Question
If $M = \left[ {\begin{array}{*{20}{c}}1&2\\2&3\end{array}} \right]$ and ${M^2} - \lambda M - {I_2} = 0$, then $\lambda = $

Answer

d
(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]$

==> $5 - \lambda = 1,\,\,8 - 2\lambda = 0,\,\,13 - 3\lambda = 1$

==> $\lambda = 4$, which satisfies all the three equations.

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