If M is any sauare matrix of order 3 over R and if M' be the transpose of M, then adj (M') - (adj M)' is equal to

Options

A. M
B. M'
C. null matrix
D. identity matrix

Answer

To solve the problem, we need to find the expression \( \text{adj}(M') - (\text{adj}(M))' \) where \( M \) is a square matrix of order 3 and \( M' \) is the transpose of \( M \).

### Step-by-Step Solution:

1. **Understanding the Adjoint of a Matrix**:
The adjoint (or adjugate) of a matrix \( A \) is defined as the transpose of the cofactor matrix of \( A \). For a square matrix \( M \), the adjoint is denoted as \( \text{adj}(M) \).

2. **Properties of the Adjoint**:
One important property of the adjoint is:
\[ \text{adj}(M') = (\text{adj}(M))' \]
This means that the adjoint of the transpose of a matrix is equal to the transpose of the adjoint of the matrix.

3. **Applying the Property**:
Given the property from step 2, we can rewrite our expression:
\[ \text{adj}(M') - (\text{adj}(M))' = (\text{adj}(M))' - (\text{adj}(M))' \]

4. **Simplifying the Expression**:
Since both terms are equal, we have:
\[ (\text{adj}(M))' - (\text{adj}(M))' = 0 \]

5. **Conclusion**:
Therefore, the expression \( \text{adj}(M') - (\text{adj}(M))' \) simplifies to:
\[ 0 \]

### Final Answer:
\[ \text{adj}(M') - (\text{adj}(M))' = 0 \]