Problem:
Compute the (pseudo)inverse of
$\left(
\begin{array}{ccc}
 0 & -5 & 4 \\
 5 & -2 & 2 \\
 2 & -1 & -4 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the inverse}: \\
 \left(
\begin{array}{ccc}
 0 & -5 & 4 \\
 5 & -2 & 2 \\
 2 & -1 & -4 \\
\end{array}
\right)^{-1} \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{Use }\text{the }\text{formula }\text{for }\text{the }\text{inverse }\text{of }\text{a }\text{3$\times $3 }\text{matrix, }A \text{which }\text{is }\text{given }\text{by }\text{the }\text{formula }A^{-1}=\frac{1}{| A| }\left(
\begin{array}{ccc}
 \left| 
\begin{array}{cc}
 a_{2,2} & a_{2,3} \\
 a_{3,2} & a_{3,3} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,3} & a_{1,2} \\
 a_{3,3} & a_{3,2} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,2} & a_{1,3} \\
 a_{2,2} & a_{2,3} \\
\end{array}
\right|  \\
 \left| 
\begin{array}{cc}
 a_{2,3} & a_{2,1} \\
 a_{3,3} & a_{3,1} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,1} & a_{1,3} \\
 a_{3,1} & a_{3,3} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,3} & a_{1,1} \\
 a_{2,3} & a_{2,1} \\
\end{array}
\right|  \\
 \left| 
\begin{array}{cc}
 a_{2,1} & a_{2,2} \\
 a_{3,1} & a_{3,2} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,2} & a_{1,1} \\
 a_{3,2} & a_{3,1} \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 a_{1,1} & a_{1,2} \\
 a_{2,1} & a_{2,2} \\
\end{array}
\right|  \\
\end{array}
\right) \text{where }a_{i,j} \text{represents }\text{the }\text{element }\text{of }A \text{in }\text{the }i^{\text{th}} \text{row }\text{and }j^{\text{th}} \text{column}: \\
  \text{= }\frac{1}{\left| 
\begin{array}{ccc}
 0 & -5 & 4 \\
 5 & -2 & 2 \\
 2 & -1 & -4 \\
\end{array}
\right| }\left(
\begin{array}{ccc}
 \left| 
\begin{array}{cc}
 -2 & 2 \\
 -1 & -4 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 4 & -5 \\
 -4 & -1 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 -5 & 4 \\
 -2 & 2 \\
\end{array}
\right|  \\
 \left| 
\begin{array}{cc}
 2 & 5 \\
 -4 & 2 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 0 & 4 \\
 2 & -4 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 4 & 0 \\
 2 & 5 \\
\end{array}
\right|  \\
 \left| 
\begin{array}{cc}
 5 & -2 \\
 2 & -1 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 -5 & 0 \\
 -1 & 2 \\
\end{array}
\right|  & \left| 
\begin{array}{cc}
 0 & -5 \\
 5 & -2 \\
\end{array}
\right|  \\
\end{array}
\right)=\frac{1}{-124}\left(
\begin{array}{ccc}
 10 & -24 & -2 \\
 24 & -8 & 20 \\
 -1 & -10 & 25 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Write }\text{the }\text{matrix }\text{in }\text{a }\text{simpler }\text{way}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }\frac{1}{124}\left(
\begin{array}{ccc}
 -10 & 24 & 2 \\
 -24 & 8 & -20 \\
 1 & 10 & -25 \\
\end{array}
\right) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/pseudoinverse_w_steps/5845.txt'}