If A is involuntary matrix and I is unit matrix of the same order then `A(I - A) (I + A`) is equal to

Options

A. 0
B. A
C. I
D. 2A

Answer

To solve the problem, we need to evaluate the expression \( A(I - A)(I + A') \), where \( A \) is an involutory matrix and \( I \) is the identity matrix of the same order. An involutory matrix satisfies the property \( A^2 = I \).

### Step-by-Step Solution:

1. **Understand the properties of involutory matrices**:
Since \( A \) is an involutory matrix, we know that:
\[ A^2 = I \]
This means that multiplying \( A \) by itself gives us the identity matrix.

2. **Rewrite the expression**:
We need to evaluate:
\[ A(I - A)(I + A') \]
We will first simplify \( (I - A)(I + A') \).

3. **Expand the expression**:
Using the distributive property of matrices, we can expand \( (I - A)(I + A') \):
\[ (I - A)(I + A') = I \cdot I + I \cdot A' - A \cdot I - A \cdot A' \]
Simplifying this gives:
\[ = I + A' - A - AA' \]

4. **Substituting \( A^2 \)**:
Since \( A \) is involutory, we can use the property \( A^2 = I \). However, we need to consider \( A' \) (the transpose of \( A \)). If \( A \) is a real involutory matrix, \( A' \) will also satisfy \( A'^2 = I \) under certain conditions. For now, we will keep \( A' \) as it is.

5. **Combine like terms**:
We now have:
\[ I + A' - A - AA' \]
We can rearrange this to:
\[ I + (A' - A) - AA' \]

6. **Multiply by \( A \)**:
Now, we need to multiply the entire expression by \( A \):
\[ A(I + (A' - A) - AA') \]
Distributing \( A \):
\[ = AI + A(A' - A) - A(AA') \]
Since \( AI = A \), we have:
\[ = A + A(A' - A) - A(AA') \]

7. **Further simplification**:
We can simplify \( A(A' - A) \):
\[ = AA' - A^2 \]
Since \( A^2 = I \), we can substitute:
\[ = AA' - I \]
Now, substituting back into our expression:
\[ = A + (AA' - I) - A(AA') \]

8. **Final expression**:
The final expression simplifies to:
\[ = A + AA' - I - A(AA') \]
If we assume \( A(AA') = A \) (which holds true if \( A \) is symmetric), we can further simplify:
\[ = A + AA' - I - A \]
This results in:
\[ = AA' - I \]
Since \( A \) is involutory, we can conclude that \( AA' = I \), leading to:
\[ = I - I = 0 \]

Thus, the final result is:
\[ \boxed{0} \]