Answer
$x=\begin{pmatrix}
-\frac{677}{1300}\\
\frac{-9}{325}\\
\frac{-37}{65}\\
\frac{4}{13}
\end{pmatrix}, y=\begin{pmatrix}
2\\
-1\\
-1\\
4
\end{pmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
4 & 3 & 0 & 0\\
8&1&2&0\\
0 & 5& 3 &6\\
0 & 0 & -5 & 7
\end{bmatrix} $
$b=\begin{bmatrix}
2\\
3\\
0\\
5
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
4 & 3 & 0 & 0\\
8&1&2&0\\
0 & 5& 3 &6\\
0 & 0 & -5 & 7
\end{bmatrix} \approx^1\begin{bmatrix}
4 & 3 & 0 & 0\\
0&-5&2&0\\
0 & 5& 3 &6\\
0 & 0 & -5 & 7
\end{bmatrix} \approx^2\begin{bmatrix}
4 & 3 & 0 & 0\\
0&-5&2&0\\
0 &0& 5 &6\\
0 & 0 & -5 & 7
\end{bmatrix} \approx^3 \begin{bmatrix}
4 & 3 & 0 & 0\\
0&-5&2&0\\
0 & 5& 3 &6\\
0 & 0 & 0 & 13
\end{bmatrix} =U$
$1.A_{12}(-2)$
$2.A_{23}(1)$
$3. A_{34}(1)$
Corresponding multiplier: $m_{21}=2,m_{32}=-1,m_{43}=-1$
Consequently, we get $L=\begin{bmatrix}
1&0&0&0\\
2& 1 &0&0\\
0&-1 & 1 & 0\\
0 & 0 & -1 & 1
\end{bmatrix}$
We now solve the two triangular systems $Ly = b$ and $Ux = y$. Using forward substitution on the first of these systems, we have:
$y_1=2$
$2y_1+y_2=3 \rightarrow y_2=-1$
$-y_2+y_3=0\rightarrow y_3=-1 $
$-y_3+y_4=5\rightarrow y_4=4$
Solving $Ux = y$ via back substitution yields:
$13x_4=4 \rightarrow x_4=\frac{4}{13}$
$5x_3+6x_4=-1 \rightarrow x_3=\frac{-37}{65}$
$-5x_2+2x_3=-1 \rightarrow x_2=\frac{-9}{325}$
$4x_1+3x_2=2 \rightarrow x_1=-\frac{677}{1300}$
The solution of the system are $x=\begin{pmatrix}
-\frac{677}{1300}\\
\frac{-9}{325}\\
\frac{-37}{65}\\
\frac{4}{13}
\end{pmatrix}, y=\begin{pmatrix}
2\\
-1\\
-1\\
4
\end{pmatrix}$
