Question:
If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3$ is equal to:
If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3$ is equal to:
Updated On: Jan 14, 2026
- $7A - I$
- $7A + I$
- $7A$
- $I - 7A$
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The Correct Option is B
Approach Solution - 1
1. Understand the problem:
Given a square matrix A satisfying \( A^2 = A \), we need to find \( (I + A)^3 \).
2. Expand \( (I + A)^3 \):
Using the binomial expansion:
\[ (I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 \]
3. Simplify using \( A^2 = A \):
Since \( A^2 = A \), then \( A^3 = A^2 = A \). Thus:
\[ (I + A)^3 = I + 3A + 3A + A = I + 7A \]
Correct Answer: (B) 7A + I
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Approach Solution -2
Given that \( A \) is a square matrix such that \( A^2 = A \). This means \( A \) is an idempotent matrix. We want to find \( (I + A)^3 \).
Let's expand \( (I + A)^3 \):
\[ (I + A)^3 = (I + A)(I + A)(I + A) \] \[ = (I + 2A + A^2)(I + A) \quad \text{(using } (I + A)^2 = I + 2A + A^2\text{)} \] \[ = (I + 2A + A)(I + A) \quad \text{(since } A^2 = A\text{)} \] \[ = (I + 3A)(I + A) \] \[ = I + 3A + A + 3A^2 \] \[ = I + 4A + 3A \quad \text{(since } A^2 = A\text{)} \] \[ = I + 7A \]Therefore, \( (I + A)^3 = I + 7A \).
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