Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A = 
$
        \begin{bmatrix}
        0 & 1 & -2 \\
        -1 & 2 & -1 \\
        2 & -4 & 3 \\
        1 & -3 & 2 \\
        \end{bmatrix}
$
U = 
$
        \begin{bmatrix}
        -1 & 2 & -1 \\
        0 & 1 & -2 \\
        0 & 0 & 1 \\
        0 & 0 & 0 \\
        \end{bmatrix}
$
L = 
$
        \begin{bmatrix}
        1 & 0 & 0 & 0\\
        0 & 1 & 0 & 0\\
        -2 & 0 & 1 & 0\\
        -1 & -1 & -1 & 0\\
        \end{bmatrix}
$
Okay so using that I need to solve the following system:
$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$
So step one is solving $Ly = b$, where $y = Ux$
So that is:
$
y_1 = 0\\
    y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$
How can we find $y_3$ in the last two equations? Because,
$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$
So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.
 A: Problem
$$
  \mathbf{A} = 
\left[
\begin{array}{rrr}
 0 & 1 & -2 \\
 -1 & 2 & -1 \\
 2 & -4 & 3 \\
 1 & -3 & 2 \\
\end{array}
\right]
$$
Associated Permutation Matrix
Don't start with a $0$ pivot element. Move the first row down. The permutation matrix interchanges the first two rows.
$$ \mathbf{P} = 
\left[
\begin{array}{cccc}
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right]
$$
Input
$$
\begin{align}
  \mathbf{P} \mathbf{A} = 
% P
\left[
\begin{array}{cccc}
 0 & \boxed{1} & 0 & 0 \\
 \boxed{1} & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
 0 & 1 & -2 \\
 -1 & 2 & -1 \\
 2 & -4 & 3 \\
 1 & -3 & 2 \\
\end{array}
\right]
%
&=
% L
\left[
\begin{array}{rrr}
 -1 & 2 & -1 \\
 0 & 1 & -2 \\
 2 & -4 & 3 \\
 1 & -3 & 2 \\
\end{array}
\right]
\end{align}
$$
Decomposition
$$
\begin{align}
  \mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% PA
\underbrace{\left[
\begin{array}{rrr}
 -1 & 2 & -1 \\
 0 & 1 & -2 \\
 2 & -4 & 3 \\
 1 & -3 & 2 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
%
&=
% L
\underbrace{\left[
\begin{array}{rrrc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 -2 & 0 & 1 & 0  \\
 -1 & -1 & -1 & 1 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{blue}{m}}
% U
\underbrace{\left[
\begin{array}{rrr}
 -1 & 2 & -1 \\
 0 & 1 & -2 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
\end{align}
$$
A: Multiplying the first row of $L$ with the first column of $U$ gives us $-1$, which is not the $(1,1)$ entry of $A$. Hence, you have made a mistake in computation of LU decomposition. 
There seems to be a missing permutation matrix being involved in your computation.
Your procedure to solve the linear system upon computing the LU decomposition is correct.
A: $$
\begin{align}
  \mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% P
\left[
\begin{array}{cccc}
 0 & \boxed{1} & 0 & 0 \\
 \boxed{1} & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
 0 & 1 & -2 \\
 -1 & 2 & -1 \\
 2 & -4 & 3 \\
 1 & -3 & 2 \\
\end{array}
\right]
%
&=
% L
\left[
\begin{array}{rrr}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 -2 & 0 & 1 & 0 \\
 -1 & -1 & -1 & 1 \\
\end{array}
\right]
% U
\left[
\begin{array}{rrr}
 -1 & 2 & -1 \\
 0 & 1 & -2 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{array}
\right]
\end{align}
$$
