Answer
See answers below
Work Step by Step
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}$
with $C_{11}=(-1)^{1+1}.1=1$
$C_{12}=(-1)^{1+2}.4=-4$
$\det(A)=a_{11}C_{11}+a_{12}C_{12}=(-1).1+(-2).(-4)=7$
b) $C_{21}=(-1)^{2+1}.(-2)=2$
$C_{22}=(-1)^{2+2}.(-1)=-1$
then $M_C=\begin{bmatrix}
C_{11} & C_{12}\\
C_{21} & C_{21}
\end{bmatrix}=\begin{bmatrix}
1 &-4\\
2 &-1
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
1 & 2\\
-4 & -1
\end{bmatrix}$
d) Because $\det(A)=7 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{7}.\begin{bmatrix}
1 & 2\\
-4 & -1
\end{bmatrix}=\begin{bmatrix}
\frac{1}{7} & \frac{2}{7}\\
\frac{-4}{7} & \frac{-1}{7}
\end{bmatrix}$
