Answer:
Option C
Explanation:
We have
$A=\begin{bmatrix}1 & 2&3 \\1 & 1&5 \\ 2& 4& 7 \end{bmatrix}$
a31=2, a32=4, a33= 7
and $A_{31}=\begin{bmatrix}2 & 3 \\1 & 5 \end{bmatrix}=10-3=7$
$A_{32}=-\begin{bmatrix}1 & 3 \\1 & 5 \end{bmatrix}=-(5-3)=-2$
$A_{33}=\begin{bmatrix}1 & 2 \\1 & 1 \end{bmatrix}=1-2=-1$
$\therefore$ $a_{31}A_{31}+a_{32}A_{32}+a_{33}A_{33}$
= 2(7)+4(-2)+7(-1)
=14-8-7=-1