Answer
$\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
0 & 1 & -12 & -15 \\
\end{matrix} \right],\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]$
Work Step by Step
A matrix in which all the diagonal elements are $1$ and all the elements lower than the principal diagonal are $0$ is known as a matrix in row echelon form.
The given matrix can be converted into the row echelon form by applying some matrix operations.
The operations to be performed are given below:
Perform the operation $-2{{R}_{1}}+{{R}_{2}}$ to get,
$\begin{align}
& \left[ \begin{matrix}
1 & -1 & 1 & 8 \\
2 & 3 & -1 & -2 \\
3 & -2 & -9 & 9 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
-2+2 & 2+3 & -2-1 & -16-2 \\
3 & -2 & -9 & 9 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
3 & -2 & -9 & 9 \\
\end{matrix} \right]
\end{align}$
Now perform the operation $-3{{R}_{1}}+{{R}_{3}}$ to get,
$\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
-3+3 & 3-2 & -3-9 & -24+9 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]$
The matrix obtained above is similar to the matrix given with missing elements.
Thus, the first solution matrix is,
$\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 5 & -3 & -18 \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]$
Now, perform the operation, ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$ to get,
$\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & \frac{1}{5}\times 5 & \frac{1}{5}\times -3 & \frac{1}{5}\times -18 \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]$
Thus, the second solution matrix is,
$\left[ \begin{matrix}
1 & -1 & 1 & 8 \\
0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\
0 & 1 & -12 & -15 \\
\end{matrix} \right]$
