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:

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

Question 2 >