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 need to analyze the given equation and apply properties of determinants and adjoint matrices. Here’s a step-by-step solution:

### Step 1: Understand the Given Equation
We are given:
\[ A \cdot \text{adj}(3A) = 5A \cdot A^T \]

### Step 2: Apply Determinant on Both Sides
Taking the determinant on both sides:
\[ \det(A \cdot \text{adj}(3A)) = \det(5A \cdot A^T) \]

### Step 3: Use Properties of Determinants
Using the properties of determinants:
1. \(\det(kA) = k^n \cdot \det(A)\) for a scalar \(k\) and matrix \(A\) of order \(n\).
2. \(\det(A^T) = \det(A)\).
3. \(\det(\text{adj}(A)) = \det(A)^{n-1}\) for an \(n \times n\) matrix \(A\).

For our case, since \(A\) is a \(3 \times 3\) matrix, we have:
- \(\det(3A) = 3^3 \cdot \det(A) = 27 \cdot \det(A)\)
- \(\det(\text{adj}(3A)) = (\det(3A))^2 = (27 \cdot \det(A))^2 = 729 \cdot \det(A)^2\)

### Step 4: Substitute Determinants
Substituting these into our determinant equation:
\[ \det(A) \cdot (729 \cdot \det(A)^2) = \det(5A) \cdot \det(A^T) \]
\[ \det(A) \cdot 729 \cdot \det(A)^2 = (5^3 \cdot \det(A)^3) \]
\[ 729 \cdot \det(A)^3 = 125 \cdot \det(A)^3 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 729 \cdot \det(A)^3 - 125 \cdot \det(A)^3 = 0 \]
Factoring out \(\det(A)^3\):
\[ \det(A)^3 (729 - 125) = 0 \]
Since \(\det(A) \neq 0\) (as \(A\) is non-singular), we have:
\[ 729 - 125 = 0 \]
Thus,
\[ \det(A)^3 = 125 \]
This simplifies to:
\[ \det(A) = \left(\frac{5}{9}\right)^3 \]

### Step 6: Find the Determinant of the Inverse
We know:
\[ \det(A^{-1}) = \frac{1}{\det(A)} \]
Thus:
\[ \det(A^{-1}) = \frac{1}{\left(\frac{5}{9}\right)^3} = \left(\frac{9}{5}\right)^3 \]

### Step 7: Find the Cube Root
Now, we need to find:
\[ \sqrt[3]{|\det(A^{-1})|} = \sqrt[3]{\left(\frac{9}{5}\right)^3} = \frac{9}{5} \]

### Final Answer
Thus, the cube root of the determinant of \(A^{-1}\) is:
\[ \frac{9}{5} \]