Q.$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Rightarrow A^{100} =$

Why This Works

The condition $A^2 = A$ is called idempotence — it means $A$ is a projection matrix. Once you square it, you get the same matrix back. This has a powerful consequence: for any integer $n \ge 1$, $A^n = A$. That’s because $A^3 = A^2 \cdot A = A \cdot A = A$, and by induction it holds for all higher powers.

So when we expand $(I - A)^3$, every power of $A$ beyond the first collapses to just $A$. That makes the algebra extremely clean — no infinite series, no diagonalization, just a simple substitution.

Step-by-Step

1. Write the expansion. $(I - A)^3 = (I - A)(I - A)(I - A)$. Expand using the binomial theorem (since $I$ and $A$ commute — $I$ commutes with everything):

2. Apply the idempotent property. Because $A^2 = A$, we have $A^3 = A^2 \cdot A = A \cdot A = A$. So $A^2 = A$ and $A^3 = A$. Substitute:

3. Simplify. The $-3A$ and $+3A$ cancel:

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

4. Compare with the given form. The problem states $(I - A)^3 = xA + I$. We just found $(I - A)^3 = I - A$. Therefore:

5. Solve for $x$. Subtract $I$ from both sides:

This is a matrix equation. Factor $A$ (careful: we cannot divide by a matrix, but we can compare coefficients if $A \neq 0$).

Rewrite as: …

We are given that $A^2 = A$. This is the defining property of an idempotent matrix — a matrix that, when multiplied by itself, gives itself back. Such matrices behave like "projections" in linear algebra. The trick here is that $I - A$ also becomes idempotent, because:

So $(I - A)^2 = I - A$ as well. This symmetry makes expansions clean.

Now, let’s work through the problem step by step.

1. Expand $(I - A)^3$ using the binomial theorem. Since matrix multiplication is distributive and associative (and $I$ commutes with any matrix), we can expand just like numbers:

But $I^3 = I$, $I^2A = A$, $IA^2 = A^2$, and $A^3 = A \cdot A^2 = A \cdot A = A^2 = A$ (using $A^2 = A$ repeatedly). So:

2. Replace $A^2$ and $A^3$ using $A^2 = A$. Since $A^2 = A$, then $A^3 = A^2 \cdot A = A \cdot A = A$ as well. So: $(I - A)^3 = I - 3A + 3A - A = I - A$ …