If $A$ is a non-singular matrix such that $(A - 2 I) (A - 4 I) = O$ , where $I$ is an identity matrix of the same order as that of $A$ . Then $\frac{1}{6} A + \frac{4}{3} A^{- 1}$ is equal to:

O

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I

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2 I

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6 I

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Solution

The correct option is B $I$
We have,
$(A - 2 I) (A - 4 I) = O$
$\Rightarrow A^{2} - 2 A - 4 A + 8 I = O$
$\Rightarrow A^{2} - 6 A + 8 I = O$
$\Rightarrow A^{- 1} (A^{2} - 6 A + 8 I) = A^{- 1} O$
$\Rightarrow A - 6 I + 8 A^{- 1} = O$
$\Rightarrow A + 8 A^{- 1} = 6 I$
$\Rightarrow \frac{1}{6} A + \frac{4}{3} A^{- 1} = I$

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