Answer
Consistent
Solution set: $\{(x_1,x_2,x_3,x_4)|x_1=-2-x_4,x_2=2-2x_4,x_3=x_4,x_4\text{ any real number}\}$
Work Step by Step
We are given the reduced row echelon form of a system of linear equations:
$\begin{bmatrix}1&0&0&1&|&-2\\0&1&0&2&|&2\\0&0&1&-1&|&0\\0&0&0&0&|&0\end{bmatrix}$
Write the system of equations corresponding to the given matrix:
$\begin{cases}
1x_1+0x_2+0x_3+1x_4=-2\\
0x_1+1x_2+0x_3+2x_4=2\\
0x_1+0x_2+1x_3-1x_4=0\\
0x_1+0x_2+0x_3+0x_4=0
\end{cases}$
$\begin{cases}
x_1+x_4=-2\\
x_2+2x_4=2\\
x_3-x_4=0\\
0=0
\end{cases}$
Because the reduced row echelon form has a row with only zeros, the system is consistent, having infinitely many solutions.
Express $x_1,x_2,x_3$ in terms of $x_4$:
$x_3-x_4=0\Rightarrow x_3=x_4$
$x_2+2x_4=2\Rightarrow x_2=2-2x_4$
$x_1+x_4=-2\Rightarrow x_1=-2-x_4$
The solution set is:
$\{(x_1,x_2,x_3,x_4)|x_1=-2-x_4,x_2=2-2x_4,x_3=x_4,x_4\text{ any real number}\}$
