Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
-1 & 3\\
-2 &1
\end{vmatrix}=1.5=5$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
1 & 2\\
-2&1
\end{vmatrix}=(-1).5=-5$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
1 &2\\
-1&3
\end{vmatrix}=1.5=5$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
-1 & 3\\
1&1
\end{vmatrix}=(-1).(-4)=4$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
0 & 2\\
1 &1
\end{vmatrix}=1.(-2)=-2$
$C_{32}=(-1)^{3+2}.\begin{vmatrix}
0 & 2\\
-1&3
\end{vmatrix}=(-1).2=-2$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
-1 & -1\\
1&-2
\end{vmatrix}=1.3=3$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
0 & 1\\
1 &-2
\end{vmatrix}=(-1).(-1)=1$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
0 & 1\\
-1&-1
\end{vmatrix}=1.1=1$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(-1)^{i+j}.M_{ij}$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{31}C_{31}=0.5+(-1)(-5)+5.1=10$
b)$M_C=\begin{bmatrix}
C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{32} \\
C_{31} & C_{32} & C_{33} \\
\end{bmatrix}=\begin{bmatrix}
5 &4 & 3\\
-5& -2&1\\
5 & -2 & 1
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
5 & -5 &5\\
4 & -2 & -2 \\
3 &1& 1
\end{bmatrix}$
d) Because $\det(A)=6 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{10}.\begin{bmatrix}
5 & -5 &5\\
4 & -2 & -2 \\
3 &1& 1
\end{bmatrix}=\begin{bmatrix}
\frac{1}{2} & \frac{-1}{2} & \frac{1}{2}\\
\frac{2}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{3}{10} &\frac{1}{10}& \frac{1}{10}
\end{bmatrix}$
