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 step by step, we need to manipulate the given equations involving the matrices A, B, and C.

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

**Step 2: Rearrange the equation.**
We can rearrange this equation to isolate terms involving C:
\[ C^2 - CB = AC \]

**Step 3: Factor out C from the left-hand side.**
Notice that we can factor C from the left-hand side:
\[ C(C - B) = AC \]

**Step 4: Multiply both sides by A.**
Since we know that \( AB = I \) (the identity matrix), we can multiply both sides of the equation by A:
\[ A(C(C - B)) = A(AC) \]
This simplifies to:
\[ A(C(C - B)) = I \]

**Step 5: Rearranging the equation.**
Now, we can write:
\[ C(C - B) = I \]
This implies:
\[ C - B = C^{-1} \]

**Step 6: Express C in terms of A and B.**
From the earlier steps, we have:
\[ C - A = AB \]
Since \( AB = I \), we can substitute:
\[ C - A = I \]

**Step 7: Find the inverse.**
Now we need to find \( (C - A)^{-1} \). Since we have established that:
\[ C - A = I \]
The inverse of the identity matrix is itself:
\[ (C - A)^{-1} = I^{-1} = I \]

Thus, the final answer is:
\[ (C - A)^{-1} = B \]