Construct ${a_{2 \times 2}}$ matrix, where
(i) ${a_{ij}} = \frac{{{{(i - 2j)}^2}}}{2}$
(ii) ${a_{ij}} = | - 2i + 3j|$

As we know, the notation, namely $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$

which indicates that A is a matrix of order $m \times n$,

also $1 \le i \le m,$ $1 \le j \le n$;$i,$ $j \in N$.

(i) Here, $A = {\left[ {{a_{ij}}} \right]_{2 \times 2}}$
$\Rightarrow$ $A = \frac{{{{(i - 2j)}^2}}}{2},1 \le i \le 2;1 \le j \le 2$

…..(i)
$\therefore$ ${a_{11}} = \frac{{{{(1 - 2)}^2}}}{2} = \frac{1}{2}$

${a_{12}} = \frac{{{{(1 - 2 \times 2)}^2}}}{2} = \frac{9}{2}$

${a_{21}} = \frac{{{{(2 - 2 \times 1)}^2}}}{2} = 0$
${a_{22}} = \frac{{{{(2 - 2 \times 2)}^2}}}{2} = 2$

Thus, $A = {\left[ {\begin{array}{cccccccccccccccccccc}{\frac{1}{2}}&{\frac{9}{2}}\\0&2\end{array}} \right]_{2 \times 2}}$

(ii) Here, $A = {\left[ {{a_{ij}}} \right]_{2 \times 2}} = | - 2i + 3j|$, $1 \le i \le 2$; $1 \le j \le 2$

$\therefore$ ${a_{11}} = | - 2 \times 1 + 3 \times 1| = 1$
${a_{12}} = | - 2 \times 1 + 3 \times 2| = 4$

${a_{21}} = | - 2 \times 2 + 3 \times 1| = 1$
${a_{22}} = | - 2 \times 2 + 3 \times 2| = 2$

$\therefore$ $A = {\left| {\begin{array}{llllllllllllllllllll}1&4\\1&2\end{array}} \right|_{2 \times 2}}$