Answer
$x=2a+4$
$y=a+1$
$z=a$
Work Step by Step
Build the augmented matrix of the system of equations;
$\begin{bmatrix}1&-3&1&|&1\\-2&1&3&|&-7\\1&-4&2&|&0\end{bmatrix}$
Bring the matrix to the row reduced echelon form:
Add $-R_1$ to $R_3$:
$\begin{bmatrix}1&-3&1&|&1\\-2&1&3&|&-7\\0&-1&1&|&-1\end{bmatrix}$
Add $2R_1$ to $R_2$:
$\begin{bmatrix}1&-3&1&|&1\\0&-5&5&|&-5\\0&-1&1&|&-1\end{bmatrix}$
Multiply $R_2$ by $-\dfrac{1}{5}$:
$\begin{bmatrix}1&-3&1&|&1\\0&1&-1&|&1\\0&-1&1&|&-1\end{bmatrix}$
Add $R_2$ to $R_3$:
$\begin{bmatrix}1&-3&1&|&1\\0&1&-1&|&1\\0&0&0&|&0\end{bmatrix}$
Add $3R_2$ to $R_1$:
$\begin{bmatrix}1&0&-2&|&4\\0&1&-1&|&1\\0&0&0&|&0\end{bmatrix}$
As Row 3 has only zeros, it means the variable $z$ is independent. Note $z=a$.
$\begin{cases}
x-2z=4\\
y-z=1
\end{cases}$
$\begin{cases}
x-2a=4\\
y-a=1
\end{cases}$
The solution is:
$x=2a+4$
$y=a+1$
$z=a$
