Class 12-NCERT Solutions-Chapter-03-Matrices-Ex 3.1

Solution:

(i) The order of the matrix:

Since the matrix has 3 rows and 4 columns, its order is \( 3 \times 4 \).

(ii) The number of elements:

Total elements = \( 3 \times 4 = 12 \).

(iii) The specific elements:

By comparing positions in the matrix:

• \( a_{13} \) (1st row, 3rd column) = 19
• \( a_{21} \) (2nd row, 1st column) = 35
• \( a_{33} \) (3rd row, 3rd column) = -5
• \( a_{24} \) (2nd row, 4th column) = 12
• \( a_{23} \) (2nd row, 3rd column) = \( \frac{5}{2} \)

Solution:

Case 1: 24 Elements

We need to find ordered pairs \( (m, n) \) such that \( mn = 24 \), where \( m, n \) are natural numbers.

The possible orders are:

\( 1 \times 24, \quad 24 \times 1 \)

\( 2 \times 12, \quad 12 \times 2 \)

\( 3 \times 8, \quad 8 \times 3 \)

\( 4 \times 6, \quad 6 \times 4 \)

Case 2: 13 Elements

We need pairs \( (m, n) \) such that \( mn = 13 \). Since 13 is a prime number, the possible orders are:

\( 1 \times 13, \quad 13 \times 1 \)

Solution:

Case 1: 18 Elements

We find pairs \( (m, n) \) such that \( mn = 18 \). The possible orders are:

\( 1 \times 18, \quad 18 \times 1 \)

\( 2 \times 9, \quad 9 \times 2 \)

\( 3 \times 6, \quad 6 \times 3 \)

Case 2: 5 Elements

Since 5 is a prime number, the possible orders are:

\( 1 \times 5, \quad 5 \times 1 \)

Solution:

A \( 2 \times 2 \) matrix is given by \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \).

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

• \( a_{11} = \frac{(1+1)^2}{2} = \frac{4}{2} = 2 \)
• \( a_{12} = \frac{(1+2)^2}{2} = \frac{9}{2} \)
• \( a_{21} = \frac{(2+1)^2}{2} = \frac{9}{2} \)
• \( a_{22} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8 \)

Thus, \( A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix} \).

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

• \( a_{11} = \frac{1}{1} = 1 \)
• \( a_{12} = \frac{1}{2} \)
• \( a_{21} = \frac{2}{1} = 2 \)
• \( a_{22} = \frac{2}{2} = 1 \)

Thus, \( A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix} \).

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

• \( a_{11} = \frac{(1+2(1))^2}{2} = \frac{9}{2} \)
• \( a_{12} = \frac{(1+2(2))^2}{2} = \frac{25}{2} \)
• \( a_{21} = \frac{(2+2(1))^2}{2} = \frac{16}{2} = 8 \)
• \( a_{22} = \frac{(2+2(2))^2}{2} = \frac{36}{2} = 18 \)

Thus, \( A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix} \).

Solution:

A \( 3 \times 4 \) matrix has 3 rows and 4 columns.

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

Calculations for elements:

• \( a_{11} = \frac{1}{2}|-3(1)+1| = 1 \), \( a_{12} = \frac{1}{2}|-3(1)+2| = \frac{1}{2} \), \( a_{13} = 0 \), \( a_{14} = \frac{1}{2} \)
• \( a_{21} = \frac{1}{2}|-3(2)+1| = \frac{5}{2} \), \( a_{22} = 2 \), \( a_{23} = \frac{3}{2} \), \( a_{24} = 1 \)
• \( a_{31} = \frac{1}{2}|-3(3)+1| = 4 \), \( a_{32} = \frac{7}{2} \), \( a_{33} = 3 \), \( a_{34} = \frac{5}{2} \)

Therefore, \( A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix} \).

(ii) \( a_{ij} = 2i - j \)

Calculations for elements:

• \( a_{11} = 2(1)-1 = 1 \), \( a_{12} = 0 \), \( a_{13} = -1 \), \( a_{14} = -2 \)
• \( a_{21} = 2(2)-1 = 3 \), \( a_{22} = 2 \), \( a_{23} = 1 \), \( a_{24} = 0 \)
• \( a_{31} = 2(3)-1 = 5 \), \( a_{32} = 4 \), \( a_{33} = 3 \), \( a_{34} = 2 \)

Therefore, \( A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix} \).

Solution:

(i) By equating corresponding elements:

\( y = 4 \), \( z = 3 \), \( x = 1 \).

(ii) By equating corresponding elements:

\( x+y = 6 \) ...(1)

\( 5+z = 5 \Rightarrow z = 0 \)

\( xy = 8 \) ...(2)

From (1), \( y = 6-x \). Substitute into (2):

\( x(6-x) = 8 \Rightarrow 6x - x^2 = 8 \Rightarrow x^2 - 6x + 8 = 0 \).

\( (x-4)(x-2) = 0 \Rightarrow x=4 \) or \( x=2 \).

If \( x=4, y=2 \). If \( x=2, y=4 \).

Therefore, \( x=4, y=2, z=0 \) or \( x=2, y=4, z=0 \).

(iii) By equating corresponding elements:

\( x+y+z = 9 \) ...(1)

\( x+z = 5 \) ...(2)

\( y+z = 7 \) ...(3)

Substitute (3) into (1): \( x + 7 = 9 \Rightarrow x = 2 \).

Substitute (2) into (1): \( 5 + y = 9 \Rightarrow y = 4 \).

Substitute values of x and y into (1): \( 2 + 4 + z = 9 \Rightarrow z = 3 \).

Therefore, \( x=2, y=4, z=3 \).

Solution:

Equating corresponding elements:

1. \( a-b = -1 \)
2. \( 2a-b = 0 \Rightarrow b = 2a \)
3. \( 2a+c = 5 \)
4. \( 3c+d = 13 \)

Substitute (2) into (1): \( a - 2a = -1 \Rightarrow -a = -1 \Rightarrow a = 1 \).

Then \( b = 2(1) = 2 \).

Substitute \( a \) into (3): \( 2(1) + c = 5 \Rightarrow c = 3 \).

Substitute \( c \) into (4): \( 3(3) + d = 13 \Rightarrow 9 + d = 13 \Rightarrow d = 4 \).

Therefore, \( a=1, b=2, c=3, d=4 \).

Answer: (C)

Solution:

A matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, \( m = n \).

Answer: (B)

Solution:

Equating corresponding elements:

\( 3x + 7 = 0 \Rightarrow x = -\frac{7}{3} \)

\( 2 - 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -\frac{2}{3} \)

Since we get two different values for \( x \), it is not possible for the matrices to be equal.

Answer: (D)

Solution:

A \( 3 \times 3 \) matrix has \( 3 \times 3 = 9 \) elements.

Each element can be either 0 or 1 (2 choices).

Total possible matrices = \( 2^9 = 512 \).