If D is determinant of order three of `Delta` is a determinant formed by the cofactors of determinant D: then

Options

A. `Delta=D^(2)`
B. `D=0"implies"Delta=0`
C. If D=27 then `Delta` is perfect cube
D. If D=27, then `Delta` is a perfect square

Answer

To solve the problem, we need to analyze the relationship between a determinant \( D \) of order 3 and the determinant \( \Delta \) formed by the cofactors of \( D \).

### Step-by-step Solution:

1. **Understanding the Determinant and Cofactor Matrix**:
Let \( D \) be a determinant of order 3, represented by a matrix \( A \) of size \( 3 \times 3 \). The cofactor matrix \( C \) of \( A \) is formed by taking the cofactors of each element of \( A \).

2. **Cofactor Matrix and its Determinant**:
The determinant of the cofactor matrix \( C \) can be expressed in terms of the determinant \( D \):
\[ \text{det}(C) = D^{n-1} \]
where \( n \) is the order of the matrix \( A \). Since \( A \) is of order 3, we have:
\[ \text{det}(C) = D^{3-1} = D^2 \]

3. **Relating \( \Delta \) and \( D \)**:
Given that \( \Delta \) is the determinant formed by the cofactors of \( D \), we can write:
\[ \Delta = D^2 \]

4. **Analyzing the Options**:
Now, we will analyze the four conditions given in the problem:
- **Option A**: \( \Delta = D^2 \) - This is correct as derived above.
- **Option B**: If \( D = 0 \), then \( \Delta = 0 \) - This is also correct since if \( D = 0 \), then \( \Delta = 0^2 = 0 \).
- **Option C**: If \( D = 27 \), then \( \Delta = 27^2 = 729 \) - This is correct as \( \Delta \) will be \( 729 \).
- **Option D**: If \( D = 27 \), then \( \Delta \) is a perfect square - This is correct because \( 729 = 27^2 \) is indeed a perfect square.

5. **Conclusion**:
All four options are correct based on our analysis.

### Final Answer:
All options A, B, C, and D are correct.