Question

If $A=\left[\begin{array}{cc}1& 3\\ 3& 4\end{array}\right]$ and $A^2-kA-5I_2=0$, then the value of $k$ is

• A. $3$
• B. $5$
• C. $7$
• D. $-7$

Answer 1

Answer: (B)

$5$

Given, $A=\left[\begin{array}{cc}1& 3\\ 3& 4\end{array}\right]$
Now, $A^2-5I_2=\left[\begin{array}{cc}1& 3\\ 3& 4\end{array}\right]\left[\begin{array}{cc}1& 3\\ 3& 4\end{array}\right]-5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}1+9& 3+12\\ 3+12& 9+16\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$
$=\left[\begin{array}{cc}5& 15\\ 15& 20\end{array}\right]=5\left[\begin{array}{cc}1& 3\\ 3& 4\end{array}\right]$
$\Rightarrow A^2-5I_2=5A$
But it is given that $A^2-5I_2=kA$
$k=5$.