Question 1, Exercise 2.2

Solutions of Question 1 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Construct a matrix $A=\left[a_{i j}\right]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{i+3 j}{2}$

Solution.

Given $a_{ij} = \dfrac{i + 3j}{2}$.

For $i = 1, j = 1$: $$a_{11} = \dfrac{1 + 3 \cdot 1}{2} = \dfrac{1 + 3}{2} = \dfrac{4}{2} = 2$$

For $i = 1, j = 2$: $$a_{12} = \dfrac{1 + 3 \cdot 2}{2} = \dfrac{1 + 6}{2} = \dfrac{7}{2}$$

For $i = 2, j = 1$: $$a_{21} = \dfrac{2 + 3 \cdot 1}{2} = \dfrac{2 + 3}{2} = \dfrac{5}{2}$$

For $i = 2, j = 2$: $$a_{22} = \dfrac{2 + 3 \cdot 2}{2} = \dfrac{2 + 6}{2} = \dfrac{8}{2} = 4$$

Using the calculated elements, the matrix $A$ is:

$$A = \left[\begin{array}{cc} 2 & \frac{7}{2} \\ \frac{5}{2} & 4 \end{array}\right]$$

Alternative Method:

Given $a_{ij}=\dfrac{i+3j}{2}$. So we have

$$\begin{align*} A&=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &=\begin{bmatrix}\frac{1+3(1)}{2} & \frac{1+3(2)}{2} \\ \frac{2+3(1)}{2} & \frac{2+3(2)}{2} \end{bmatrix} \\ &=\begin{bmatrix}2 & \frac{7}{2} \\ \frac{5}{2} & 4 \end{bmatrix} \end{align*}$$

Construct a matrix $A=\left[a_{i j}\right]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{i \times j}{2}$

Solution. Given $a_{ij}=\dfrac{i \times j}{2}$. So we have $$\begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &= \begin{bmatrix} \dfrac{1 \times 1}{2} & \dfrac{1 \times 2}{2} \\ \dfrac{2 \times 1}{2} & \dfrac{2 \times 2}{2} \end{bmatrix} \\ &= \begin{bmatrix} \dfrac{1}{2} & 1 \\ 1 & 2 \end{bmatrix} \end{align*}$$

Construct a matrix $A=\left[a_{i j}\right]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{i}{j}$

Solution.

Construct a matrix $A=\left[a_{ij}\right]$ of order $2 \times 2$ for which $a_{ij} = \dfrac{i}{j}$: $$\begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &= \begin{bmatrix} \dfrac{1}{1} & \dfrac{1}{2} \\ \dfrac{2}{1} & \dfrac{2}{2} \end{bmatrix} \\ &= \begin{bmatrix} 1 & \dfrac{1}{2} \\ 2 & 1 \end{bmatrix} \end{align*}$$

Construct a matrix $A=\left[a_{i j}\right]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{2 i-3 j}{3}$

Solution.

Given $a_{ij} = \frac{2i - 3j}{3}$, we need to find the matrix $A$:

$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

First, we calculate each element $a_{ij}$:

$$\begin{aligned} a_{11} &= \frac{2 \cdot 1 - 3 \cdot 1}{3} = \frac{2 - 3}{3} = \frac{-1}{3} = -\frac{1}{3}, \\ a_{12} &= \frac{2 \cdot 1 - 3 \cdot 2}{3} = \frac{2 - 6}{3} = \frac{-4}{3} = -\frac{4}{3}, \\ a_{21} &= \frac{2 \cdot 2 - 3 \cdot 1}{3} = \frac{4 - 3}{3} = \frac{1}{3}, \\ a_{22} &= \frac{2 \cdot 2 - 3 \cdot 2}{3} = \frac{4 - 6}{3} = \frac{-2}{3} = -\frac{2}{3}. \end{aligned}$$

Thus, the matrix $A$ is:

$$A = \begin{bmatrix} -\frac{1}{3} & -\frac{4}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix}$$

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