Question
Given $A=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]$ and $1=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $A^2=9 A+m l$. Find $m$
✓
Answer
$A^2=9 A+M I$
$\Rightarrow A^2-9 A=m l$
Now, $A^2=A A$
$=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right] \\ =\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]$
Substituting $A^2$ in $(1),$ we have
$A^2-9 A=m l$
$\Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-9\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]=m\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-\left[\begin{array}{cc}18 & 0 \\ -9 & 63\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}-14 & 0 \\ 0 & -14\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right]$
$\Rightarrow m=-14$
$\Rightarrow A^2-9 A=m l$
Now, $A^2=A A$
$=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right] \\ =\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]$
Substituting $A^2$ in $(1),$ we have
$A^2-9 A=m l$
$\Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-9\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]=m\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-\left[\begin{array}{cc}18 & 0 \\ -9 & 63\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}-14 & 0 \\ 0 & -14\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right]$
$\Rightarrow m=-14$
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