Answer
See solution
Work Step by Step
\[
L=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-1 & 1 & 0 \\
2 & 0 & 1
\end{array}\right]
\]
Lower Triangular matrix
\[
U=\left[\begin{array}{ccc}
4 & 3 & -5 \\
0 & -2 & 2 \\
0 & 0 & 2
\end{array}\right]
\]
Upper Triangular Matrix
\[
\mathbf{b}=\left[\begin{array}{c}
2 \\
-4 \\
6
\end{array}\right]
\]
Given b for $\mathbf{A x}=\mathbf{b}$
\[
L \quad \mathbf{b}]=\left[\begin{array}{cccc}
1 & 0 & 0 & 2 \\
-1 & 1 & 0 & -4 \\
2 & 0 & 1 & 6
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 2
\end{array}\right]
\]
\[
\begin{array}{c}
y=\left[\begin{array}{c}
2 \\
-2 \\
2
\end{array}\right] \\
{\left[\begin{array}{cc}
U \quad \mathrm{y}
\end{array}\right]=\left[\begin{array}{cccc}
4 & 3 & -5 & 2 \\
0 & -2 & 2 & -2 \\
0 & 0 & 2 & 2
\end{array}\right]=\left[\begin{array}{cccc}
4 & 3 & -5 & 2 \\
0 & -2 & 2 & -2 \\
0 & 0 & 1 & 1
\end{array}\right]}
\end{array}
\]
Next solve $U x=y,$ using back-substitution
\[
\left[\begin{array}{ll}
U & \mathbf{y}
\end{array}\right]=\left[\begin{array}{cccc}
4 & 3 & 0 & 7 \\
0 & -2 & 0 & -2 \\
0 & 0 & 1 & 1
\end{array}\right]=\left[\begin{array}{cccc}
4 & 3 & 0 & 7 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 1
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 0 & 1 / 4 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 1
\end{array}\right]
\]
Reducing to echelon form
\[
\mathbf{x}=\left[\begin{array}{c}
1 / 4 \\
2 \\
1
\end{array}\right]
\]