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# If $$A$$ is a square matrix, such that $$A^2 = A$$ , then $$(1 + A )^3 - 7A$$ is equal to:

$(A) \: A \qquad\qquad (B) \: I - A \qquad\qquad (C) \: 1 \qquad\qquad (D) \: 3A$

Toolbox:
• $(a+b)^3=a^3+b^3+3ab(a+b)$
Step 1: Given
$A^2=A$
$A^3=A^2.A$
$\;\;\;=A.A$ [given$A^2=A$]
$\;\;\;\;=A^2$
$(1+A)^3-7A=(1+3A+3A^2+A^3)-7A.$
we know that $(a+b)^3=a^3+b^3+3ab(a+b)$
$(1+A)^3=1+A^3+3A(1+A)$
$(1+A)^3-7A=(1+3A+3A^2+A^3)-7A$
Step 2: Replace $A^2=A$
$\;\;\;\;\;=(1+3A+3A+A.A)-7A.$
$\;\;\;\;\;=(1+6A+A^2)-7A.$
$\;\;\;\;\;=(1+6A+A)-7A.$
$\;\;\;\;\;=(1+7A)-7A.$
$\;\;\;\;\;=1.$
Hence (C) is the right option.
edited Mar 20, 2013