Answer
The solution is, $ x=2t+4,y=t+1,z=t $
Work Step by Step
Consider the given system of equations
$\begin{align}
& x-3y+z=1 \\
& -2x+y+3z=-7 \\
& x-4y+2z=0
\end{align}$
Therefore, in matrix form the system of equations can be written as below:
$ AX=b $
Where
$ A=\left[ \begin{array}{*{35}{r}}
1 & -3 & 1 \\
-2 & 1 & 3 \\
1 & -4 & 2 \\
\end{array} \right];b=\left[ \begin{array}{*{35}{r}}
1 \\
-7 \\
0 \\
\end{array} \right];X=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & -3 & 1 & 1 \\
-2 & 1 & 3 & -7 \\
1 & -4 & 2 & 0 \\
\end{array} \right]$
Apply elementary row operation on $ A $ to convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}+2{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-{{R}_{1}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & -3 & 1 & 1 \\
0 & -5 & 5 & -5 \\
0 & -1 & 1 & -1 \\
\end{array} \right]$
Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}-\frac{1}{5}{{R}_{2}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & -3 & 1 & 1 \\
0 & -5 & 5 & -5 \\
0 & 0 & 0 & 0 \\
\end{array} \right]$
Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $ A $ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=2$
