Let A be a square matrix of order 3 such that det. `(A)=1/3`, then the value of det. `("adj. "A^(-1))` is

Options

A. `1//9`
B. `1//3`
C. `3`
D. `9`

Explanation

To find the value of \( \text{det}(\text{adj}(A^{-1})) \) given that \( \text{det}(A) = \frac{1}{3} \), we can follow these steps:

### Step 1: Understand the relationship between the determinant of a matrix and its adjugate.
The determinant of the adjugate of a matrix \( A \) is given by the formula:
\[ \text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1} \]
where \( n \) is the order of the matrix \( A \).

### Step 2: Apply the formula to the inverse of the matrix.
Since we need to find \( \text{det}(\text{adj}(A^{-1})) \), we can modify the formula:
\[ \text{det}(\text{adj}(A^{-1})) = (\text{det}(A^{-1}))^{n-1} \]

### Step 3: Find the determinant of the inverse of matrix \( A \).
We know that:
\[ \text{det}(A^{-1}) = \frac{1}{\text{det}(A)} \]
Given \( \text{det}(A) = \frac{1}{3} \), we can find \( \text{det}(A^{-1}) \):
\[ \text{det}(A^{-1}) = \frac{1}{\frac{1}{3}} = 3 \]

### Step 4: Substitute \( n \) and calculate \( \text{det}(\text{adj}(A^{-1})) \).
Since \( A \) is a \( 3 \times 3 \) matrix, \( n = 3 \):
\[ \text{det}(\text{adj}(A^{-1})) = (\text{det}(A^{-1}))^{3-1} = (\text{det}(A^{-1}))^2 \]
Substituting the value we found:
\[ \text{det}(\text{adj}(A^{-1})) = (3)^2 = 9 \]

### Final Answer:
Thus, the value of \( \text{det}(\text{adj}(A^{-1})) \) is \( 9 \).

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