If A, B, and C are three square matrices of the same order, then `AB=AC implies B=C`. Then

Options

A. `|A| ne 0`
B. A is invertible
C. A may be orthogonal
D. A is symmetric

Answer

To solve the problem, we need to prove that if \( AB = AC \) for square matrices \( A \), \( B \), and \( C \) of the same order, then \( B = C \) under the condition that \( A \) is invertible.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ AB = AC \]

2. **Subtract \( AC \) from both sides:**
\[ AB - AC = 0 \]

3. **Factor out \( A \) from the left side:**
\[ A(B - C) = 0 \]

4. **Assume \( A \) is invertible:**
If \( A \) is invertible, it has an inverse denoted as \( A^{-1} \).

5. **Multiply both sides of the equation \( A(B - C) = 0 \) by \( A^{-1} \):**
\[ A^{-1}(A(B - C)) = A^{-1}(0) \]

6. **Using the property of the inverse matrix:**
\[ (A^{-1}A)(B - C) = 0 \]
This simplifies to:
\[ I(B - C) = 0 \]
where \( I \) is the identity matrix.

7. **Since \( I \) is the identity matrix, we can simplify further:**
\[ B - C = 0 \]

8. **Thus, we conclude:**
\[ B = C \]

### Conclusion:
If \( A \) is invertible, then from the equation \( AB = AC \), we can conclude that \( B = C \).