Question

$A=\left[\begin{array}{ccc}1& -3& -4\\ -1& 3& 4\\ 1& -3& -4\end{array}\right]$ and $A^2=\lambda I$ then $\lambda =$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
  
• D. $-2$

Answer 1

Answer: (A)

$0$

$A=\left[\begin{array}{ccc}1& -3& -4\\ -1& 3& 4\\ 1& -3& -4\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1& -3& -4\\ -1& 3& 4\\ 1& -3& -4\end{array}\right]\left[\begin{array}{ccc}1& -3& -4\\ -1& 3& 4\\ 1& -3& -4\end{array}\right]$
$=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$
But given $A^2=\lambda I$
$\Rightarrow \lambda =0$