Answer
See answer below
Work Step by Step
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
3&-1 &2\\
6&-1 & 1\\
-3 & 5&2
\end{bmatrix} \approx^1\begin{bmatrix}
3&-1 &2\\
0&1 & -3\\
-3 & 5&2
\end{bmatrix} \approx^2 \begin{bmatrix}
3&-1 &2\\
0&1 & -3\\
0 & 4&4
\end{bmatrix} \approx^3 \begin{bmatrix}
3&-1 &2\\
0&1 & -3\\
0 & 0&16
\end{bmatrix}$
$1.A_{12}(-2)$
$2.A_{13}(1)$
$3.A_{23}(-4)$
Corresponding multiplier: $m_{21}=2,m_{31}=-1,m_{32}=4$
Consequently, we get $L=\begin{bmatrix}
1&0&0\\
2& 1&0\\
-1 & 4 & 1
\end{bmatrix}$
We leave it as an exercise to verify that $LU = A$
$\begin{bmatrix}
1&0&0\\
2& 1&0\\
-1 & 4 & 1
\end{bmatrix}\begin{bmatrix}
3&-1 &2\\
0&1 & -3\\
0 & 0&16
\end{bmatrix}= \begin{bmatrix}
3&-1 &2\\
6&-1 &1\\
-3& 5&2
\end{bmatrix}=A$
Hence, the resulted matrices are correct
