LU-decomposition of A I have:
$A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
and I'm asked to LU-decomposition A, then solve $Ax=0$.
What I did:
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 18 & 6 & -12
\end{bmatrix}$
$L=\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}$
And because:
$Ax=b\hspace{12mm}L(Ux)=b$
$Ly=b\hspace{12mm}Ux=y$
I got:
$Ly=0$
$
\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y_1
\\
y_2
\\
y_3
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
Using this step:
$1)$ $R2=R2-2\times R1$
I got $y=0$
Then solving:
$Ux=y$
$
\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 18 & 6 & -12
\end{bmatrix}
\begin{bmatrix}
x_1
\\
x_2
\\
x_3
\\
x_4
\\
x_5
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
And using this steps:
$1)$ $R1=R1+2\times R2$
$2)$ $R3=\frac{R3}{6}$
$3)$ $R3=R3-R2$
I got:
$
\begin{cases}
x_2=2x_1+8x_3+5x_4
\\
x_4=-3x_3+2x_5
\\
x_1\,,\,x_3\,,\,x_5\quad free
\end{cases}
$
And I chose free variables to be $x_1=1$ , $x_3=1$ , $x_5=1$ so I got:
$
\begin{cases}
x_1=1
\\
x_2=5
\\
x_3=1
\\
x_4=-1
\\
x_5=1
\end{cases}
$
Just wondering if I have done it right? And if not, then how should I do it?
Thanks in advance.
 A: $A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
$3)$ $R3=R3-6\times R2$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$
$L=\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 6 & 1
\end{bmatrix}$
Solving $x$ from $Ax=0$:
$Ly=0$
$
\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y_1
\\
y_2
\\
y_3
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
\Longrightarrow
\begin{cases}
y_1=0
\\
2y_1+y_2=0
\\
y_1+6y_2+y_3=0
\end{cases}
\Longrightarrow
\begin{cases}
y_1=0
\\
y_2=0
\\
y_3=0
\end{cases}
\Longrightarrow
y=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
=0
$
$Ux=y$
$Ux=0$
$
\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x_1
\\
x_2
\\
x_3
\\
x_4
\\
x_5
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
And using this steps (to get echelon form):
$1)$ $R1=\frac{R1}{2}$
$2)$ $R1=R1+R2$
$3)$ $R2=\frac{R2}{3}$
And since the pivot columns of the matrix are $1$ and $3$, so the basic variables are $x_1$ and $x_3$. The remaining variables, $x_2\,,\,x_4$ and $x_5$ must be free.
I got:
$
\begin{cases}
x_1=\frac{1}{2}x_2-4x_3-\frac{5}{2}x_4
\\
x_3=-\frac{1}{3}x_4+\frac{2}{3}x_5
\\
x_2\,,\,x_4\,,\,x_5\quad free
\end{cases}
$
And I chose free variables to be $x_2=2$ , $x_4=6$ , $x_5=3$ so I got:
$
\begin{cases}
x_1=-14
\\
x_2=2
\\
x_3=0
\\
x_4=6
\\
x_5=3
\end{cases}
\Longrightarrow
x=\begin{bmatrix}
-14
\\
2
\\
0
\\
6
\\
3
\end{bmatrix}
=\begin{bmatrix}
-14 & 2 & 0 & 6 & 3
\end{bmatrix}^T
$
And I check it out by calculating:
$\Rightarrow L\times U$ and I got the $A$ so L and U matrices are correct
$\Rightarrow A\times x$ and I got $0$ so it $x$ values are correct, since it is true when $0=0$
Here is a photo of what I have done with the coefficients of $x_3$ and $x_4$:

