Question

If $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$ then $A^2+xA+yI=0$ for $(x,y)=$_______

• A. $(4,-1)$
• B. $(1,3)$
• C. $(-4,1)$
• D. $(-1,3)$

Answer 1

Answer: (C)

$A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$
The characteristic equation of ' $A$ ' is
$|A-\lambda I|=0$
$\left|\begin{array}{cc}3-\lambda & 2\\ 1& 1-\lambda \end{array}\right|=0$
$(3-\lambda )(1-\lambda )-2=0$
$3-\lambda -3\lambda +{\lambda }^2-2=0$
$\left({\lambda }^2-4\lambda +1\right)=0$...(i)
By Caylay-Hamilton theorem : Every square matrix satisfied its characteristic equation, then put $(\lambda =A)$ is in Eq. (i)
$A^2-4A+I=0$
On comparing with $A^2+xA+yI=0$
$\Rightarrow x=-4,y=1$