If A is non-singular and $(A-2I) (A - 4I)=O$, then $\frac{1}{6}A+\frac{4}{3}A^{-1}=$
Answer & explanation
Correct answer: option 1
We have,
$(A - 2I) (A-4I)=O$
$⇒A^2-2A-4A + 8I =O$
$⇒A^2-6A + 8I =O$
$⇒ A^{-1} (A^2-6A+8I) = A^{-1}O$
$⇒A - 6I +8A^{-1}=O$
$⇒A+8A^{-1}=6I⇒\frac{1}{6}A+\frac{4}{3}A^{-1}=I$