Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
1 & 5\\
-1 &2
\end{vmatrix}=1.7=7$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
-3 & 0\\
-1 &2
\end{vmatrix}=(-1).(-6)=6$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
-3 & 0\\
1 &5
\end{vmatrix}=1.(-15)=-15$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
2 & 5\\
0 &2
\end{vmatrix}=(-1).4=-4$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
2 & 0\\
0 &2
\end{vmatrix}=1.4=4$
$C_{32}=(-1)^{3+2}.\begin{vmatrix}
2 & 0\\
2 &5
\end{vmatrix}=(-1).10=-10$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
2 & 1\\
0 &-1
\end{vmatrix}=1.(-2)=-2$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
2 & -3\\
0 &-1
\end{vmatrix}=(-1).(-2)=2$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
2 & -3\\
2&1
\end{vmatrix}=1.8=8$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(-1)^{i+j}.M_{ij}$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{31}C_{31}=2.7+2.6+0(-15)=26$
b)$M_C=\begin{bmatrix}
C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{32} \\
C_{31} & C_{32} & C_{33} \\
\end{bmatrix}=\begin{bmatrix}
7 & -4 & -2\\
6 & 4 & 2 \\
-15 & -10 & 8
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
7 & 6 & -15\\
-4 & 4 & -10 \\
-2 & 2 & 8
\end{bmatrix}$
d) Because $\det(A)=26 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{26}.\begin{bmatrix}
7 & 6 & -15\\
-4 & 4 & -10 \\
-2 & 2 & 8
\end{bmatrix}=\begin{bmatrix}
\frac{7}{26} & \frac{-2}{13} & \frac{-15}{26}\\
\frac{-2}{13} & \frac{2}{13} & \frac{-5}{13} \\
\frac{-1}{13} & \frac{1}{13} & \frac{4}{13}
\end{bmatrix}$
