Answer
$x=\begin{pmatrix}
-\frac{1}{12}\\
\frac{1}{3}\\
\frac{1}{2}
\end{pmatrix}, y=\begin{pmatrix}
1\\
-3\\
-2
\end{pmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
2 & 2 & 1\\
6&3&-1\\
-4 & 2 & 2
\end{bmatrix} $
$b=\begin{bmatrix}
1\\
0\\
2
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
2 & 2 & 1\\
6&3&-1\\
-4 & 2 & 2
\end{bmatrix} \approx^1\begin{bmatrix}
2 & 2 & 1\\
0&-3&-4\\
0 & 6 &4
\end{bmatrix} \approx^2 \begin{bmatrix}
2 & 2 & 1\\
0&-3&-4\\
0 & 0 &-4
\end{bmatrix} =U$
$1.A_{12}(-3),A_{13}(2)$
$2.A_{23}(2)$
Corresponding multiplier: $m_{21}=3,m_{31}=-2,m_{32}=-2$
Consequently, we get $L=\begin{bmatrix}
1&0&0\\
3& 1 &0\\
-2&-2 & 1
\end{bmatrix}$
We now solve the two triangular systems $Ly = b$ and $Ux = y$. Using forward substitution on the first of these systems, we have:
$y_1=1$
$3y_1+y_2=0 \rightarrow 3.1+y_2=0 \rightarrow y_2=-3$
$-2y_1-2y_2+y_3=2 \rightarrow y_3=-2$
Solving $Ux = y$ via back substitution yields:
$-4x_3=-2 \rightarrow x_3=\frac{1}{2}$
$-3x_2-4x_3=-3 \rightarrow x_2=\frac{1}{3}$
$2x_1+2x_2+x_3=1 \rightarrow x_1=-\frac{1}{12}$
The solution of the system are $x=\begin{pmatrix}
-\frac{1}{12}\\
\frac{1}{3}\\
\frac{1}{2}
\end{pmatrix}, y=\begin{pmatrix}
1\\
-3\\
-2
\end{pmatrix}$
