The value of `|(i^m,i^(m+1),i^(m+2)),(i^(m+5),i^(m+4),i^(m+3)),(i^(m+6),i^(m+7),i^(m+8))|` , when `i=sqrt-1`, is

Options

A. 1 if m is a multiple of 4
B. 0 for all real m
C. `-1` if m is a multiple of 3
D. 2

Answer

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

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} i^m & i^{m+1} & i^{m+2} \\ i^{m+5} & i^{m+4} & i^{m+3} \\ i^{m+6} & i^{m+7} & i^{m+8} \end{vmatrix} \]

### Step 2: Expand the Determinant
We will expand the determinant using the first row:
\[ D = i^m \begin{vmatrix} i^{m+4} & i^{m+3} \\ i^{m+7} & i^{m+8} \end{vmatrix} - i^{m+1} \begin{vmatrix} i^{m+5} & i^{m+3} \\ i^{m+6} & i^{m+8} \end{vmatrix} + i^{m+2} \begin{vmatrix} i^{m+5} & i^{m+4} \\ i^{m+6} & i^{m+7} \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} i^{m+4} & i^{m+3} \\ i^{m+7} & i^{m+8} \end{vmatrix} = i^{m+4} \cdot i^{m+8} - i^{m+3} \cdot i^{m+7} = i^{2m+12} - i^{2m+10} = i^{2m+10}(i^2 - 1) = i^{2m+10}(-1) = -i^{2m+10} \]

2. For the second determinant:
\[ \begin{vmatrix} i^{m+5} & i^{m+3} \\ i^{m+6} & i^{m+8} \end{vmatrix} = i^{m+5} \cdot i^{m+8} - i^{m+3} \cdot i^{m+6} = i^{2m+13} - i^{2m+9} = i^{2m+9}(i^4 - 1) = i^{2m+9}(0) = 0 \]

3. For the third determinant:
\[ \begin{vmatrix} i^{m+5} & i^{m+4} \\ i^{m+6} & i^{m+7} \end{vmatrix} = i^{m+5} \cdot i^{m+7} - i^{m+4} \cdot i^{m+6} = i^{2m+12} - i^{2m+10} = i^{2m+10}(i^2 - 1) = -i^{2m+10} \]

### Step 4: Substitute Back into the Determinant
Substituting these back into the expression for \( D \):
\[ D = i^m (-(-i^{2m+10})) - i^{m+1}(0) + i^{m+2}(-i^{2m+10}) \]
\[ D = i^m i^{2m+10} + 0 - i^{m+2} i^{2m+10} = i^{3m+10} - i^{3m+10} = 0 \]

### Conclusion
Thus, the value of the determinant is:
\[ \boxed{0} \]