Problem:
Convert the following matrix to reduced row echelon form:
$\left(
\begin{array}{cc}
 5 & 3 \\
 -9 & -6 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Do row reduction}: \\
 \left(
\begin{array}{cc}
 5 & 3 \\
 -9 & -6 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{Swap }\text{row }1 \text{with }\text{row }2: \\
 \left(
\begin{array}{cc}
 -9 & -6 \\
 5 & 3 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }\frac{5}{9}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
 \left(
\begin{array}{cc}
 -9 & -6 \\
 0 & -\frac{1}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{row }2 \text{by }-3: \\
 \left(
\begin{array}{cc}
 -9 & -6 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }6\, \times \, \text{(row }2) \text{to }\text{row }1: \\
 \left(
\begin{array}{cc}
 -9 & 0 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Divide }\text{row }1 \text{by }-9: \\
 \left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{This }\text{matrix }\text{is }\text{now }\text{in }\text{reduced }\text{row }\text{echelon }\text{form.$\backslash $nAll }\text{non-zero }\text{rows }\text{are }\text{above }\text{rows }\text{of }\text{all }\text{zeros }\text{(there }\text{are }\text{no }\text{rows }\text{of }\text{all }\text{zeros):} \\
 \left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Each }\text{pivot }\text{is }\text{1 }\text{and }\text{is }\text{strictly }\text{to }\text{the }\text{right }\text{of }\text{every }\text{pivot }\text{above }\text{it}: \\
 \left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Each }\text{pivot }\text{is }\text{the }\text{only }\text{non-zero }\text{entry }\text{in }\text{its }\text{column}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/row_reduce_w_steps/3750.txt'}