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