If A, B and C are square matrices of same order and I is an identity matrix of the same order, such that `C^(2)=CB+AC and AB=I`, then `(C-A)^(-1)` is equal to

Options

A. `C-A`
B. `C-B`
C. `B-C`
D. `B+I`

Answer

To solve the problem, we need to find the inverse of the matrix \( C - A \) given the conditions \( C^2 = CB + AC \) and \( AB = I \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ C^2 = CB + AC \]

2. **Rearrange the equation**:
Move \( CB \) to the left side:
\[ C^2 - CB = AC \]

3. **Factor out \( C \)**:
From the left side, factor out \( C \):
\[ C(C - B) = AC \]

4. **Add and subtract \( AB \)**:
To manipulate the equation, add and subtract \( AB \):
\[ C(C - B) = AC + AB - AB \]
This simplifies to:
\[ C(C - B) = A(C - B) \]

5. **Factor out \( C - B \)**:
Now, we can factor out \( C - B \):
\[ C(C - B) - A(C - B) = 0 \]
This can be factored as:
\[ (C - B)(C - A) = 0 \]

6. **Use the property of matrices**:
Since \( AB = I \), we know that \( A \) and \( B \) are inverses of each other. Thus, we can replace \( AB \) with \( I \):
\[ (C - B)(C - A) = I \]

7. **Conclusion about inverses**:
Since the product of \( (C - B) \) and \( (C - A) \) equals the identity matrix \( I \), we conclude that:
\[ (C - A)^{-1} = C - B \]

### Final Answer:
Thus, the inverse of \( C - A \) is:
\[ (C - A)^{-1} = C - B \]