Answer
See answer below
Work Step by Step
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
5&2 &1\\
-10&-2 & 3\\
15& 2&-3
\end{bmatrix} \approx^1\begin{bmatrix}
5&2 &1\\
0&2 & 5\\
15& 2&-3
\end{bmatrix} \approx^2\begin{bmatrix}
5&2 &1\\
0&2 & 5\\
0& -4&-6
\end{bmatrix} \approx^3\begin{bmatrix}
5&2 &1\\
0&2 & 5\\
0& -4&-6
\end{bmatrix} \approx^4 \begin{bmatrix}
5&2 &1\\
0&2 & 5\\
0& 0&4
\end{bmatrix}$
$1.A_{12}(2)$
$2.A_{13}(-3)$
$3.A_{23}(2)$
Corresponding multiplier: $m_{21}=-2,m_{31}=3,m_{32}=-2$
Consequently, we get $L=\begin{bmatrix}
1&0&0\\
-2& 1&0\\
3 & -2 & 1
\end{bmatrix}$
We leave it as an exercise to verify that $LU = A$
$\begin{bmatrix}
1&0&0\\
-2& 1&0\\
3 & -2 & 1
\end{bmatrix}\begin{bmatrix}
5&2 &1\\
0&2 & 5\\
0& 0&4
\end{bmatrix}=\begin{bmatrix}
5&2 &1\\
-10&-2 & 3\\
15& 2&-3
\end{bmatrix}=A$
Hence, the resulted matrices are correct
