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The matrix $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$ satisfies which one of the following polynomial equations?
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Verified Answer
The correct answer is:
$A^{2}-3 A-2 I=0$
Given that, $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$
$\therefore \quad A^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{cc}1+4 & 2+4 \\ 2+4 & 4+4\end{array}\right]$
$=\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]$
Let $\mathrm{A}^{2}+\mathrm{xA}+\mathrm{yI}=0$ where $\mathrm{x}$ and $\mathrm{y}$ are constant.
$\Rightarrow\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]+\left[\begin{array}{cc}x & 2 x \\ 2 x & 2 x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}5+x+y & 6+2 x \\ 6+2 x & 8+2 x+y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
So, $\begin{aligned} 6+2 \mathrm{x}=0 & \Rightarrow \mathrm{x}=-3 \\ 5+\mathrm{x}+\mathrm{y}=0 & \Rightarrow \mathrm{y}=-5-\mathrm{x}=-2 \\ & \Rightarrow \mathrm{A}^{2}-3 \mathrm{~A}-2 \mathrm{I}=0 \end{aligned}$
$\therefore \quad A^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{cc}1+4 & 2+4 \\ 2+4 & 4+4\end{array}\right]$
$=\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]$
Let $\mathrm{A}^{2}+\mathrm{xA}+\mathrm{yI}=0$ where $\mathrm{x}$ and $\mathrm{y}$ are constant.
$\Rightarrow\left[\begin{array}{ll}5 & 6 \\ 6 & 8\end{array}\right]+\left[\begin{array}{cc}x & 2 x \\ 2 x & 2 x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}5+x+y & 6+2 x \\ 6+2 x & 8+2 x+y\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
So, $\begin{aligned} 6+2 \mathrm{x}=0 & \Rightarrow \mathrm{x}=-3 \\ 5+\mathrm{x}+\mathrm{y}=0 & \Rightarrow \mathrm{y}=-5-\mathrm{x}=-2 \\ & \Rightarrow \mathrm{A}^{2}-3 \mathrm{~A}-2 \mathrm{I}=0 \end{aligned}$
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