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If $A$ satisfies the equation $x^{3}-5 x^{2}+4 x+\lambda=0$ then $A^{-1}$ exists if

(a) $\lambda \neq 1$

(b) $\lambda \neq 2$

(c) $\lambda \neq-1$

(d) $\lambda \neq 0$


(d) $\lambda \neq 0$

$A$ satisfies $x^{3}-5 x^{2}+4 x+\lambda=0$

$\Rightarrow A^{3}-5 A^{2}+4 A=-\lambda$

Assuming $A^{-1}$ exists, we get

$A^{-1}\left(A^{3}-5 A^{2}+4 A\right)=-\lambda A^{-1}$

$\Rightarrow A^{2}-5 A+4=-A^{-1} \lambda$

$\Rightarrow A^{-1}=\frac{-\left(A^{2}-5 A+4\right)}{\lambda}$

Thus, $\mathrm{A}^{-1}$ exists if $\lambda \neq 0$.

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