We are given a square matrix $A$ such that $A^2 = A$, and we need to find the value of $m$ in the equation:

$(I + A)^3 - 18A = I + mA$

Step 1: Expand $(I + A)^3$

Using the binomial expansion for $(I + A)^3$, we have:

$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$

Since $I^3 = I$, $I^2 = I$, and $A^2 = A$, this simplifies to:

$(I + A)^3 = I + 3A + 3A + A^3 = I + 6A + A^3$

Now, because $A^2 = A$, we can simplify $A^3$ as follows:

$A^3 = A \cdot A^2 = A \cdot A = A^2 = A$

Thus:

$(I + A)^3 = I + 6A + A$

$(I + A)^3 = I + 7A$

Step 2: Substitute into the original equation

We substitute $(I + A)^3 = I + 7A$ into the given equation $(I + A)^3 - 18A = I + mA$:

$I + 7A - 18A = I + mA$

Simplify the left-hand side:

$I + (7A - 18A) = I + mA$

$I - 11A = I + mA$

Step 3: Compare terms

Now, compare the coefficients of $I$ and $A$ on both sides:

• The constant term: $I = I$ (which is already balanced).
• The coefficient of $A$: $-11A = mA$.

Thus, we have:

$m = -11$

Final Answer:

The value of $m$ is $\boxed{-11}$.

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Would you like further details or clarification?

Here are 5 related questions you can explore:

1. How would you prove that $A^2 = A$ implies $A$ is idempotent?
2. What happens if $A$ is a diagonal matrix with eigenvalues 0 or 1?
3. How does the result change if we replace $(I + A)^3$ with higher powers, such as $(I + A)^n$?
4. Can this type of matrix $A$ be invertible? Under what conditions?
5. What are some practical applications of idempotent matrices in linear algebra?

Tip: When solving matrix equations, always try to simplify using properties like $A^2 = A$ (idempotent property) before expanding complex expressions.