Answer
$x=\begin{bmatrix}
-10\\
18\\
2
\end{bmatrix}$
Work Step by Step
Write the system in the matrix form:
$\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
2 &-1 & 1
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2 \\
x_3
\end{bmatrix}=\begin{bmatrix}
12\\
24\\
-36
\end{bmatrix}$
Find the inverse for matrix $A=\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
2 &-1 & 1
\end{bmatrix}$:
$\begin{bmatrix}
1 & 1 & 2| 1& 0&0\\
1 & 2 & -1|0 & 1 &0\\
2 &-1 & 1 | 0 & 0 &1
\end{bmatrix} \approx^1\begin{bmatrix}
1 & 1 & 2| 1& 0&0\\
0 & 1 & -3|-1 & 1 &0\\
0 &-3 & -3 | -2 & 0 &1
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & 1 & 2| 1& 0&0\\
0 & 1 & -3|-1 & 1 &0\\
0 &0 & -12 | -5 &3 &1
\end{bmatrix} \approx^3\begin{bmatrix}
1 & 0 & 5| 2& -1&0\\
0 & 1 & -3|-1 & 1 &0\\
0 &0 & 1 | \frac{5}{12} &-\frac{1}{4} &-\frac{1}{12}
\end{bmatrix} \approx^4\begin{bmatrix}
1 & 0 & 0| -\frac{1}{12}& \frac{1}{4}&\frac{5}{12}\\
0 & 1 &0|\frac{1}{4} & \frac{1}{4} &-\frac{1}{4}\\
0 &0 & 1 | \frac{5}{12} &-\frac{1}{4} &-\frac{1}{12}
\end{bmatrix}$
Hence, $A^{-1}=\begin{bmatrix}
-\frac{1}{12}&\frac{1}{4}& \frac{5}{12}\\
\frac{1}{4} & \frac{1}{4} & -\frac{1}{4}\\
\frac{5}{12} & -\frac{1}{4} & -\frac{1}{12}
\end{bmatrix}$
Using $x=A^{-1}b$ to find x:
$x=\begin{bmatrix}
-\frac{1}{12}&\frac{1}{4}& \frac{5}{12}\\
\frac{1}{4} & \frac{1}{4} & -\frac{1}{4}\\
\frac{5}{12} & -\frac{1}{4} & -\frac{1}{12}
\end{bmatrix}.\begin{bmatrix}
12\\
24 \\
-36
\end{bmatrix}=\begin{bmatrix}
-10\\
18\\
2
\end{bmatrix}$
