MCQ
If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if$:$
  • A
    $\lambda\neq1$
  • B
    $\lambda\neq2$
  • C
    $\lambda\neq-1$
  • $\lambda\neq0$

Answer

Correct option: D.
$\lambda\neq0$
A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus, $A^{-1}$ exists if $\lambda\neq0$.

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