If $A$ is non-singular and $(A - 2I)(A - 4I) - 0$, then $\frac{1}{6}\ A + \frac{4}{3}A^{- 1}$ Is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
We have,
$(A - 2I)(A - 4I) = 0$
$\Rightarrow A^{2} - 2A - 4A + 8I = 0$
$\Rightarrow A^{2} - 6A + 8I = 0$
${\Rightarrow A}^{- 1}\left( A^{2} - 6A + 8I \right) = A^{- 1}0$
$\Rightarrow A - 6I + 8A^{- 1} = 0$
$\Rightarrow A + 8A^{- 1} = 6I$
$\Rightarrow \frac{1}{6}A + \frac{4}{3}A^{- 1} = 1$

Q. If $A$ is non-singular and $(A - 2I)(A - 4I) - 0$, then $\frac{1}{6}\ A + \frac{4}{3}A^{- 1}$ Is equal to

Explanation

Reviewed by Dr. Abubacker

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