The determinant
`Delta=|(1,1+i,i),(1+i,i,1),(i,1,1+i)|` equals

Options

A. `7+4i`
B. `-7+4i`
C. `-7-4i`
D. `2(i-1)`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 1 & 1+i & i \\ 1+i & i & 1 \\ i & 1 & 1+i \end{vmatrix} \]

we will follow a systematic approach.

### Step 1: Write the determinant
We start by writing the determinant in a clearer format:

\[ \Delta = \begin{vmatrix} 1 & 1+i & i \\ 1+i & i & 1 \\ i & 1 & 1+i \end{vmatrix} \]

### Step 2: Apply column operations
We can simplify the determinant by adding all columns together. We will replace the first column with the sum of all three columns.

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

Calculating the new first column:

- First row: \(1 + (1+i) + i = 1 + 1 + i + i = 2 + 2i\)
- Second row: \((1+i) + i + 1 = 1 + i + i + 1 = 2 + 2i\)
- Third row: \(i + 1 + (1+i) = i + 1 + 1 + i = 2 + 2i\)

Now the determinant becomes:

\[ \Delta = \begin{vmatrix} 2 + 2i & 1+i & i \\ 2 + 2i & i & 1 \\ 2 + 2i & 1 & 1+i \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \(2 + 2i\) from the first column:

\[ \Delta = (2 + 2i) \begin{vmatrix} 1 & 1+i & i \\ 1 & i & 1 \\ 1 & 1 & 1+i \end{vmatrix} \]

### Step 4: Perform row operations
Next, we can simplify the determinant further by performing row operations. We will replace the second and third rows by subtracting the first row from them:

\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1 \]

Calculating the new rows:

- Second row: \((1 - 1, i - (1+i), 1 - i) = (0, -1, 1-i)\)
- Third row: \((1 - 1, 1 - (1+i), (1+i) - i) = (0, -i, 1)\)

Now the determinant becomes:

\[ \Delta = (2 + 2i) \begin{vmatrix} 1 & 1+i & i \\ 0 & -1 & 1-i \\ 0 & -i & 1 \end{vmatrix} \]

### Step 5: Calculate the determinant of the 2x2 matrix
The determinant can now be calculated using the first column:

\[ \Delta = (2 + 2i) \cdot 1 \cdot \begin{vmatrix} -1 & 1-i \\ -i & 1 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} -1 & 1-i \\ -i & 1 \end{vmatrix} = (-1)(1) - (-i)(1-i) = -1 + i(1-i) = -1 + i - i^2 = -1 + i + 1 = i \]

### Step 6: Substitute back into the determinant
Now substituting back, we have:

\[ \Delta = (2 + 2i)(i) = 2i + 2i^2 = 2i - 2 = -2 + 2i \]

### Final Result
Thus, the value of the determinant \(\Delta\) is:

\[ \Delta = -2 + 2i \]