Let A a non singular square matrix of order `3xx3`. Then `|adjA|` is equal to

Options

A. `|A|`
B. `|A|^(2)`
C. `|A|^(3)`
D. `3|A|`

Answer

To find the value of \(|\text{adj} A|\) for a non-singular square matrix \(A\) of order \(3 \times 3\), we can use the properties of determinants and adjoint matrices.

### Step-by-step Solution:

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

2. **Taking Determinants**:
Taking the determinant on both sides of the equation \(A \cdot \text{adj} A = \det(A) \cdot I\), we have:
\[ \det(A \cdot \text{adj} A) = \det(\det(A) \cdot I) \]

3. **Using the Determinant Property**:
The property of determinants states that:
\[ \det(AB) = \det(A) \cdot \det(B) \]
Thus, we can rewrite the left side:
\[ \det(A) \cdot \det(\text{adj} A) = \det(A) \cdot \det(I) \]

4. **Determinant of Identity Matrix**:
The determinant of the identity matrix \(I\) of order \(n\) is \(1\):
\[ \det(I) = 1 \]
Therefore, the equation simplifies to:
\[ \det(A) \cdot \det(\text{adj} A) = \det(A) \]

5. **Dividing by \(\det(A)\)**:
Since \(A\) is non-singular, \(\det(A) \neq 0\). We can divide both sides by \(\det(A)\):
\[ \det(\text{adj} A) = 1 \]

6. **Using the Formula for Adjoint Determinant**:
For a square matrix of order \(n\), the determinant of the adjoint matrix is given by:
\[ \det(\text{adj} A) = \det(A)^{n-1} \]
In our case, \(n = 3\):
\[ \det(\text{adj} A) = \det(A)^{3-1} = \det(A)^2 \]

7. **Conclusion**:
Since we have established that \(\det(\text{adj} A) = \det(A)^2\), and knowing that \(\det(A)\) is non-zero for a non-singular matrix, we conclude:
\[ |\text{adj} A| = \det(A)^2 \]

### Final Result:
Thus, the value of \(|\text{adj} A|\) is equal to \(\det(A)^2\).