If `A=[a_("ij")]_(4xx4)`, such that `a_("ij")={(2",","when "i=j),(0",","when "i ne j):}`, then `{("det (adj (adj A))")/(7)}` is (where `{*}` represents fractional part function)

Options

A. `1//7`
B. `2//7`
C. `3//7`
D. none of these

Explanation

To solve the problem, we need to analyze the matrix \( A \) and compute the required expression step by step.

### Step 1: Define the Matrix \( A \)

Given that \( A = [a_{ij}]_{4 \times 4} \) where:
- \( a_{ij} = 2 \) when \( i = j \) (diagonal elements)
- \( a_{ij} = 0 \) when \( i \neq j \) (off-diagonal elements)

The matrix \( A \) can be written as:
\[ A = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} \]

### Step 2: Calculate the Determinant of \( A \)

The determinant of a diagonal matrix is the product of its diagonal elements. Therefore, we have:
\[ \text{det}(A) = 2 \times 2 \times 2 \times 2 = 2^4 = 16 \]

### Step 3: Calculate the Adjoint of \( A \)

The adjoint of a matrix \( A \) is given by:
\[ \text{adj}(A) = \text{det}(A) \cdot A^{-1} \]
For a diagonal matrix, the adjoint can also be computed as:
\[ \text{adj}(A) = \begin{bmatrix} \text{det}(A_{11}) & 0 & 0 & 0 \\ 0 & \text{det}(A_{22}) & 0 & 0 \\ 0 & 0 & \text{det}(A_{33}) & 0 \\ 0 & 0 & 0 & \text{det}(A_{44}) \end{bmatrix} \]
Where \( A_{ii} \) is the matrix obtained by removing the \( i \)-th row and \( i \)-th column from \( A \). Each of these determinants is \( 2^3 = 8 \) (since they are \( 3 \times 3 \) diagonal matrices with diagonal entries 2).

Thus, we have:
\[ \text{adj}(A) = \begin{bmatrix} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \end{bmatrix} \]

### Step 4: Calculate the Determinant of \( \text{adj}(A) \)

The determinant of \( \text{adj}(A) \) is:
\[ \text{det}(\text{adj}(A)) = 8 \times 8 \times 8 \times 8 = 8^4 = 4096 \]

### Step 5: Calculate the Adjoint of \( \text{adj}(A) \)

Using the property of determinants:
\[ \text{det}(\text{adj}(A)) = \text{det}(A)^{n-1} \]
where \( n \) is the order of the matrix. Here \( n = 4 \), so:
\[ \text{det}(\text{adj}(\text{adj}(A))) = \text{det}(\text{adj}(A))^{4-1} = 4096^3 \]

Calculating \( 4096^3 \):
\[ 4096 = 2^{12} \implies 4096^3 = (2^{12})^3 = 2^{36} \]

### Step 6: Find the Fractional Part of \( \frac{\text{det}(\text{adj}(\text{adj}(A)))}{7} \)

Now we calculate:
\[ \frac{2^{36}}{7} \]
To find the fractional part, we need to compute \( 2^{36} \mod 7 \).

Using Fermat's Little Theorem:
\[ 2^6 \equiv 1 \mod 7 \]
Thus:
\[ 2^{36} = (2^6)^6 \equiv 1^6 \equiv 1 \mod 7 \]
So, we have:
\[ \frac{2^{36}}{7} = k + \frac{1}{7} \quad \text{for some integer } k \]
The fractional part is:
\[ \left\{ \frac{2^{36}}{7} \right\} = \frac{1}{7} \]

### Final Answer

The fractional part of \( \frac{\text{det}(\text{adj}(\text{adj}(A)))}{7} \) is:
\[ \boxed{\frac{1}{7}} \]