Answer
Consistent
Solution set: $\{(x_1,x_2,x_3,x_4)|x_1=2-4x_4,x_2=3-x_3-3x_4,x_3,x_4\text{ any real numbers}\}$
Work Step by Step
We are given the reduced row echelon form of a system of linear equations:
$\begin{bmatrix}1&0&0&4&|&2\\0&1&1&3&|&3\\0&0&0&0&|&0\end{bmatrix}$
Write the system of equations corresponding to the given matrix:
$\begin{cases}
1x_1+0x_2+0x_3+4x_4=2\\
0x_1+1x_2+1x_3+3x_4=3\\
0x_1+0x_2+0x_3+0x_4=0
\end{cases}$
$\begin{cases}
x_1+4x_4=2\\
x_2+x_3+3x_4=3\\
0=0
\end{cases}$
Because the reduced row echelon form has a row with only zeros, the system is consistent, having infinitely many solutions.
Express $x_1,x_2$ in terms of $x_3,x_4$:
$x_2+x_3+3x_4=3\Rightarrow x_2=3-x_3-3x_4$
$x_1+4x_4=2\Rightarrow x_1=2-4x_4$
The solution set is:
$\{(x_1,x_2,x_3,x_4)|x_1=2-4x_4,x_2=3-x_3-3x_4,x_3,x_4\text{ any real numbers}\}$
