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

0

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- 1

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Solution

The correct option is D $- 1$
Given $A$ is a square matrix which is nilpotent
ie $A^{m} = 0.... (1)$
now, given that
$(1 - A)^{n} = I + A + A^{2} . . . . + A^{m - 1}$
multiplying both sides by $(1 - A)$ , we get
$(I - A) (I - A)^{n} = (1 - A) (I + A + A^{2} + . . . + A^{m - 1})$
$(I - A) (I - A)^{n} = I + A + A^{2} . . . + A^{m - 1}$
$- A - A^{2} - A^{3} . . . . - A^{m - 1} - A^{m}$
$(I - A) (I - A)^{n} = I - A^{n}$
$(I - A) (I - A)^{n} = I$ [from(1)]
$\Rightarrow (I - A^{n}) = (1 - A)^{- 1}$
[ $∵ (I - A)$ is the inverse of $(I - A)^{n}$ ]
$∴ n = - 1$

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