Question

If $A=\left(\begin{array}{cc}1& 2\\ 2& 3\end{array}\right)$ and $A^2-kA-I_2=O$, where $I_2$ is the $2\times 2$ identity matrix, then what is the value of $k$?

• A. $4$
• B. $-4$
• C. $8$
• D. $-8$

Answer 1

The correct option is A $4$

Given $A=\left(\begin{array}{cc}1& 2\\ 2& 3\end{array}\right)A^2-kA-I_2=0$

$\to A^2=\left(\begin{array}{cc}1& 2\\ 2& 3\end{array}\right)\left(\begin{array}{cc}1& 2\\ 2& 3\end{array}\right)=\left(\begin{array}{cc}5& 8\\ 8& 10\end{array}\right)$

$A^2-kA-I_2=\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-k\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}4& 8\\ 8& 12\end{array}\right]-k\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]=0$

$=\left[\begin{array}{cc}4-k& 8-2k\\ 8-k& 12-3k\end{array}\right]=0$

$\therefore 4-k=0\Rightarrow k=4$

$12-3k=0\Rightarrow k=4$

$\therefore k=4$.