MATRICES — Maths STD 12 Science — Question

1. We have, $\text{A}=[\text{a}_\text{ij}]_{2\times2}$

Such that, $\text{a}_\text{ij}=\frac{(\text{i}-2\text{j})^2}{2};$ where $1\leq\text{i}\leq2;1\leq\text{j}\leq2\ ....(\text{i})$

$\therefore\ \text{a}_{11}=\frac{(1-2)^2}{2}=\frac{1}{2}$

$\text{a}_{12}=\frac{(1-2\times2)^2}{2}=\frac{9}{2}$

$\text{a}_{21}=\frac{(2-2\times1)^2}{2}=0$

$\text{a}_{22}=\frac{(2-2\times2)^2}{2}=2$

So, $\text{A}=\begin{bmatrix}\frac{1}{2}&\frac{9}{2}\\0&2\end{bmatrix}$

2. We have, $\text{A}=[\text{a}_\text{ij}]_{2\times2}$

Such that, $\text{a}_\text{ij}=|-2\text{i}+3\text{j}|;$ where $1\leq\text{i}\leq2;1\leq\text{j}\leq2$

$\therefore\ \text{a}_{11}=|-2\times1+3\times1|=1$

$\text{a}_{12}=|-2\times1+3\times2|=4$

$\text{a}_{21}=|-2\times2+3\times1|=-1$

$\text{a}_{22}=|-2\times2+3\times2|=2$

$\therefore\ \text{A}=\begin{bmatrix}1&4\\-1&2\end{bmatrix}$