Answer
$${A_{21}} = 8,\,\,\,{A_{22}} = 16,\,\,\,{A_{23}} = 0$$
Work Step by Step
$$\eqalign{
& \left[ {\matrix{
2 & { - 1} & 4 \cr
3 & 0 & 1 \cr
{ - 2} & 1 & 4 \cr
} } \right] \cr
& {\rm{The\, cofactor\, of\, }}{a_{ij}},{\rm{ written\, }}{A_{ij}},{\rm{ is\, defined \,as\, follows}} \cr
& {A_{ij}} = {\left( { - 1} \right)^{i + j}} \cdot {M_{ij}} \cr
& {\rm{The\, cofactors\, in \,the\, second \,row\, are\, }}{A_{21}},\,\,\,{A_{22}},{\rm{ }}{A_{23}} \cr
& {A_{21}} = {\left( { - 1} \right)^{2 + 1}} \cdot {M_{21}} \cr
& {A_{21}} = \left( { - 1} \right)\left| {\matrix{
{ - 1} & 4 \cr
1 & 4 \cr
} } \right| \cr
& {A_{21}} = - \left( { - 4 - 4} \right) \cr
& {A_{21}} = 8 \cr
& \cr
& {A_{22}} = {\left( { - 1} \right)^{2 + 2}} \cdot {M_{21}} \cr
& {A_{22}} = \left| {\matrix{
2 & 4 \cr
{ - 2} & 4 \cr
} } \right| \cr
& {A_{22}} = 8 + 8 \cr
& {A_{22}} = 16 \cr
& \cr
& {A_{23}} = {\left( { - 1} \right)^{2 + 3}} \cdot {M_{21}} \cr
& {A_{23}} = \left( { - 1} \right)\left| {\matrix{
2 & { - 1} \cr
{ - 2} & 1 \cr
} } \right| \cr
& {A_{23}} = - \left( {2 - 2} \right) \cr
& {A_{23}} = 0 \cr
& \cr
& {\rm{The\, cofactors\, of\, the\, second\, row \,are:}} \cr
& {A_{21}} = 8,\,\,\,{A_{22}} = 16,\,\,\,{A_{23}} = 0 \cr} $$
