Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
2 &1\\
7 &-1
\end{vmatrix}=1.(-9)=-9$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
-3 & 5\\
7&-1
\end{vmatrix}=(-1).(-32)=32$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
-3&5\\
2&1
\end{vmatrix}=1.(-10)=-10$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
1 & 1\\
0&-1
\end{vmatrix}=(-1).(-1)=1$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
2 & 5\\
0 &-1
\end{vmatrix}=1.(-2)=-2$
$C_{32}=(-1)^{3+2}.\begin{vmatrix}
2 & 5\\
1&1
\end{vmatrix}=(-1).(-3)=3$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
1 &2\\
0&7
\end{vmatrix}=1.7=7$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
2&-3\\
0 &7
\end{vmatrix}=(-1).14=-14$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
2& -3\\
1&2
\end{vmatrix}=1.7=7$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(-1)^{i+j}.M_{ij}$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{31}C_{31}=2.(-9)+1.32+0.(-13)=14$
b)$M_C=\begin{bmatrix}
C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{32} \\
C_{31} & C_{32} & C_{33} \\
\end{bmatrix}=\begin{bmatrix}
-9 &1 &7\\
32& -2&-14\\
-13 & 3 & 7
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
-9 & 32 &-13\\
1 & -2 & 3 \\
7 &-14& 7
\end{bmatrix}$
d) Because $\det(A)=6 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{14}.\begin{bmatrix}
-9 & 32 &-13\\
1 & -2 & 3 \\
7 &-14& 7
\end{bmatrix}=\begin{bmatrix}
\frac{-9}{14} & \frac{16}{7} & \frac{-13}{14}\\
\frac{1}{14} & \frac{-1}{7} & \frac{3}{14} \\
\frac{1}{2} &-1& \frac{1}{2}
\end{bmatrix}$
