If `Delta=|(i^(n),i^(n+1),i^(n+2)),(i^(n+5),i^(n+4),i^(n+3)),(i^(n+6),i^(n+7),i^(n+8))|` where `i=sqrt(-1)` , then its value is

Options

A. `0AA n in R`
B. 1 if `n =4k`
C. `-i` if `n =3k`
D. none

Answer

To solve the determinant \( \Delta = \begin{vmatrix} i^n & i^{n+1} & i^{n+2} \\ i^{n+5} & i^{n+4} & i^{n+3} \\ i^{n+6} & i^{n+7} & i^{n+8} \end{vmatrix} \), where \( i = \sqrt{-1} \), we can follow these steps:

### Step 1: Factor out \( i^n \)
Notice that each element in the determinant can be expressed in terms of \( i^n \):
- The first row becomes \( i^n, i^{n+1} = i^n \cdot i, i^{n+2} = i^n \cdot i^2 \).
- The second row becomes \( i^{n+5} = i^n \cdot i^5, i^{n+4} = i^n \cdot i^4, i^{n+3} = i^n \cdot i^3 \).
- The third row becomes \( i^{n+6} = i^n \cdot i^6, i^{n+7} = i^n \cdot i^7, i^{n+8} = i^n \cdot i^8 \).

Thus, we can factor out \( i^n \) from each row:
\[ \Delta = i^n \cdot \begin{vmatrix} 1 & i & i^2 \\ i^5 & i^4 & i^3 \\ i^6 & i^7 & i^8 \end{vmatrix} \]

### Step 2: Simplify the determinant
Next, we can simplify the determinant:
\[ \Delta = i^n \cdot \begin{vmatrix} 1 & i & -1 \\ i & 1 & -i \\ -1 & -i & 1 \end{vmatrix} \]

### Step 3: Perform row operations
Now, we can perform row operations to simplify the determinant further. We can add the first column to the second column:
\[ \Delta = i^n \cdot \begin{vmatrix} 1 & 0 & -1 \\ i & 1 & -i \\ -1 & -i & 1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Now, we can calculate the determinant. The first column remains unchanged, while the second column has been modified:
\[ \Delta = i^n \cdot \begin{vmatrix} 1 & 0 & -1 \\ i & 1 & -i \\ -1 & -i & 1 \end{vmatrix} \]

The determinant can be calculated using the formula for a 3x3 determinant:
\[ \Delta = i^n \cdot \left( 1 \cdot (1 \cdot 1 - (-i)(-1)) - 0 + (-1)(i \cdot (-i) - 1 \cdot (-1)) \right) \]

Calculating the determinant:
\[ = i^n \cdot (1 - i) = i^n \cdot (1 - i) \]

### Step 5: Check for linear dependence
Notice that if we add the first column to the third column, we can see that the first column will become \( 0 \):
\[ \Delta = i^n \cdot \begin{vmatrix} 1 & i & 0 \\ i & 1 & 0 \\ -1 & -i & 0 \end{vmatrix} \]

Since one column is now all zeros, the determinant evaluates to zero:
\[ \Delta = 0 \]

### Final Result
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]