Let A be a non - singular matrix of order 3 such that `Aadj (3A)=5A A^(T)`, then `root3(|A^(-1)|)` is equal to

Answer

To solve the problem, we will follow these steps:

1. **Understanding the given equation**: We have the equation \( A \cdot \text{adj}(3A) = 5A A^T \).

2. **Taking the determinant of both sides**: We will apply the determinant to both sides of the equation.

\[ \det(A \cdot \text{adj}(3A)) = \det(5A A^T) \]

3. **Using properties of determinants**:
- The determinant of a product of matrices is the product of their determinants.
- The determinant of the adjugate of a matrix \( B \) is given by \( \det(\text{adj}(B)) = \det(B)^{n-1} \), where \( n \) is the order of the matrix.
- For a scalar \( k \) and a matrix \( B \), \( \det(kB) = k^n \det(B) \).

Applying these properties, we have:

\[ \det(A) \cdot \det(\text{adj}(3A)) = \det(5) \cdot \det(A) \cdot \det(A^T) \]

Since \( \det(A^T) = \det(A) \), we can simplify this to:

\[ \det(A) \cdot \det(\text{adj}(3A)) = 5 \cdot \det(A) \cdot \det(A) \]

4. **Calculating \( \det(\text{adj}(3A)) \)**:
\[ \det(\text{adj}(3A)) = \det(3A)^{2} = (3^3 \cdot \det(A))^2 = 27 \cdot \det(A)^2 \]

5. **Substituting back**:
\[ \det(A) \cdot (27 \cdot \det(A)^2) = 5 \cdot \det(A)^2 \]

Simplifying this gives:
\[ 27 \cdot \det(A)^3 = 5 \cdot \det(A)^2 \]

6. **Dividing both sides by \( \det(A)^2 \)** (since \( A \) is non-singular, \( \det(A) \neq 0 \)):
\[ 27 \cdot \det(A) = 5 \]

7. **Solving for \( \det(A) \)**:
\[ \det(A) = \frac{5}{27} \]

8. **Finding \( \det(A^{-1}) \)**:
\[ \det(A^{-1}) = \frac{1}{\det(A)} = \frac{27}{5} \]

9. **Finding \( \sqrt[3]{|\det(A^{-1})|} \)**:
\[ \sqrt[3]{|\det(A^{-1})|} = \sqrt[3]{\frac{27}{5}} = \frac{\sqrt[3]{27}}{\sqrt[3]{5}} = \frac{3}{\sqrt[3]{5}} \]

Thus, the final answer is:
\[ \sqrt[3]{|\det(A^{-1})|} = \frac{3}{\sqrt[3]{5}} \]