If a square matrix A satisfies $A^{2} = A,$ then:

{(I + A)}^{2} = I + 3 A

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{(I + A)}^{3} = I + 4 A

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{(I + A)}^{n} = I + (2^{n} - 1) A

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{(I + A)}^{n} = I + 2^{n} A

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Solution

The correct options are
A ${(I + A)}^{n} = I + (2^{n} - 1) A$
B ${(I + A)}^{2} = I + 3 A$
We can write
$(I + A)^{2} = I^{2} + A^{2} + 2 I A = I^{2} + A + 2 A = I^{2} + 3 A$
( $∵ A^{2} = A$ and $I A = A$ )
For $(I + A)^{n} = a_{1} I^{n} + a_{2} I^{n - 1} A + . . . . . a_{n + 1} A^{n}$
We know $I^{m} A^{n} = A$ and $a_{1}, a_{2} . . .$ are binomial coefficients and $A^{n} = A$
so, $(I + A)^{n} = I^{n} + (a_{2} + a_{3} + . . . . . a_{n + 1}) A$ ( $∵ a_{1} = 1$ )
The sum of binomial coefficients $= 2^{n}$
$a_{1} + a_{2} + . . . + a_{n + 1} = 2^{n}$
$\Rightarrow a_{2} + . . . + a_{n + 1} = 2^{n} - 1$
$(I + A)^{n} = I + (2^{n} - 1) A$

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