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If A is non-singular and $(A-2I) (A - 4I)=O$, then $\frac{1}{6}A+\frac{4}{3}A^{-1}=$ |
$I$ $O$ $2I$ $6I$ |
$I$ |
We have, $(A - 2I) (A-4I)=O$ $⇒A^2-2A-4A + 8I =O$ $⇒A^2-6A + 8I =O$ $⇒ A^{-1} (A^2-6A+8I) = A^{-1}O$ $⇒A - 6I +8A^{-1}=O$ $⇒A+8A^{-1}=6I⇒\frac{1}{6}A+\frac{4}{3}A^{-1}=I$ |
