CBSE Class 12 Maths Chapter 3: Matrices NCERT Solutions

The order of a matrix is defined by the number of rows ($m$) and the number of columns ($n$), denoted as $m \times n$. The total number of elements in a matrix is the product of its order, i.e., $m \times n$.

Given that a matrix has 28 elements, we need to find pairs of natural numbers ($m, n$) such that $m \times n = 28$. The pairs of factors for 28 are:

$$m\; \backslash times\; n\; =\; 28$$

The possible pairs $(m, n)$ are $(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)$.

Therefore, the possible orders for a matrix with 28 elements are $1 \times 28$, $2 \times 14$, $4 \times 7$, $7 \times 4$, $14 \times 2$, and $28 \times 1$.

Now, consider a matrix with 13 elements. We need to find pairs of natural numbers ($m, n$) such that $m \times n = 13$. Since 13 is a prime number, its only factors are 1 and 13.

$$m\; \backslash times\; n\; =\; 13$$

The possible pairs $(m, n)$ are $(1, 13)$ and $(13, 1)$.

Hence, the possible orders for a matrix with 13 elements are $1 \times 13$ and $13 \times 1$.

The given matrix is $A = \begin{bmatrix} a & 1 & x \ 2 & \sqrt{3} & x^2 - y \ 0 & 5 & \frac{-2}{5} \end{bmatrix}$.

(i) To find the order of the matrix, we count the number of rows and columns. There are 3 rows and 3 columns in matrix $A$. Therefore, the order of the matrix $A$ is $3 \times 3$.

(ii) The number of elements in a matrix of order $m \times n$ is $m \times n$. For matrix $A$, the order is $3 \times 3$. So, the number of elements is $3 \times 3 = 9$.

(iii) The element $a_{ij}$ represents the element in the $i$-th row and $j$-th column. We need to find $a_{23}$, $a_{31}$, and $a_{12}$.

• $a_{23}$: This is the element in the 2nd row and 3rd column. Looking at the matrix, the element is $x^2 - y$.
• $a_{31}$: This is the element in the 3rd row and 1st column. Looking at the matrix, the element is $0$.
• $a_{12}$: This is the element in the 1st row and 2nd column. Looking at the matrix, the element is $1$.

So, $a_{23} = x^2 - y$, $a_{31} = 0$, and $a_{12} = 1$.

We need to construct a $2 \times 2$ matrix $A = [a_{ij}]$, where $1 \le i \le 2$ and $1 \le j \le 2$. The general form of a $2 \times 2$ matrix is $A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$.

(i) Given $a_{ij} = \frac{(i-2j)^2}{2}$

We calculate each element:

• $a_{11} = \frac{(1 - 2 \times 1)^2}{2} = \frac{(1 - 2)^2}{2} = \frac{(-1)^2}{2} = \frac{1}{2}$
• $a_{12} = \frac{(1 - 2 \times 2)^2}{2} = \frac{(1 - 4)^2}{2} = \frac{(-3)^2}{2} = \frac{9}{2}$
• $a_{21} = \frac{(2 - 2 \times 1)^2}{2} = \frac{(2 - 2)^2}{2} = \frac{0^2}{2} = 0$
• $a_{22} = \frac{(2 - 2 \times 2)^2}{2} = \frac{(2 - 4)^2}{2} = \frac{(-2)^2}{2} = \frac{4}{2} = 2$

Therefore, the matrix is $A = \begin{bmatrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{bmatrix}$.

(ii) Given $a_{ij} = |-2i + 3j|$

We calculate each element:

• $a_{11} = |-2 \times 1 + 3 \times 1| = |-2 + 3| = |1| = 1$
• $a_{12} = |-2 \times 1 + 3 \times 2| = |-2 + 6| = |4| = 4$
• $a_{21} = |-2 \times 2 + 3 \times 1| = |-4 + 3| = |-1| = 1$
• $a_{22} = |-2 \times 2 + 3 \times 2| = |-4 + 6| = |2| = 2$

Therefore, the matrix is $A = \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$.

We need to construct a $3 \times 2$ matrix $A = [a_{ij}]$, where $1 \le i \le 3$ and $1 \le j \le 2$. The general form of a $3 \times 2$ matrix is $A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \ a_{31} & a_{32} \end{bmatrix}$.

The formula for the elements is given by $a_{ij} = e^{ix} \sin jx$. We calculate each element:

• $a_{11} = e^{1 \cdot x} \sin(1 \cdot x) = e^x \sin x$
• $a_{12} = e^{1 \cdot x} \sin(2 \cdot x) = e^x \sin 2x$
• $a_{21} = e^{2 \cdot x} \sin(1 \cdot x) = e^{2x} \sin x$
• $a_{22} = e^{2 \cdot x} \sin(2 \cdot x) = e^{2x} \sin 2x$
• $a_{31} = e^{3 \cdot x} \sin(1 \cdot x) = e^{3x} \sin x$
• $a_{32} = e^{3 \cdot x} \sin(2 \cdot x) = e^{3x} \sin 2x$

Substituting these values into the matrix form, we get: