We are given the equation $$(I-A)^3 = xA + I$$. Using the binomial expansion for matrices (since $I$ and $A$ commute, i.e., $IA = AI$), we have: $$(I-A)^3 = I^3 - 3I^2A + 3IA^2 - A^3$$ Since $I^n = I$ and $IA = AI = A$, this simplifies to: $$I - 3A + 3A^2 - A^3 = xA + I$$
We are given that $A^2 = A$. We can use this to simplify higher powers of $A$: $$A^3 = A^2 \cdot A = A \cdot A = A^2 = A$$ Substituting $A^2 = A$ and $A^3 = A$ into the expanded equation: $$I - 3A + 3(A) - (A) = xA + I$$
Simplify the left side of the equation: $$I - 3A + 3A - A = xA + I$$ $$I - A = xA + I$$ Subtracting $I$ from both sides: $$-A = xA$$ Comparing the coefficients of $A$, we get: $$x = -1$$
Final Answer: -1
AI generated content. Review strictly for academic accuracy.