Answer
See answer below
Work Step by Step
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
2& -3 & 1 & 2\\
4& -1 &1& 1\\
-8& 2& 2 & -5\\
6&1&5&2
\end{bmatrix} \approx^1\begin{bmatrix}
2& -3 & 1 & 2\\
0& 5 &-1& 3\\
0& -10&6 & 3\\
0&10&2&4
\end{bmatrix} \approx^2\begin{bmatrix}
2& -3 & 1 & 2\\
0& 5 &-1& 3\\
0& 0&4 & -3\\
0&0&4&2
\end{bmatrix} \approx^3\begin{bmatrix}
2& -3 & 1 & 2\\
0& 5 &-1& 3\\
0& 0&4 & -3\\
0&0&0&5
\end{bmatrix} $
$1.A_{12}(-2),A_{13}(4),A_{14}(-3)$
$2.A_{23}(2),A_{24}(-2)$
$3.A_{34}(-1)$
Corresponding multiplier: $m_{21}=2,m_{31}=-4,m_{41}(3),m_{32}=1,m_{42}(2),m_{43}(1)$
Consequently, we get $L=\begin{bmatrix}
1&0&0&0\\
2& 1&0 & 0\\
-4 & -2 & 1 & 0\\
3 &2 &1 &1
\end{bmatrix}$
We leave it as an exercise to verify that $LU = A$
$\begin{bmatrix}
1&0&0&0\\
2& 1&0 & 0\\
-4 & -2 & 1 & 0\\
3 &2 &1 &1
\end{bmatrix} \begin{bmatrix}
2& -3 & 1 & 2\\
0& 5 &-1& 3\\
0& 0&4 & -3\\
0&0&0&5
\end{bmatrix} =\begin{bmatrix}
2& -3 & 1 & 2\\
4& -1 &1& 1\\
-8& 2& 2 & -5\\
6&1&5&2
\end{bmatrix} =A$
Hence, the resulted matrices are correct
