Problem:
Find the characteristic polynomial of the following matrix:
$\left(
\begin{array}{cc}
 -8 & -6 \\
 -1 & -6 \\
\end{array}
\right)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the characteristic polynomial of the matrix }M\text{ with respect to the variable }\lambda : \\
 M=\left(
\begin{array}{cc}
 -8 & -6 \\
 -1 & -6 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{To }\text{find }\text{the }\text{characteristic }\text{polynomial }\text{of }\text{a }\text{matrix, }\text{subtract }\text{a }\text{variable }\text{multiplied }\text{by }\text{the }\text{identity }\text{matrix }\text{and }\text{take }\text{the }\text{determinant}: \\
 | M-\lambda  \mathbb{I}|  \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{lll}
 | M-\lambda  \mathbb{I}|  & = & \left| 
\begin{array}{cc}
 -8 & -6 \\
 -1 & -6 \\
\end{array}
-\lambda  
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right|  \\
 \text{} & = & \left| 
\begin{array}{cc}
 -8 & -6 \\
 -1 & -6 \\
\end{array}
-
\begin{array}{cc}
 \lambda  & 0 \\
 0 & \lambda  \\
\end{array}
\right|  \\
\end{array}
 \\
  \text{= }\left| 
\begin{array}{cc}
 -\lambda -8 & -6 \\
 -1 & -\lambda -6 \\
\end{array}
\right|  \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{along }\text{the }\text{diagonals }\text{and }\text{subtract}: \\
 (-\lambda -8) (-\lambda -6)-(-6)\, (-1) \\
\end{array}
 \\
 
\begin{array}{l}
 (-\lambda -8) (-\lambda -6)=\lambda ^2+14 \lambda +48: \\
  \text{= }48+14 \lambda +\lambda ^2-(-6)\, (-1) \\
\end{array}
 \\
 
\begin{array}{l}
 -6 (-1)=6: \\
  \text{= }\left(\lambda ^2+14 \lambda +48\right)-6 \\
\end{array}
 \\
 
\begin{array}{l}
 \left(\lambda ^2+14 \lambda +48\right)-6=\lambda ^2+14 \lambda +42: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} &  \text{= }\lambda ^2+14 \lambda +42 \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/characteristic_polynomial_w_steps/4930.txt'}