We need to evaluate the expression $(A-I)^{3}-A$. Using the binomial expansion formula $(x-y)^{3} = x^{3} - 3x^{2}y + 3xy^{2} - y^{3}$, we expand $(A-I)^{3}$:
$$ (A-I)^{3} = A^{3} - 3A^{2}I + 3AI^{2} - I^{3} $$Since $I$ is the identity matrix, $AI = IA = A$ and $I^{n} = I$. Thus:
$$ (A-I)^{3} = A^{3} - 3A^{2} + 3A - I $$We are given that $A^{2} = A$. This implies $A^{3} = A^{2} \cdot A = A \cdot A = A^{2} = A$. Substituting these into the expanded expression:
$$ (A-I)^{3} = A - 3A + 3A - I $$ $$ (A-I)^{3} = A - I $$Now, substitute this result back into the original expression $(A-I)^{3} - A$:
$$ (A-I)^{3} - A = (A - I) - A $$ $$ = -I $$Final Answer: -I
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