We are given that $A$ is a square matrix such that $A^2 = A$, which means $A$ is idempotent.

We need to compute $(I - A)^3 + A$, where $I$ is the identity matrix.

Step-by-Step Solution:

1. Expand $(I - A)^3$ using the binomial theorem:

$(I - A)^3 = (I - A)(I - A)(I - A)$

We can calculate this incrementally:

$(I - A)^2 = (I - A)(I - A) = I^2 - 2A + A^2$

Since $A^2 = A$:

$(I - A)^2 = I - 2A + A = I - A$

Now, multiply by $(I - A)$ again:

$(I - A)^3 = (I - A)(I - A)^2 = (I - A)(I - A) = I - A$

Thus:

$(I - A)^3 = I - A$

2. Now add $A$ to the expression:

$(I - A)^3 + A = (I - A) + A$

The terms $-A$ and $+A$ cancel out:

$(I - A) + A = I$

Thus, the expression simplifies to $I$.

Final Answer:

The value of $(I - A)^3 + A$ is $\boxed{I}$.

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Would you like further details or have any questions?

Here are five related questions to explore this concept further:

1. What are the properties of an idempotent matrix?
2. How does the binomial expansion work for matrix expressions?
3. Can a matrix be both idempotent and invertible?
4. What happens if $A^2 = A$ but $A \neq A^2$?
5. How does this result change for non-square matrices?

Tip: In matrix algebra, idempotent matrices simplify many computations due to their property $A^2 = A$, which helps reduce higher powers of the matrix to simpler forms.