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Question

# If A satisfies the equation ${x}^{3}-5{x}^{2}+4x+\lambda =0$ then A−1 exists if (a) $\lambda \ne 1$ (b) $\lambda \ne 2$ (c) $\lambda \ne -1$ (d) $\lambda \ne 0$

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Solution

## $\left(\mathrm{d}\right)\mathrm{\lambda }\ne 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}A\mathrm{satisfies}{x}^{3}-5{x}^{2}+4x+\lambda \mathit{}=0.\phantom{\rule{0ex}{0ex}}⇒{A}^{3}-5{A}^{\mathit{2}}+4A=-\lambda \phantom{\rule{0ex}{0ex}}\mathrm{Assuming}{A}^{-1}\mathrm{exists},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{A}^{-1}\left({A}^{3}-5{A}^{2}+4A\right)=-\lambda {A}^{-1}\phantom{\rule{0ex}{0ex}}⇒{A}^{2}-5A+4=-{A}^{-1}\mathrm{\lambda }\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=\frac{-\left({A}^{2}-5A+4\right)}{\lambda }\phantom{\rule{0ex}{0ex}}\mathrm{Thus},{\mathrm{A}}^{-1}\mathrm{exists}\mathrm{if}\mathrm{\lambda }\ne 0.$

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