Answer :
Certainly! Let's explore this step by step.
Given:
- \( A \) is a square matrix such that \( A^2 = A \).
- We need to find the expression for \( (1 + A)^3 - 7A \).
First, let's understand the property of \( A \):
- Since \( A^2 = A \), \( A \) is known as an idempotent matrix.
Now, let's simplify the expression \( (1 + A)^3 - 7A \) step by step.
1. We will use the binomial expansion to expand \( (1 + A)^3 \).
[tex]\[ (1 + A)^3 = 1 + 3A + 3A^2 + A^3 \][/tex]
2. We know from the given property of the matrix \( A \) that \( A^2 = A \). Therefore:
[tex]\[ A^2 = A \quad \text{and} \quad A^3 = A^2 \cdot A = A \cdot A = A \][/tex]
3. Substituting these results back into the binomial expansion:
[tex]\[ (1 + A)^3 = 1 + 3A + 3A + A = 1 + 3A + 3A + A \][/tex]
4. Combine like terms:
[tex]\[ (1 + A)^3 = 1 + 7A \][/tex]
Now, we need to find \( (1 + A)^3 - 7A \):
[tex]\[ (1 + A)^3 - 7A = (1 + 7A) - 7A \][/tex]
5. Simplify by subtracting \( 7A \):
[tex]\[ (1 + 7A) - 7A = 1 + 7A - 7A = 1 \][/tex]
Therefore, the expression \( (1 + A)^3 - 7A \) simplifies to:
[tex]\[ \boxed{1} \][/tex]
Given:
- \( A \) is a square matrix such that \( A^2 = A \).
- We need to find the expression for \( (1 + A)^3 - 7A \).
First, let's understand the property of \( A \):
- Since \( A^2 = A \), \( A \) is known as an idempotent matrix.
Now, let's simplify the expression \( (1 + A)^3 - 7A \) step by step.
1. We will use the binomial expansion to expand \( (1 + A)^3 \).
[tex]\[ (1 + A)^3 = 1 + 3A + 3A^2 + A^3 \][/tex]
2. We know from the given property of the matrix \( A \) that \( A^2 = A \). Therefore:
[tex]\[ A^2 = A \quad \text{and} \quad A^3 = A^2 \cdot A = A \cdot A = A \][/tex]
3. Substituting these results back into the binomial expansion:
[tex]\[ (1 + A)^3 = 1 + 3A + 3A + A = 1 + 3A + 3A + A \][/tex]
4. Combine like terms:
[tex]\[ (1 + A)^3 = 1 + 7A \][/tex]
Now, we need to find \( (1 + A)^3 - 7A \):
[tex]\[ (1 + A)^3 - 7A = (1 + 7A) - 7A \][/tex]
5. Simplify by subtracting \( 7A \):
[tex]\[ (1 + 7A) - 7A = 1 + 7A - 7A = 1 \][/tex]
Therefore, the expression \( (1 + A)^3 - 7A \) simplifies to:
[tex]\[ \boxed{1} \][/tex]
