Answer
Consistent
Solution set: $\left\{(x,y,z)|x=\frac{6}{13}-\frac{4}{13}z,y=\frac{4}{13}-\frac{7}{13}z,z\text{ is any real number}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x-3y-z=0\\
3x+2y+2z=2\\
x+5y+3z=2
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
2&-3&-1&|&0\\3&2&2&|&2\\1&5&3&|&2\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
Interchange $R_1$ and $R_3$
$\begin{bmatrix}
1&5&3&|&2\\3&2&2&|&2\\2&-3&-1&|&0\end{bmatrix}$
$R_2=-3r_1+r_2$
$\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\2&-3&-1&|&0\end{bmatrix}$
$R_3=-2r_1+r_3$
$\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\0&-13&-7&|&-4\end{bmatrix}$
$R_3=-r_2+r_3$
$\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\0&0&0&|&0\end{bmatrix}$
$R_2=-\dfrac{1}{13}r_2$
$\begin{bmatrix}1&5&3&|&2\\0&1&\frac{7}{13}&|&\frac{4}{13}\\0&0&0&|&0\end{bmatrix}$
As the last row contains only zeros, the system is consistent, with infinitely many solutions.
Write the corresponding system:
$\begin{cases}
x+5y+3z=2\\
y+\frac{7}{13}z=\frac{4}{13}
\end{cases}$
Express $x,y$ in terms of $z$:
$y+\frac{7}{13}z=\frac{4}{13}\Rightarrow y=\frac{4}{13}-\frac{7}{13}z$\\
$x+5y+3z=2\Rightarrow x=2-5\left(\frac{4}{13}-\frac{7}{13}z\right)-3z=\frac{4}{13}-\frac{7}{13}z$
The solutions set is:
$\left\{(x,y,z)|x=\frac{6}{13}-\frac{4}{13}z,y=\frac{4}{13}-\frac{7}{13}z,z\text{ is any real number}\right\}$
