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यदि $M = \left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right]$ और ${M^2} - \lambda M - {I_2} = 0$, तब $\lambda = $
$-2$
$2$
$-4$
$4$
Solution
(d) ${M^2} – \lambda M – {I_2} = 0$
$ \Rightarrow \,\,\left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}\lambda &{2\lambda }\\{2\lambda }&{3\lambda }\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = O$
$ \Rightarrow \,\,\left[ {\begin{array}{*{20}{c}}5&8\\8&{13}\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}\lambda &{2\lambda }\\{2\lambda }&{3\lambda }\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = O$
$ \Rightarrow \,\,\left[ {\begin{array}{*{20}{c}}{5 – \lambda }&{8 – 2\lambda }\\{8 – 2\lambda }&{13 – 3\lambda }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
$5 – \lambda = 1,\,\,8 – 2\lambda = 0,\,\,13 – 3\lambda = 1$
$\lambda = 4$, जो तीनों समीकरणों को सन्तुष्ट करता है।
