Question
Show that the matrix $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ satisfies the equation $A^3 - 4A^2 + A = 0.$
✓
Answer
Given, $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^3=\text{A}^2.\text{A}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}14+12&21+24\\8+7&12+14\end{bmatrix}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
Hence, $\text{A}^3-4\text{A}^2+\text{A}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$ =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
So, $ \text{A}^3-4\text{A}+\text{A}=0$
$\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^3=\text{A}^2.\text{A}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}14+12&21+24\\8+7&12+14\end{bmatrix}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
Hence, $\text{A}^3-4\text{A}^2+\text{A}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$ =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
So, $ \text{A}^3-4\text{A}+\text{A}=0$
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