Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
-1 & 4\\
1 &7
\end{vmatrix}=1.(-11)=-11$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
-1 & 2\\
1&7
\end{vmatrix}=(-1).(-9)=9$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
-1 &2\\
-1&4
\end{vmatrix}=1.(-2)=-2$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
3 & 4\\
5&7
\end{vmatrix}=(-1).1=-1$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
1 & 2\\
5 &7
\end{vmatrix}=1.(-3)=-3$
$C_{32}=(-1)^{3+2}.\begin{vmatrix}
1 & 2\\
3&4
\end{vmatrix}=(-1).(-2)=2$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
3 & -1\\
5&1
\end{vmatrix}=1.8=8$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
1 & -1\\
5 &1
\end{vmatrix}=(-1).6=-6$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
1 & -1\\
3&-1
\end{vmatrix}=1.2=2$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(-1)^{i+j}.M_{ij}$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{31}C_{31}=1.(-11)+3.9+5.(-2)=6$
b)$M_C=\begin{bmatrix}
C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{32} \\
C_{31} & C_{32} & C_{33} \\
\end{bmatrix}=\begin{bmatrix}
-11 & -1 & 9\\
9& -3 & -6 \\
-2 & 2 & 2
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
-11 & 9 &-2\\
-1 & -3 & 2 \\
8 & -6& 2
\end{bmatrix}$
d) Because $\det(A)=6 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{6}.\begin{bmatrix}
-11 & 9 &-2\\
-1 & -3 & 2 \\
8 & -6& 2
\end{bmatrix}$=\begin{bmatrix}
\frac{-11}{6} & \frac{3}{2} & \frac{-1}{3}\\
\frac{-1}{6} & \frac{-1}{2} & \frac{1}{3} \\
\frac{4}{3} &-1& \frac{1}{3}
\end{bmatrix}$
