Answer
See answers below
Work Step by Step
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}$
with $C_{11}=(-1)^{1+1}.5=5$
$C_{12}=(-1)^{1+2}.4=-4$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}=3.5+1.(-4)=11$
b) $C_{21}=(-1)^{2+1}.1=-1$
$C_{22}=(-1)^{2+2}.3=3$
then $M_C=\begin{bmatrix}
C_{11} & C_{12}\\
C_{21} & C_{21}
\end{bmatrix}=\begin{bmatrix}
5 &-4\\
-1 &3
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
5 & -1\\
-4 & 3
\end{bmatrix}$
d) Because $\det(A)=11 \ne 0$ we are able to find $A^{-1}$ by using theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{11}.\begin{bmatrix}
5 & -1\\
-4 & 3
\end{bmatrix}=\begin{bmatrix}
\frac{5}{11} & \frac{-1}{11}\\
\frac{-4}{11} & \frac{3}{11}
\end{bmatrix}$
