Question
construct a $2×2$ matrix $A = [a_{ij}]$ whose elemants $a_{ij}$ are given by $a_{ij}$ $=\begin{cases}\frac{|-3\text{i}+\text{j}|}{2},&\text{if i }\neq\text{j}\\{i}+\text{j})^2,&\text{if i }=\text{j}\end{cases}$

Answer

Let us the matrix be $\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22} \end{bmatrix}$ For the entries $a_{12}$ and $a_{21}$ we have i $\neq$ j, so by the Condition we have $\text{a}_{12}=\frac{|-3\text{i}+\text{j}|}{2},\text{ a}_{21}=\frac{|-3\text{i}+\text{j}|}{2}$ $\Rightarrow\text{a}_{12}=\frac{|-3+2|}{2},\text{ a}_{21}=\frac{|-3.2+1|}{2}$ $\Rightarrow\text{a}_{12}=\frac{|-1|}{2},\text{ a}_{21}=\frac{|-5|}{2}$ $\Rightarrow\text{a}_{12}=\frac{1}{2},\text{ a}_{21}=\frac{5}{2}$ For the entries $a_{11}$ and $a_{22}$ we have i = j, so by the given condition we have $\Rightarrow\text{a}_{11}=(\text{i + j})^2, \text{ a}_{22}=(\text{i + j})^2$ $\Rightarrow\text{a}_{11}=(1+1)^2,\text{ a}_{22}=(2+2)^2$ $\Rightarrow\text{a}_{11}=4,\text{ a}_{22}=16$ So, $\text{A}=\begin{bmatrix}4&\frac{1}{2}\\\frac{5}{2}&16 \end{bmatrix}$

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