Answer
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 5 & -10 & -5 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right],\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 1 & -2 & -1 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]$
Work Step by Step
A matrix in which all the diagonal elements are $1$ and the all the elements lower than the principal diagonal are $0$ is called a matrix in row-echelon form.
The given matrix can be converted into the row-echelon form by performing some matrix operations. This is done as follows:
Perform the operation $-2{{R}_{1}}+{{R}_{2}}$ to get
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
-2+2 & 4+1 & -6-4 & -8+3 \\
-3 & 4 & -1 & -2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 5 & -10 & -5 \\
-3 & 4 & -1 & -2 \\
\end{matrix} \right]$
Now perform the operation $3{{R}_{1}}+{{R}_{3}}$ to get
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 5 & -10 & -5 \\
3-3 & -6+4 & 9-1 & 12-2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 5 & -10 & -5 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]$
The matrix obtained above is similar to the matrix provided with missing elements.
Thus, the first solution matrix is given by,
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 5 & -10 & -5 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]$
Now perform the operation $\frac{1}{5}{{R}_{2}}$ to get
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & \frac{1}{5}\times 5 & \frac{1}{5}\times (-10) & \frac{1}{5}\times (-5) \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 1 & -2 & -1 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]$
Thus, the second solution matrix is given by,
$\left[ \begin{matrix}
1 & -2 & 3 & 4 \\
0 & 1 & -2 & -1 \\
0 & -2 & 8 & 10 \\
\end{matrix} \right]$
