then \[{{A}^{2}}+xA+I=0\] for \[(x,y)\] is
A) \[(-4,1)\]
B) \[(-1,3)\]
C) \[(4,-1)\]
D) \[(1,3)\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]\] The characteristic equation of ‘A’ is \[|A-\lambda I|=0\] \[\left| \begin{matrix} 3-\lambda & 2 \\ 1 & 1-\lambda \\ \end{matrix} \right|=0\] \[(3-\lambda )(1-\lambda )-2=0\] \[3-\lambda -3\lambda +{{\lambda }^{2}}-2=0\] \[({{\lambda }^{2}}-4\lambda +1)=0\] 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 \[{{A}^{2}}+xA+yI=0\] \[\Rightarrow \] \[x=-4,y=1\]
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