Answer
See answers below
Work Step by Step
$C_{11}=(-1)^{1+1}.\begin{vmatrix}
1 & -1 &1\\
1 & -1 & -1\\
1 & 1 &-1
\end{vmatrix}=1.1.2=2$
$C_{21}=(-1)^{2+1}.\begin{vmatrix}
1 & 1 &1\\
1 & -1 & -1\\
1 & 1 &-1
\end{vmatrix}=(-1).1.2=-2$
$C_{31}=(-1)^{3+1}.\begin{vmatrix}
1 & 1 &1\\
1 & -1 & 1\\
1 & 1 &-1
\end{vmatrix}=1.1.0=0$
$C_{41}=(-1)^{4+1}.\begin{vmatrix}
1 & 1 &1\\
1 & -1 & -1\\
1 & 1 &-1
\end{vmatrix}=(-1).1.2=-2$
$C_{12}=(-1)^{1+2}.\begin{vmatrix}
1 & 1 &1\\
1 & -1 & -1\\
1 & 1 &-1
\end{vmatrix}=(-1).1.2=-2$
$C_{22}=(-1)^{2+2}.\begin{vmatrix}
1 & 1 &1\\
1 & -1 & -1\\
-1 & 1 &-1
\end{vmatrix}=1.1.2=2$
$C_{42}=(-1)^{4+2}.\begin{vmatrix}
1 & 1 &1\\
-1 & -1 & 1\\
1 & -1 &-1
\end{vmatrix}=1.1.2=2$
$C_{13}=(-1)^{1+3}.\begin{vmatrix}
-1 & 1 &1\\
1 & 1 & -1\\
-1 & 1 &-1
\end{vmatrix}=(-1).1.0=0$
$C_{32}=(-1)^{2+3}.\begin{vmatrix}
1 & 1 &1\\
-1 &-1 & 1\\
-1 & 1 &-1
\end{vmatrix}=(-1).1.0=0$
$C_{23}=(-1)^{2+3}.\begin{vmatrix}
1 & 1 &1\\
1 & 1 & -1\\
-1 & 1 &-1
\end{vmatrix}=(-1).1.0=0$
$C_{33}=(-1)^{3+3}.\begin{vmatrix}
1 & 1 &1\\
-1 & 1 & 1\\
-1 & 1 &-1
\end{vmatrix}=(-2).1.1=-2$
$C_{43}=(-1)^{4+3}.\begin{vmatrix}
1 & 1 &1\\
-1 & 1 & 1\\
1 & 1 &-1
\end{vmatrix}=(-1).1.(-2)=2$
$C_{14}=(-1)^{4+1}.\begin{vmatrix}
-1 & 1 &-1\\
1 & 1 & -1\\
-1 & 1 &1
\end{vmatrix}=(-)1.(-1).2=2$
$C_{24}=(-1)^{2+4}.\begin{vmatrix}
1 & 1 &1\\
1 & 1 & -1\\
-1 & 1 &1
\end{vmatrix}=1.1.2=2$
$C_{3}=(-1)^{3+4}.\begin{vmatrix}
1 & 1 &1\\
-1 & 1 & -1\\
-1 & 1 &1
\end{vmatrix}=(-1).1.2=-2$
$C_{44}=(-1)^{4+4}.\begin{vmatrix}
1 & 1 &1\\
-1 & 1 & -1\\
1 & 1 &-1
\end{vmatrix}=1.1.0=0$
a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}+a_{41}C_{41}=1.2-1(-2)+1.0-1.(-2)=6$
b)$M_C=\begin{bmatrix}
2 &-2 & 0 &2\\
-2& 2&0&2\\
0 &0 & -2 &-2\\
-2 &2 &2 & 0
\end{bmatrix}$
c) $adj(A)=\begin{bmatrix}
2 &-2 & 0 &-2\\
-2& 2&0&2\\
0 &0 & -2 &2\\
-2 &2 &2 & 0
\end{bmatrix}$
d) Because $\det(A)=6 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{6}.\begin{bmatrix}
2 &-2 & 0 &-2\\
-2& 2&0&2\\
0 &0 & -2 &2\\
-2 &2 &2 & 0
\end{bmatrix}=\begin{bmatrix}
\frac{1}{3} & \frac{-1}{3} & 0 & \frac{-1}{3}\\
\frac{-1}{3} & \frac{1}{3} & 0 & \frac{1}{3}\\
0 &0& \frac{-1}{3} & \frac{1}{3}\\
\frac{1}{3} & \frac{1}{3} & \frac{-1}{3} &0
\end{bmatrix}$
