If A is `3xx3` non-singular matrix such that `A^(-1)+(adj.A)=0," then det "(A)=`

Options

A. 1
B. `-1`
C. 0
D. `1//2`

Answer

To solve the problem, we need to find the determinant of a non-singular \(3 \times 3\) matrix \(A\) given that:

\[ A^{-1} + \text{adj}(A) = 0 \]

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ A^{-1} + \text{adj}(A) = 0 \]
This can be rearranged to:
\[ A^{-1} = -\text{adj}(A) \]

2. **Using the Formula for Inverse**:
We know that the inverse of a matrix \(A\) can be expressed as:
\[ A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) \]
Substituting this into our rearranged equation gives:
\[ \frac{1}{\det(A)} \cdot \text{adj}(A) = -\text{adj}(A) \]

3. **Factoring Out \(\text{adj}(A)\)**:
We can factor out \(\text{adj}(A)\) (assuming \(\text{adj}(A) \neq 0\)):
\[ \text{adj}(A) \left( \frac{1}{\det(A)} + 1 \right) = 0 \]
Since \(A\) is non-singular, \(\text{adj}(A) \neq 0\). Therefore, we can set the term in parentheses to zero:
\[ \frac{1}{\det(A)} + 1 = 0 \]

4. **Solving for \(\det(A)\)**:
Rearranging gives:
\[ \frac{1}{\det(A)} = -1 \]
Taking the reciprocal:
\[ \det(A) = -1 \]

5. **Conclusion**:
Thus, the determinant of the matrix \(A\) is:
\[ \det(A) = -1 \]

### Final Answer:
\[ \det(A) = -1 \]