Answer
Consistent
Solution: $\left\{\left(2,-1,1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-2y+3z=7\\
2x+y+z=4\\
-3x+2y-2z=-10
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
1&-2&3&|&7\\2&1&1&|&4\\-3&2&-2&|&-10\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=-2r_1+r_2$
$\begin{bmatrix}
1&-2&3&|&7\\0&5&-5&|&-10\\-3&2&-2&|&-10\end{bmatrix}$
$R_3=3r_1+R_3$
$\begin{bmatrix}1&-2&3&|&7\\0&5&-5&|&-10\\0&-4&7&|&11\end{bmatrix}$
$R_2=\dfrac{1}{5}r_2$
$\begin{bmatrix}1&-2&3&|&7\\0&1&-1&|&-2\\0&-4&7&|&11\end{bmatrix}$
$R_1=2r_2+r_1$
$\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&-4&7&|&11\end{bmatrix}$
$R_3=4r_2+r_3$
$\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&0&3&|&3\end{bmatrix}$
$R_3=\dfrac{1}{3}r_3$
$\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&0&1&|&1\end{bmatrix}$
$R_1=-r_3+r_1$
$\begin{bmatrix}1&0&0&|&2\\0&1&-1&|&-2\\0&0&1&|&1\end{bmatrix}$
$R_2=-r_3+r_2$
$\begin{bmatrix}1&0&0&|&2\\0&1&0&|&-1\\0&0&1&|&1\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
1x+0y+0z=2\\
0x+1y+0z=-1\\
0x+0y+1z=1
\end{cases}$
$\begin{cases}
x=2\\
y=-1\\
z=1
\end{cases}$
The system is consistent. The solution set is:
$\left\{\left(2,-1,1\right)\right\}$
