Question

If $A=\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$ and $A^2-4A-nI=0$, then ${}^{\prime }{n}^{\prime }$ is equal to :

• A. $3$
• B. $-3$
• C. $\frac{1}{3}$
• D. $\frac{-1}{3}$

Answer 1

The correct option is B $-3$

$A^2=A.A$
$=\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$ $\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$ $=\left[\begin{array}{cc}5& -4\\ -4& 5\end{array}\right]$
Now, $A^2-4A=\left[\begin{array}{cc}5& -4\\ -4& 5\end{array}\right]$ $-4\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$
$=\left[\begin{array}{cc}5& -4\\ -4& 5\end{array}\right]-\left[\begin{array}{cc}8& -4\\ -4& 8\end{array}\right]$
$=\left[\begin{array}{cc}-3& 0\\ 0& -3\end{array}\right]$ $=-3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=-3I$
$\Rightarrow A^2-4A+3I=0$, comparing this with $A^2-4A-nI=0$, we get
$\Rightarrow n=-3$