Answer
$y=\begin{pmatrix}
1\\
2\\
-5
\end{pmatrix}$
$x=\begin{pmatrix}
3\\
-1\\
-1
\end{pmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & -3 & 5\\
3&2&2\\
2 & 5 & 2
\end{bmatrix} $
$b=\begin{bmatrix}
1\\
5\\
-1
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1 & -3 & 5\\
3&2&2\\
2 & 5 & 2
\end{bmatrix} \approx^1\begin{bmatrix}
1 & -3 & 5\\
0&11&-13\\
0 & 11 & -8
\end{bmatrix} \approx^2 \approx\begin{bmatrix}
1 & -3 & 5\\
0&11&-13\\
0 & 0 &5
\end{bmatrix}=U$
$1.A_{12}(-3),A_{13}(-2)$
$2.A_{23}(-1)$
Corresponding multiplier: $m_{21}=3,m_{31}=2,m_{32}=1$
Consequently, we get $L=\begin{bmatrix}
1&0&0\\
3& 1 &0\\
2 & 1 & 1
\end{bmatrix}$
We now solve the two triangular systems $Ly = b$ and $Ux = y$. Using forward substitution on the first of these systems, we have:
$y_1=1$
$3y_1+y_2=5 \rightarrow 3.1+y_2=5 \rightarrow y_2=2$
$2y_1+y_2+y_3=-1 \rightarrow y_3=-5$
Solving $Ux = y$ via back substitution yields:
$5x_3=-5 \rightarrow x_3=-1$
$11x_2-13x_3=2=-7 \rightarrow x_2=-1$
$x_2-3x_2+5x_3=1 \rightarrow x_1=3$
The solution of the system are $y=\begin{pmatrix}
1\\
2\\
-5
\end{pmatrix}, x=\begin{pmatrix}
3\\
-1\\
-1
\end{pmatrix}$
