Answer
Consistent
Solution set: $\{(1,3,-2)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-y+z=-4\\
2x-3y+4z=-15\\
5x+y-2z=12
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
1&-1&1&|&-4\\2&-3&4&|&-15\\5&1&-2&|&12\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=-2r_1+r_2$
$\begin{bmatrix}
1&-1&1&|&-4\\0&-1&2&|&-7\\5&1&-2&|&12\end{bmatrix}$
$R_3=-5r_1+r_3$
$\begin{bmatrix}1&-1&1&|&-4\\0&-1&2&|&-7\\0&6&-7&|&32\end{bmatrix}$
$R_2=-r_2$
$\begin{bmatrix}1&-1&1&|&-4\\0&1&-2&|&7\\0&6&-7&|&32\end{bmatrix}$
$R_3=-6r_2+r_3$
$\begin{bmatrix}1&-1&1&|&-4\\0&1&-2&|&7\\0&0&5&|&-10\end{bmatrix}$
$R_3=\dfrac{1}{5}r_3$
$\begin{bmatrix}1&-1&1&|&-4\\0&1&-2&|&7\\0&0&1&|&-2\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
x-y+z=-4\\
y-2z=7\\
z=-2
\end{cases}$
Solve the system by back substitution:
$z=-2$
$y-2z=7$
$y-2(-2)=7$
$y=3$
$x-y+z=-4$
$x-3+(-2)=-4$
$x-5=-4$
$x=1$
The solution is:
$\{(1,3,-2)\}$
