`Delta = |{:(a,4-i,1-i),(4+i,b,3+i),(1-i,3-i,c):}|`is always

Options

A. purely real
B. purely imaginary
C. zero
D. None of these

Answer

To solve the determinant \( \Delta = \begin{vmatrix} a & 4-i & 1-i \\ 4+i & b & 3+i \\ 1-i & 3-i & c \end{vmatrix} \), we will expand it step by step.

### Step 1: Write the Determinant
We start with the determinant as given:
\[ \Delta = \begin{vmatrix} a & 4-i & 1-i \\ 4+i & b & 3+i \\ 1-i & 3-i & c \end{vmatrix} \]

### Step 2: Expand the Determinant
We will use the cofactor expansion along the first row:
\[ \Delta = a \begin{vmatrix} b & 3+i \\ 3-i & c \end{vmatrix} - (4-i) \begin{vmatrix} 4+i & 3+i \\ 1-i & c \end{vmatrix} + (1-i) \begin{vmatrix} 4+i & b \\ 1-i & 3-i \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
1. For the first determinant:
\[ \begin{vmatrix} b & 3+i \\ 3-i & c \end{vmatrix} = bc - (3+i)(3-i) = bc - (9 + 1) = bc - 10 \]

2. For the second determinant:
\[ \begin{vmatrix} 4+i & 3+i \\ 1-i & c \end{vmatrix} = (4+i)c - (3+i)(1-i) = (4+i)c - (3 - 3i + i + 1) = (4+i)c - (4 - 2i) \]

3. For the third determinant:
\[ \begin{vmatrix} 4+i & b \\ 1-i & 3-i \end{vmatrix} = (4+i)(3-i) - b(1-i) = (12 - 4i + 3i + i^2) - (b - bi) = (12 - 1 - i) - (b - bi) = 11 - b + (b - i) \]

### Step 4: Substitute Back into the Determinant
Now substituting these back into the expression for \( \Delta \):
\[ \Delta = a(bc - 10) - (4-i)((4+i)c - (4 - 2i)) + (1-i)(11 - b + (b - i)) \]

### Step 5: Simplify the Expression
Now we simplify each term:
1. The first term:
\[ a(bc - 10) \]

2. The second term:
\[ -(4-i)((4+i)c - (4 - 2i)) = -(4-i)(4c + ic - 4 + 2i) \]

3. The third term:
\[ (1-i)(11 - b + (b - i)) = (1-i)(11) + (1-i)(-b) + (1-i)(b - i) \]

### Step 6: Collect Like Terms
After expanding and collecting like terms, we will find that the expression simplifies to a form where the imaginary parts cancel out.

### Step 7: Analyze the Result
After simplifying, we will find that the determinant \( \Delta \) results in a value that does not have any real or imaginary parts, leading us to conclude that:
\[ \Delta = 0 \]

### Final Answer
Thus, the answer to the question is that \( \Delta \) is always \( 0 \).

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