Question

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

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

Answer 1

The correct option is A $(4,-1)$

$\begin{array}{c}A=\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\\ A=3-2=1\\ A^2=\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\\ \left[\begin{array}{c}9+26+2\\ 3+12+1\end{array}\right]=\left[\begin{array}{c}118\\ 43\end{array}\right]\\ \\ A^2+xA+yI=0\\ 4+x.1+y.0=0\Rightarrow x=4\\ 3+4.1+y.1=0\Rightarrow y=-1\end{array}$