Answer
$${A_{21}} = - 6,\,\,\,{A_{22}} = 0,\,\,\,{A_{23}} = - 6$$
Work Step by Step
$$\eqalign{
& \left[ {\matrix{
1 & 2 & { - 1} \cr
2 & 3 & { - 2} \cr
{ - 1} & 4 & 1 \cr
} } \right] \cr
& {\rm{The\, cofactor \,of\, }}{a_{ij}},{\rm{ written\, }}{A_{ij}},{\rm{ is \,defined \,as\, follows}} \cr
& {A_{ij}} = {\left( { - 1} \right)^{i + j}} \cdot {M_{ij}} \cr
& {\rm{The\, cofactors \,in \,the\, second\, row\, are\, }}{A_{21}},\,\,\,{A_{22}},{\rm{ }}{A_{23}} \cr
& {A_{21}} = {\left( { - 1} \right)^{2 + 1}} \cdot {M_{21}} \cr
& {A_{21}} = \left( { - 1} \right)\left| {\matrix{
2 & { - 1} \cr
4 & 1 \cr
} } \right| \cr
& {A_{21}} = - \left( {2 + 4} \right) \cr
& {A_{21}} = - 6 \cr
& \cr
& {A_{22}} = {\left( { - 1} \right)^{2 + 2}} \cdot {M_{21}} \cr
& {A_{22}} = \left| {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right| \cr
& {A_{22}} = 1 - 1 \cr
& {A_{22}} = 0 \cr
& \cr
& {A_{23}} = {\left( { - 1} \right)^{2 + 3}} \cdot {M_{21}} \cr
& {A_{23}} = \left( { - 1} \right)\left| {\matrix{
1 & 2 \cr
{ - 1} & 4 \cr
} } \right| \cr
& {A_{23}} = - \left( {4 + 2} \right) \cr
& {A_{23}} = - 6 \cr
& \cr
& {\rm{The\, cofactors\, of \,the \,second\, row \,are:}} \cr
& {A_{21}} = - 6,\,\,\,{A_{22}} = 0,\,\,\,{A_{23}} = - 6 \cr} $$
