Given \[{{a}_{i\text{ }j}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| -3i\text{ }+\text{ }j \right|\]
Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 3}}\]
So, the elements in a \[3\times 4\]matrix are
\[{{a}_{11}},\text{ }{{a}_{12}},\text{ }{{a}_{13}},\text{ }{{a}_{14,}}~{{a}_{21}},\text{ }{{a}_{22}},\text{ }{{a}_{23}},\text{ }{{a}_{24}},\text{ }{{a}_{31,~}}{{a}_{32,~}}{{a}_{33,~}}{{a}_{34}}\]
\[A\text{ }=\]
\[{{a}_{11}}~=\]
\[{{a}_{12}}~=\]
\[{{a}_{13}}~=\]
\[{{a}_{14}}~=\]
\[{{a}_{21}}~=\]
\[{{a}_{22}}~=\]
\[{{a}_{23}}~=\]
\[{{a}_{24\text{ }=}}\]
\[{{a}_{31}}~=\]
\[{{a}_{32}}~=\]
\[{{a}_{33}}~=\]
\[{{a}_{34}}~=\]
Substituting these values in matrix A we get,
\[A\text{ }=\]
Multiplying by negative sign we get,