Problem:
Compute the rank of
$\left(
\begin{array}{cccc}
 -5 & -10 & 10 & 4 \\
 -8 & -10 & -3 & -1 \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the rank of the matrix}: \\
 M=\left(
\begin{array}{cccc}
 -5 & -10 & 10 & 4 \\
 -8 & -10 & -3 & -1 \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{Reduce }\text{matrix }M \text{to }\text{row }\text{echelon }\text{form}: \\
 \left(
\begin{array}{cccc}
 -5 & -10 & 10 & 4 \\
 -8 & -10 & -3 & -1 \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Swap }\text{row }1 \text{with }\text{row }2: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 -5 & -10 & 10 & 4 \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{5}{8}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & -\frac{15}{4} & \frac{95}{8} & \frac{37}{8} \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\text{row }1 \text{from }\text{row }3: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & -\frac{15}{4} & \frac{95}{8} & \frac{37}{8} \\
 0 & 12 & 13 & -9 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{3}{4}\, \times \, \text{(row }1) \text{from }\text{row }4: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & -\frac{15}{4} & \frac{95}{8} & \frac{37}{8} \\
 0 & 12 & 13 & -9 \\
 0 & \frac{3}{2} & \frac{17}{4} & \frac{11}{4} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Swap }\text{row }2 \text{with }\text{row }3: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & 12 & 13 & -9 \\
 0 & -\frac{15}{4} & \frac{95}{8} & \frac{37}{8} \\
 0 & \frac{3}{2} & \frac{17}{4} & \frac{11}{4} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }\frac{5}{16}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & 12 & 13 & -9 \\
 0 & 0 & \frac{255}{16} & \frac{29}{16} \\
 0 & \frac{3}{2} & \frac{17}{4} & \frac{11}{4} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{1}{8}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & 12 & 13 & -9 \\
 0 & 0 & \frac{255}{16} & \frac{29}{16} \\
 0 & 0 & \frac{21}{8} & \frac{31}{8} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{14}{85}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
 \left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & 12 & 13 & -9 \\
 0 & 0 & \frac{255}{16} & \frac{29}{16} \\
 0 & 0 & 0 & \frac{304}{85} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{There }\text{are }4 \text{nonzero }\text{rows }\text{in }\text{the }\text{row }\text{echelon }\text{form }\text{of }\text{the }\text{matrix, }\text{so }\text{the }\text{rank }\text{is }4: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \text{rank}\left(
\begin{array}{cccc}
 -5 & -10 & 10 & 4 \\
 -8 & -10 & -3 & -1 \\
 -8 & 2 & 10 & -10 \\
 -6 & -6 & 2 & 2 \\
\end{array}
\right) \text{= }\text{rank}\left(
\begin{array}{cccc}
 -8 & -10 & -3 & -1 \\
 0 & 12 & 13 & -9 \\
 0 & 0 & \frac{255}{16} & \frac{29}{16} \\
 0 & 0 & 0 & \frac{304}{85} \\
\end{array}
\right)=4 \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/rank_and_nullity_w_steps/6633.txt'}