If A and B are non - singular matrices of order `3xx3`, such that `A=(adjB)` and `B=(adjA)`, then det `(A)+det(B)` is equal to (where `det(M)` represents the determinant of matrix M and adj M represents the adjoint matrix of matrix M)

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the relationships between the matrices A and B given that \( A = \text{adj}(B) \) and \( B = \text{adj}(A) \).

### Step-by-Step Solution:

1. **Understanding the Adjoint Matrix**:
The adjoint of a matrix \( M \), denoted as \( \text{adj}(M) \), is defined such that:
\[ M \cdot \text{adj}(M) = \det(M) I \]
where \( I \) is the identity matrix.

2. **Applying the Adjoint Property**:
Given \( A = \text{adj}(B) \), we can substitute this into the adjoint property:
\[ B \cdot A = \det(B) I \]
Substituting \( A \) gives:
\[ B \cdot \text{adj}(B) = \det(B) I \]

3. **Using the Adjoint of the Adjoint**:
We know that:
\[ \text{adj}(\text{adj}(M)) = \det(M)^{n-1} M \]
For a \( 3 \times 3 \) matrix, this becomes:
\[ \text{adj}(\text{adj}(A)) = \det(A)^2 A \]
Since \( B = \text{adj}(A) \), we can write:
\[ A = \det(B)^{2} B \]

4. **Finding Determinants**:
Now we can find the determinants of both sides:
\[ \det(A) = \det(\det(B)^2 B) = \det(B)^2 \det(B) = \det(B)^3 \]
This implies:
\[ \det(A) = \det(B)^3 \]

5. **Substituting Back**:
Similarly, using \( A = \text{adj}(B) \):
\[ \det(B) = \det(\det(A)^2 A) = \det(A)^2 \det(A) = \det(A)^3 \]
This gives:
\[ \det(B) = \det(A)^3 \]

6. **Setting Up the Equations**:
We now have two equations:
\[ \det(A) = \det(B)^3 \quad \text{and} \quad \det(B) = \det(A)^3 \]

7. **Letting \( x = \det(A) \) and \( y = \det(B) \)**:
We can rewrite the equations as:
\[ x = y^3 \quad \text{and} \quad y = x^3 \]

8. **Solving the System**:
Substitute \( y = x^3 \) into \( x = y^3 \):
\[ x = (x^3)^3 = x^9 \]
This gives:
\[ x^9 - x = 0 \implies x(x^8 - 1) = 0 \]
Thus, \( x = 0 \) or \( x^8 = 1 \). Since \( A \) and \( B \) are non-singular, \( x \neq 0 \). Therefore, \( x = 1 \) (the only real solution).

9. **Finding \( \det(B) \)**:
If \( \det(A) = 1 \), then substituting back gives:
\[ \det(B) = \det(A)^3 = 1^3 = 1 \]

10. **Final Calculation**:
Now we can find \( \det(A) + \det(B) \):
\[ \det(A) + \det(B) = 1 + 1 = 2 \]

### Conclusion:
Thus, the final answer is:
\[ \det(A) + \det(B) = 2 \]