Question

Using coafactors of the elements of second row, evaluate
$\Delta =\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$

Answer 1

Given $\Delta =\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$
Coafactors of the elements of second row
$A_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}3& 8\\ 2& 3\end{array}\right|=-(9-16)=7[\because A_{ij}=(-1{)}^{i+j}{M}_{ij}]A_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}5& 8\\ 1& 3\end{array}\right|=15-8=7$
and $A_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}5& 3\\ 1& 2\end{array}\right|=-(10-3)=-7$
Now, expansion of $\Delta$ using cofactors of elements of second row is given by
$\Delta =a_{21}{A}_{21}+a_{22}{A}_{22}+a_{23}{A}_{23}=2\times 7+0\times 7+1(-7)=14-7=7$