Answer
Consistent
Solution set: $\left\{\left(8,2,0\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-y=6\\
2x-3z=16\\
2y+z=4
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
1&-1&0&|&6\\2&0&-3&|&16\\0&2&1&|&4\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=-2r_1+r_2$
$\begin{bmatrix}
1&-1&0&|&6\\0&2&-3&|&4\\0&2&1&|&4\end{bmatrix}$
$R_3=-r_2+r_3$
$\begin{bmatrix}1&-1&0&|&6\\0&2&-3&|&4\\0&0&4&|&0\end{bmatrix}$
$R_3=\dfrac{1}{4}r_3$
$\begin{bmatrix}1&-1&0&|&6\\0&2&-3&|&4\\0&0&1&|&0\end{bmatrix}$
$R_2=\dfrac{1}{2}r_2$
$\begin{bmatrix}1&-1&0&|&6\\0&1&-\frac{3}{2}&|&2\\0&0&1&|&0\end{bmatrix}$
$R_2=\dfrac{3}{2}r_3+r_2$
$\begin{bmatrix}1&-1&0&|&6\\0&1&0&|&2\\0&0&1&|&0\end{bmatrix}$
$R_1=-r_2+r_1$
$\begin{bmatrix}1&0&0&|&8\\0&1&0&|&2\\0&0&1&|&0\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
1x+0y+0z=8\\
0x+1y+0z=2\\
0x+0y+1z=0
\end{cases}$
$\begin{cases}
x=8\\
y=2\\
z=0
\end{cases}$
The system is consistent. The solution set is:
$\left\{\left(8,2,0\right)\right\}$
