Suppose `I+A` is non-singular. Let `B=(l+A)^(-1)` and C=l-A, then ……. (where l,A,O are identity, square and null matrices of order n respectively)

Options

A. BC=CB
B. BC=O
C. BC=l
D. none of these

Explanation

To solve the problem, we need to find the products \( BC \) and \( CB \) where \( B = (I + A)^{-1} \) and \( C = I - A \). We will show that \( BC = CB \).

### Step-by-Step Solution:

1. **Define the matrices**:
- Let \( B = (I + A)^{-1} \)
- Let \( C = I - A \)

2. **Calculate \( BC \)**:
\[ BC = (I + A)^{-1}(I - A) \]
We can use the expansion of \( (I + A)^{-1} \) using the Neumann series:
\[ (I + A)^{-1} = I - A + A^2 - A^3 + \ldots \]
Thus,
\[ BC = (I - A + A^2 - A^3 + \ldots)(I - A) \]

3. **Distributing \( (I - A) \)**:
\[ BC = (I - A)(I) + (I - A)(A^2) - (I - A)(A^3) + \ldots \]
- The first term gives \( I - A \).
- The second term gives \( A^2 - A^3 \).
- The third term gives \( -A^3 + A^4 \).
- Continuing this pattern, we can see that the series will yield:
\[ BC = I - 2A + 2A^2 - 2A^3 + \ldots \]

4. **Calculate \( CB \)**:
\[ CB = (I - A)(I + A)^{-1} \]
Using the same expansion for \( (I + A)^{-1} \):
\[ CB = (I - A)(I - A + A^2 - A^3 + \ldots) \]

5. **Distributing \( (I - A) \)**:
\[ CB = (I - A)(I) + (I - A)(-A) + (I - A)(A^2) - (I - A)(A^3) + \ldots \]
- The first term gives \( I - A \).
- The second term gives \( -A + A^2 \).
- The third term gives \( A^2 - A^3 \).
- Continuing this pattern, we can see that the series will yield:
\[ CB = I - 2A + 2A^2 - 2A^3 + \ldots \]

6. **Conclusion**:
Since both \( BC \) and \( CB \) yield the same result:
\[ BC = CB \]
Therefore, we conclude that \( BC = CB \).