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Q.

If $| A | \neq 0$ and (A–2I) (A–3I) = 0 then $A^{- 1} =$

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a

$\frac{A - 5 I}{6}$

b

$\frac{5 I - A}{5}$

c

$\frac{5 A - I}{6}$

d

$\frac{5 I - A}{6}$

answer is D.

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Detailed Solution

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$\begin{array}{rcl} (A - 3 I) (A - 3 I) & = & 0 \\ A^{2} - 5 AI + 6 I & = & 0 \\ A^{- 1} (A^{2} - 5 A + 6 I) & = & 0 \\ A^{- 1} A A - 5 A^{- 1} A + 6 A^{- 1} I & = & 0 \\ A - 5 I + 6 A^{- 1} & = & 0 \\ 6 A^{- 1} & = & 5 I - A \\ A^{- 1} & = & \frac{5 I - A}{6} \end{array}$

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