Question

Given $A=\left[\begin{array}{cc}2& 0\\ -1& 7\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ and $A^2=9A+mI$. Find $m.$

Answer 1

Given: $A=\left[\begin{array}{cc}2& 0\\ -1& 7\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore A^2=A\cdot A=\left[\begin{array}{cc}2& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}2& 0\\ -1& 7\end{array}\right]$

$=\left[\begin{array}{cc}4& 0\\ -2-7& 0+49\end{array}\right]$

$=\left[\begin{array}{cc}4& 0\\ -9& 49\end{array}\right]$

Now, $9A+mI=9\left[\begin{array}{cc}2& 0\\ -1& 7\end{array}\right]+m\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}18& 0\\ -9& 63\end{array}\right]+\left[\begin{array}{cc}m& 0\\ 0& m\end{array}\right]$

$=\left[\begin{array}{cc}18+m& 0\\ -9& 63+m\end{array}\right]$

Now, According to question,
$A^2=9A+mI$

$\left[\begin{array}{cc}4& 0\\ -9& 49\end{array}\right]=\left[\begin{array}{cc}18+m& 0\\ -9& 63+m\end{array}\right]$

$4=18+m\Rightarrow m=-14$
$49=63+m\Rightarrow m=-14$
Hence $m=-14.$