Construct a_ 2 2 matrix, where, 1. a_ij= (i-2j)^2 2 2. a_ij=|-2i+3j|

1. We have, A=[a_ij]_ 2 2 Such that, a_ij= (i-2j)^2 2 ; where 1 2;1 2 ....(i) a_ 11 =(1-2)^2 2 =1 2 a_ 12 =(1-2 2)^2 2 =9 2 a_ 21 =(2-2 1)^2 2 =0 a_ 22 =(2-2 2)^2 2 =2 So, A= 1 2 &9 2 0&2 2. We have, A=[a_ij]_ 2 2 Such that, a_ij=|-2i+3j|; where 1 2;1 2 a_ 11 =|-2 1+3 1|=1 a_ 12 =|-2 1+3 2|=4 a_ 21 =|-2 2+3 1|=-1 a_ 22 =|-2 2+3 2|=2 A= 1&4 -1&2

Construct a_ 2 2 matrix, where, 1. a_ij= (i-2j)^2 2 2. a_ij=|-2i+…

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Question

Construct $a_{2 \times 2}$ matrix, where,

$\text{a}_\text{ij}=\frac{(\text{i}-2\text{j})^2}{2}$
$\text{a}_\text{ij}=|-2\text{i}+3\text{j}|$

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Answer

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

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

Such that, $\text{a}_\text{ij}=|-2\text{i}+3\text{j}|;$ where $1\leq\text{i}\leq2;1\leq\text{j}\leq2$
$\therefore\ \text{a}_{11}=|-2\times1+3\times1|=1$
$\text{a}_{12}=|-2\times1+3\times2|=4$
$\text{a}_{21}=|-2\times2+3\times1|=-1$
$\text{a}_{22}=|-2\times2+3\times2|=2$
$\therefore\ \text{A}=\begin{bmatrix}1&4\\-1&2\end{bmatrix}$

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