If $A=\left[\begin{array}{ccc}5 & 6 & 3 \\ -4 & 3 & 2 \\ -4 & -7 & 3\end{array}\right]$, then cofactors of all elements of second row are respectively.
$-39,3,11$
$-39,27,11$
$39,-3-11$
$-39,-27,11$
Solution
$A=\left[\begin{array}{ccc}
5 & 6 & 3 \\
-4 & 3 & 2 \\
-4 & -7 & 3
\end{array}\right]$
Cofactors of elements in second row are
Cofactor of -4
Cofactor of 3
$\begin{aligned}
& =(-1)^{2+1}\left|\begin{array}{cc}
6 & 3 \\
-7 & 3
\end{array}\right|=-(18+21)=-39 \\
& =(-1)^{2+2}\left|\begin{array}{cc}
5 & 3 \\
-4 & 3
\end{array}\right|=15+12=27 \\
& =(-1)^{2+3}\left|\begin{array}{cc}
5 & 6 \\
-4 & -7
\end{array}\right|=-(-35+24)=11
\end{aligned}$