Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
1 & 5\\
2 &3
\end{vmatrix}=1.(-7)=-7$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
3 & -1\\
2&3
\end{vmatrix}=(-1).11=-11$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
3 &-1\\
3&5
\end{vmatrix}=1.16=16$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
2 & 5\\
0 &3
\end{vmatrix}=(-1).6=-6$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
-2 & -1\\
0 &3
\end{vmatrix}=1.(-6)=-6$
$C_{32}=(-1)^{3+2}.\begin{vmatrix}
-2 & -1\\
2 &5
\end{vmatrix}=(-1).8=-8$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
2 & 1\\
0 &2
\end{vmatrix}=1.4=4$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
-2 & 3\\
0 &0
\end{vmatrix}=(-1).(-4)=4$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
-2 & 3\\
2&1
\end{vmatrix}=1.4=4$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(-1)^{i+j}.M_{ij}$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{31}C_{31}=-2.(-7)+2.(-1)+0.16=12$
b)$M_C=\begin{bmatrix}
C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{32} \\
C_{31} & C_{32} & C_{33} \\
\end{bmatrix}=\begin{bmatrix}
-7 & -6 & 4\\
-11 & -6 & 4 \\
16 & 8 & 4
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
-7 & -11 &16\\
-6 & -6 & 8 \\
4 & 4 & 4
\end{bmatrix}$
d) Because $\det(A)=12 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{12}.\begin{bmatrix}
-7 & -11 &16\\
-6 & -6 & 8 \\
4 & 4 & 4
\end{bmatrix}=\begin{bmatrix}
\frac{-7}{12} & \frac{-11}{12} & \frac{4}{3}\\
\frac{-1}{2} & \frac{-1}{2} & \frac{2}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{bmatrix}$
