Q.If $A$ is square matrix such that $A^2 = A$, show that $(I + A)^3 = 7A + I$.

Concept: Idempotent Matrix Property — a matrix satisfying $A^2 = A$ is idempotent.

Step 1: Expand $(I + A)^3$ using the binomial theorem (since $I$ and $A$ commute):

Step 2: Use $A^2 = A$ to simplify $A^2$ and $A^3$: …

We start with the given condition: $A^2 = A$. A matrix satisfying this is called idempotent — it's a projection-like matrix where once you apply it, applying it again does nothing. This property is the engine of the entire simplification.

The expression we need to simplify is $(I + A)^3$. Since $I$ and $A$ commute (the identity commutes with everything), we can expand this using the binomial theorem, just like with numbers.

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

For any two commuting matrices $X$ and $Y$, $(X+Y)^3 = X^3 + 3X^2Y + 3XY^2 + Y^3$. Here $X = I$ and $Y = A$, so:

Step 2: Simplify powers of $I$ and $A$.

• $I^3 = I$, $I^2 = I$, and $I A = A$ (since identity times any matrix is that matrix).
• For $A$: we know $A^2 = A$. Then $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$. So every power $A^n$ for $n \ge 1$ is just $A$.

Thus the expansion becomes:

Step 3: Combine like terms.

That's exactly $7A + I$, which is the same as $I + 7A$ (addition is commutative). …

Method: Simplifying matrix powers using a defining relation (idempotency)

When a matrix satisfies a special relation such as $A^2=A$ (idempotent), you use that relation to collapse every higher power back to $A$, rather than multiplying repeatedly. Because the identity $I$ commutes with every matrix, ordinary binomial expansion is also allowed.

Steps

Step 1: Expand using the binomial theorem — valid because $I$ commutes with $A$.

The scalar-like expansion works since $IA=AI=A$.

Step 2: Reduce each power using the defining relation. …

Common Mistakes

Mistake 1: Not reducing $A^3$ (and higher powers) using $A^2=A$.

Why it's wrong: leaving $A^2$ and $A^3$ in the expansion prevents you from reaching $I+7A$. Correct approach: use $A^2=A\Rightarrow A^3=A$ to replace every power by $A$.

Mistake 2: Doubting the binomial expansion because "matrices don't commute".

Why it's wrong: here the two "terms" are $I$ and $A$, and $I$ commutes with everything, so $(I+A)^3$ expands exactly like $(1+a)^3$. Correct approach: expand normally, since one factor is the identity. …