Answer
See explanation
Work Step by Step
\[
L=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-3 & 1 & 0 \\
4 & -1 & 1
\end{array}\right]
\]
Lower Triangular matrix
\[
U=\left[\begin{array}{ccc}
2 & -1 & 2 \\
0 & -3 & 4 \\
0 & 0 & 1
\end{array}\right]
\]
Upper Triangular Matrix
\[
\mathbf{b}=\left[\begin{array}{l}
1 \\
0 \\
4
\end{array}\right]
\]
$\mathbf{A x}=\mathbf{b}$
\[
L \mathrm{b}]=\left[\begin{array}{cccc}
1 & 0 & 0 & 1 \\
-3 & 1 & 0 & 0 \\
4 & -1 & 1 & 4
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 3 \\
0 & -1 & 1 & 0
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 3
\end{array}\right]
\]
\[
\begin{array}{c}
y=\left[\begin{array}{c}
1 \\
3 \\
3
\end{array}\right] \\
{\left[\begin{array}{cc}
U & \mathbf{y}
\end{array}\right]=\left[\begin{array}{cccc}
2 & -1 & 2 & 1 \\
0 & -3 & 4 & 3 \\
0 & 0 & 1 & 3
\end{array}\right]=\left[\begin{array}{cccc}
2 & -1 & 0 & -5 \\
0 & -3 & 0 & -9 \\
0 & 0 & 1 & 3
\end{array}\right]}
\end{array}
\]
Next solve $U x=y,$ using back-substitution
\[
\left[\begin{array}{cc}
U & \mathbf{y}
\end{array}\right]=\left[\begin{array}{cccc}
2 & -1 & 0 & -5 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 3
\end{array}\right]=\left[\begin{array}{cccc}
2 & 0 & 0 & -2 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 3
\end{array}\right]
\]
Reducing to echelon form
\[
\mathbf{x}=\left[\begin{array}{c}
-1 \\
3 \\
3
\end{array}\right]
\]
