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}
1& -1 & 2 & 3\\
2& 0 &3 & -4\\
3& -1& 7 & 8\\
1&3&4&5
\end{bmatrix} \approx^1\begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
3& -1& 7 & 8\\
1&3&4&5
\end{bmatrix} \approx^2\begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
0&2& 1 & -1\\
1&3&4&5
\end{bmatrix} \approx^3\begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
0&2& 1 & -1\\
0&4&2&2
\end{bmatrix} \approx^4 \begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
0&0&2 &9\\
0&0&4&22
\end{bmatrix} \approx^5 \begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
0&0&2 &9\\
0&0&0&4
\end{bmatrix} $
$1.A_{12}(-2)$
$2.A_{13}(-3)$
$3.A_{14}(-1)$
$4.A_{23}(2),A_{24}(-2)$
$4.A_{34}(-2)$
Corresponding multiplier: $m_{21}=2,m_{31}=3,m_{41}(1),m_{32}=1,m_{42}(2),m_{43}(2)$
Consequently, we get $L=\begin{bmatrix}
1&0&0&0\\
2& 1&0 & 0\\
3 & 1 & 1 & 0\\
1 &2 &2 &1
\end{bmatrix}$
We leave it as an exercise to verify that $LU = A$
$\begin{bmatrix}
1&0&0&0\\
2& 1&0 & 0\\
3 & 1 & 1 & 0\\
1 &2 &2 &1
\end{bmatrix} \begin{bmatrix}
1& -1 & 2 & 3\\
0& 2 &-1 & 10\\
0&0&2 &9\\
0&0&0&4
\end{bmatrix} =\begin{bmatrix}
1& -1 & 2 & 3\\
2& 0 &3 & -4\\
3& -1& 7 & 8\\
1&3&4&5
\end{bmatrix}=A$
Hence, the resulted matrices are correct
