(\left[\begin{array}{cc}

a-b & 2 a+c \\

2 a-b & 3 c+d

\end{array}\right]=\left[\begin{array}{cc}

-1 & 5 \\

0 & 13

\end{array}\right]\)

Equating the corresponding elements of the two matrices, we get

a – b = – 1 ………… (1)

2a + c = 5 ……….. (2)

2a – b = 0 ……….. (3)

3c + d = 13 ……….. (4)

(3) – (1) ⇒ a = 1

(3) ⇒ b = 2

(2) ⇒ c = 3

(4) ⇒ d = 4

Question 8.

A = [a]_{m x n} is a square matrix, if

(a) m n

(c) m = n

(d) None of these

Solution:

(c) m = n

For a square matrix

Number of rows = Number of columns

∴ m = n

Question 9.

Which of the given values of x and y make the following pair of matrices equal?

\(\left[\begin{array}{cc}

3 x+7 & 5 \\

y+1 & 2-3 x

\end{array}\right],\left[\begin{array}{cc}

0 & y-2 \\

8 & 4

\end{array}\right]\)

(a) x= \(\frac { -1 }{ 3 }\), y = 7

(b) Not possible to find

(c) y = 7, x = \(\frac { -2 }{ 3 }\)

(d) x = \(\frac { -1 }{ 3 }\), y = \(\frac { – 2 }{ 3 }\)

Solution:

(b) For equal matrices, the corresponding elements are equal. Equating the corre-spondingelements, we get x = \(\frac { -7 }{ 3 }\), y = 7, y = 7 and x = \(\frac { -2 }{ 3 }\)

x has 2 values which is not possible.

Question 10.

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

(a) 27

(b) 18

(c) 81

(d) 512

Solution:

(d) 512

There are 9 elements and each element can be either 0 or 1.

The 9 places can be filled in 29 ways.

∴ Total number of ways = 29 = 512