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Solution:(a)A=\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]Find Eigen values of \operatorname{det}(A-d I)=0\begin{array}{c}A-\lambda I=\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]-\left[\begin{array}{ll}\lambda & 0 \\0 & \lambda\end{array}\right] \\A-\lambda I=\left[\begin{array}{cc}0-\lambda & -1 \\1 & 0-\lambda\end{array}\right] \\\operatorname{det}(A-\lambda I)=0\left[\begin{array}{cc}0-\lambda & -1 \\1 & 0-\lambda\end{array} \mid=0\right. \\(0-\lambda)(0-\lambda)+1=0 \\\lambda^{2}+1=0 \\\lambda^{2}=-1 \Rightarrow \lambda=\sqrt{-1} \\\lambda= \pm i\end{array}Eigen values ore \lambda= \pm iFind Eigen noefer corres ponding Eigen value \lambda= \pm \grave{l}, \lambda=i\begin{array}{c}{\left[\begin{array}{cc}0-\lambda & -1 \\1 & 0-\lambda\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}0 \\0\end{array}\right]} \\{\left[\begin{array}{cc}0-i & -1 \\1 & 0-i\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}0 \\0\end{array}\right]} \\-x_{1} i-x_{2}=0 \\x_{2}=-i x_{1}\end{array}Eigen veetar v_{1}=\left[\begin{array}{r}1 \\ -i\end{array}\right]CS Scanned with CamScannerلy\begin{array}{l}\lambda= \pm i, \quad V=\left[\begin{array}{r}1 \\-i\end{array}\right] \\{\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]\left[\begin{array}{l}1 \\0\end{array}\right]=\left[\begin{array}{c}0 \\-1\end{array}\right]}\end{array}contrsEquillibrium = contreCS Scanned with CamScanner ...