Answer
Consistent
Solution set: $\left\{\left(-3,\dfrac{1}{2},1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+2y-z=-3\\
2x-4y+z=-7\\
-2x+2y-3z=4
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
1&2&-1&|&-3\\2&-4&1&|&-7\\-2&2&-3&|&4\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=-2r_1+r_2$
$\begin{bmatrix}
1&2&-1&|&-3\\0&-8&3&|&-1\\-2&2&-3&|&4\end{bmatrix}$
$R_3=2r_1+r_3$
$\begin{bmatrix}1&2&-1&|&-3\\0&-8&3&|&-1\\0&6&-5&|&-2\end{bmatrix}$
$R_2=-\dfrac{1}{8}$
$\begin{bmatrix}1&2&-1&|&-3\\0&1&-\frac{3}{8}&|&\frac{1}{8}\\0&6&-5&|&-2\end{bmatrix}$
$R_3=-6r_2+r_3$
$\begin{bmatrix}1&2&-1&|&-3\\0&1&-\frac{3}{8}&|&\frac{1}{8}\\0&0&-\frac{11}{4}&|&-\frac{11}{4}\end{bmatrix}$
$R_3=-\dfrac{4}{11}r_3$
$\begin{bmatrix}1&2&-1&|&-3\\0&1&-\frac{3}{8}&|&\frac{1}{8}\\0&0&1&|&1\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
x+2y-z=-3\\
y-\frac{3}{8}z=\frac{1}{8}\\
z=1
\end{cases}$
Solve the system by back substitution:
$z=1$
$y-\frac{3}{8}z=\frac{1}{8}$
$y-\frac{3}{8}=\frac{1}{8}$
$y=\frac{1}{2}$
$x+2y-z=-3$
$x+2\left(\dfrac{1}{2}\right)-1=-3$
$x+1-1=-3$
$x=-3$
The system is consistent. The solution is:
$\left\{\left(-3,\dfrac{1}{2},1\right)\right\}$
