Answer
Consistent
Solution set: $\left\{\left(x,y,z\right)|x=\dfrac{4z+1}{3},y=\dfrac{6-5z}{3},z\text{ is any real number}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+y-z=4\\
-x+y+3z=1
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
2&1&-1&|&4\\-1&1&3&|&1\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_1=r_2+r_1$
$\begin{bmatrix}
1&2&2&|&5\\-1&1&3&|&1\end{bmatrix}$
$R_2=r_1+r_2$
$\begin{bmatrix}1&2&2&|&5\\0&3&5&|&6\end{bmatrix}$
$R_2=\dfrac{1}{3}r_2$
$\begin{bmatrix}1&2&2&|&5\\0&1&\frac{5}{3}&|&2\end{bmatrix}$
$R_1=-2r_2+r_1$
$\begin{bmatrix}1&0&-\frac{4}{3}&|&1\\0&1&\frac{5}{3}&|&2\end{bmatrix}$
The system has two equations and 3 variables; therefore it has infinitely many solutions.
Write the corresponding system of equations:
$\begin{cases}
x-\dfrac{4}{3}z=1\\
y+\frac{5}{3}z=2
\end{cases}$
Express $x,y$ in terms of $z$:
$y+\frac{5}{3}z=2\Rightarrow y=2-\frac{5}{3}z=\dfrac{6-5z}{3}$
$x-\dfrac{4}{3}z=1\Rightarrow x=\dfrac{4}{3}z+1=\dfrac{4z+1}{3}$
The system is consistent. The solution set is:
$\left\{\left(x,y,z\right)|x=\dfrac{4z+1}{3},y=\dfrac{6-5z}{3},z\text{ is any real number}\right\}$
