(i) Given \[{{a}_{i\text{ }j}}~=\text{ }i\text{ }\times \text{ }j\]

Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\text{ }\times \text{ }3}}\]

So, the elements in a \[2\text{ }\times \text{ }3\]matrix are

\[\begin{array}{*{35}{l}}

[{{a}_{11}},\text{ }{{a}_{12}},\text{ }{{a}_{13}},\text{ }{{a}_{21}},\text{ }{{a}_{22}},\text{ }{{a}_{23}}]  \\

~  \\

\end{array}\]

\[{{a}_{11}}~=\text{ }1\text{ }\times \text{ }1\text{ }=\text{ }1\]

\[{{a}_{12}}~=\text{ }1\text{ }\times \text{ }2\text{ }=\text{ }2\]

\[{{a}_{13}}~=\text{ }1\text{ }\times \text{ }3\text{ }=\text{ }3\]

\[{{a}_{21}}~=\text{ }2\text{ }\times \text{ }1\text{ }=\text{ }2\]

\[{{a}_{22}}~=\text{ }2\text{ }\times \text{ }2\text{ }=\text{ }4\]

\[{{a}_{23}}~=\text{ }2\text{ }\times \text{ }3\text{ }=\text{ }6\]

Substituting these values in matrix A we get,

(ii) Given \[{{a}_{i\text{ }j~}}=\text{ }2i\text{ }-\text{ }j\]

Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 3}}\]

So, the elements in a \[2\text{ }\times \text{ }3\] matrix are

\[{{a}_{11}},\text{ }{{a}_{12}},\text{ }{{a}_{13}},\text{ }{{a}_{21}},\text{ }{{a}_{22}},\text{ }{{a}_{23}}\]

\[{{a}_{11}}~=\text{ }2\text{ }\times \text{ }1\text{ }-\text{ }1\text{ }=\text{ }2\text{ }-\text{ }1\text{ }=\text{ }1\]

\[{{a}_{12}}~=\text{ }2\text{ }\times \text{ }1\text{ }-\text{ }2\text{ }=\text{ }2\text{ }-\text{ }2\text{ }=\text{ }0\]

\[{{a}_{13}}~=\text{ }2\text{ }\times \text{ }1\text{ }-\text{ }3\text{ }=\text{ }2\text{ }-\text{ }3\text{ }=\text{ }\text{ }1\]

\[{{a}_{21}}~=\text{ }2\text{ }\times \text{ }2\text{ }-\text{ }1\text{ }=\text{ }4\text{ }-\text{ }1\text{ }=\text{ }3\]

\[{{a}_{22}}~=\text{ }2\text{ }\times \text{ }2\text{ }-\text{ }2\text{ }=\text{ }4\text{ }-\text{ }2\text{ }=\text{ }2\]

\[{{a}_{23}}~=\text{ }2\text{ }\times \text{ }2\text{ }-\text{ }3\text{ }=\text{ }4\text{ }-\text{ }3\text{ }=\text{ }1\]

Substituting these values in matrix A we get,