Question

Find the value of $k$ if $M=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ and $M^2-kM-I_2=0$.

Answer 1

Given, $M=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$
$M^2=M\cdot M$
$=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$
$=\left[\begin{array}{cc}1\times 1+2\times 2& 1\times 2+2\times 3\\ 2\times 1+3\times 2& 2\times 2+3\times 3\end{array}\right]$
$=\left[\begin{array}{cc}1+4& 2+6\\ 2+6& 4+9\end{array}\right]=\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]$

Now, $M^2-kM-I_2=0$
$\Rightarrow M^2-I_2=kM$
$\Rightarrow \left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=k\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}4& 8\\ 8& 12\end{array}\right]=k\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$
$\Rightarrow 4\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]=k\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$
On comparing both sides, we get
$\Rightarrow k=4$.