Problem:
Convert the following matrix to reduced row echelon form:
$\left(
\begin{array}{ccccccc}
 0 & -6 & -6 & 8 & -1 & -6 & 5 \\
 8 & 3 & -8 & 8 & 8 & 8 & 7 \\
 7 & 0 & -6 & -2 & 1 & -6 & 1 \\
 0 & 4 & 9 & -7 & 0 & 8 & -2 \\
\end{array}
\right)$.
Answer:
$\left(
\begin{array}{ccccccc}
 1 & 0 & 0 & 0 & \frac{387}{1355} & \frac{678}{1355} & \frac{1071}{1355} \\
 0 & 1 & 0 & 0 & \frac{1016}{1355} & \frac{1864}{1355} & -\frac{497}{1355} \\
 0 & 0 & 1 & 0 & \frac{9}{542} & \frac{323}{271} & \frac{277}{542} \\
 0 & 0 & 0 & 1 & \frac{1219}{2710} & \frac{1593}{1355} & \frac{1987}{2710} \\
\end{array}
\right)$
{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/row_reduce/21532.txt'}