Answer
Consistent
Solution set: $\left\{\left(-3,2,1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+y=-4\\
-2y+4z=0\\
3x-2z=-11
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
2&1&0&|&-4\\0&-2&4&|&0\\3&0&-2&|&-11\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_3=-r_1+r_3$
$\begin{bmatrix}
2&1&0&|&-4\\0&-2&4&|&0\\1&-1&-2&|&-7\end{bmatrix}$
Interchange $R_1$ and $R_3$
$\begin{bmatrix}1&-1&-2&|&-7\\0&-2&4&|&0\\2&1&0&|&-4\end{bmatrix}$
$R_3=-2r_1+r_3$
$\begin{bmatrix}1&-1&-2&|&-7\\0&-2&4&|&0\\0&3&4&|&10\end{bmatrix}$
$R_2=-\dfrac{1}{2}r_2$
$\begin{bmatrix}1&-1&-2&|&-7\\0&1&-2&|&0\\0&3&4&|&10\end{bmatrix}$
$R_1=-r_2+r_3$
$\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&3&4&|&10\end{bmatrix}$
$R_3=-3r_2+r_1$
$\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&0&10&|&10\end{bmatrix}$
$R_3=\dfrac{1}{10}r_3$
$\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&0&1&|&1\end{bmatrix}$
$R_1=4r_3+r_1$
$\begin{bmatrix}1&0&0&|&-3\\0&1&-2&|&0\\0&0&1&|&1\end{bmatrix}$
$R_2=2r_3+r_2$
$\begin{bmatrix}1&0&0&|&-3\\0&1&0&|&2\\0&0&1&|&1\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
1x+0y+0z=-3\\
0x+1y+0z=2\\
0x+0y+1z=1
\end{cases}$
$\begin{cases}
x=-3\\
y=2\\
z=1
\end{cases}$
The system is consistent. The solution set is:
$\left\{\left(-3,2,1\right)\right\}$
