If A and B are any two square matrices of the same order than ,

Options

A. `(AB)'=A'B'`
B. adj ( AB) = adj( A) adj (B)
C. (AB) '=B'A'
D. AB = O `rArr ` A = O or B =O

Answer

To solve the problem regarding the properties of square matrices A and B of the same order, we will analyze the given statements and determine which ones are correct.

### Step-by-Step Solution:

1. **Understanding the Transpose of a Product of Matrices**:
We start with the property of the transpose of the product of two matrices. The property states that:
\[ (AB)^T = B^T A^T \]
This means that when we take the transpose of the product of two matrices A and B, it is equal to the product of the transpose of B and the transpose of A, in that order.

**Hint**: Remember the rule for transposing the product of matrices.

2. **Applying the Property to the Given Matrices**:
Given the matrices A and B, we can apply the property:
\[ (AB)^T = B^T A^T \]
This indicates that if we take the transpose of the product AB, we can rearrange the order of multiplication when we take the transpose.

**Hint**: Focus on how the order of multiplication changes when taking the transpose.

3. **Adjoint of Matrices**:
The adjoint of a matrix is denoted as adj(A). One important property of adjoints is:
\[ \text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A) \]
This means that the adjoint of the product of two matrices is equal to the product of the adjoint of B and the adjoint of A, in that order.

**Hint**: Recall the property of adjoints when dealing with the product of matrices.

4. **Analyzing the Options**:
Now, let's analyze the options provided in the question:
- **Option 1**: \( (AB)^T = B^T A^T \) (Correct)
- **Option 2**: \( \text{adj}(A) \cdot \text{adj}(B) = \text{adj}(AB) \) (Correct)
- **Option 3**: \( (A^T B^T) = (B A)^T \) (Correct)
- **Option 4**: If \( AB = 0 \), then either \( A = 0 \) or \( B = 0 \) (This is not necessarily true; both can be non-zero matrices resulting in a zero product).

**Hint**: Carefully check each option against the properties of matrices discussed.

5. **Conclusion**:
Based on the analysis, options 1, 2, and 3 are correct. Option 4 is not necessarily true as both matrices can be non-zero and still yield a product of zero.

### Final Answer:
The correct options are 1, 2, and 3.