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A square matrix A is said to be nilpotent of index m. If A^m = 0 now, if for this A, $(I - A)^{n} = I + A + A^{2} + A^{3} + \dots + A^{m - 1}$ , then n is equal to

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-1

-m

answer is D.

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Detailed Solution

Let $B = I + A + A^{2} + \dots . A^{m - 1}$
$\begin{array}{l} \Rightarrow B (I - A) = (I - A) (I + A + A^{2} + \dots + A^{m - 1}) \\ \Rightarrow B (I - A) = I - A^{m} = I (∵ A^{m} = 0) \Rightarrow B = (I - A)^{- 1} \Rightarrow n = - 1 \end{array}$

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