Question
If $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda,\mu$ so that $\text{A}^2=\lambda\text{A}+\mu\text{I}$

Answer

Given: $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^2=\lambda\text{A}=\mu\text{I}$
$\Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\lambda\begin{bmatrix}2&3\\1&2\end{bmatrix}+\mu\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda&3\lambda\\1\lambda&2\lambda\end{bmatrix}+\begin{bmatrix}\mu&0\\0&\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda+0\\\lambda+0&2\lambda+\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda\\\lambda&2\lambda+\mu\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 7=2\lambda+\mu\ \dots(1)$
$ 12=3\lambda$
$\Rightarrow\lambda=\frac{12}{3}=4$
Putting the value of $\lambda$ in eq. (1), we get
$7=2(4)+\mu$
$\Rightarrow7-8=\mu$
$\therefore\ \mu=-1$

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