Question

If $A=\begin{bmatrix} {3}&{2}\\ {1}&{1} \end{bmatrix} $ 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)

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