If z = `|{:(1 , 1 + 2i, - 5i),(1 - 2i,-3,5+3i),(5i,5-3i,7):}|`, then `(i= sqrt(-1))`

Options

A. z is purely real
B. z is purely imaginary
C. z + `bar(z)` = 0
D. `(z - bar(z) )` i is purely imaginary

Answer

To solve the determinant \( z = \begin{vmatrix} 1 & 1 + 2i & -5i \\ 1 - 2i & -3 & 5 + 3i \\ 5i & 5 - 3i & 7 \end{vmatrix} \), we will follow these steps:

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

\[ z = \begin{vmatrix} 1 & 1 + 2i & -5i \\ 1 - 2i & -3 & 5 + 3i \\ 5i & 5 - 3i & 7 \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand this determinant using the first row:

\[ z = 1 \cdot \begin{vmatrix} -3 & 5 + 3i \\ 5 - 3i & 7 \end{vmatrix} - (1 + 2i) \cdot \begin{vmatrix} 1 - 2i & 5 + 3i \\ 5i & 7 \end{vmatrix} - 5i \cdot \begin{vmatrix} 1 - 2i & -3 \\ 5i & 5 - 3i \end{vmatrix} \]

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

1. For \( \begin{vmatrix} -3 & 5 + 3i \\ 5 - 3i & 7 \end{vmatrix} \):
\[ = (-3)(7) - (5 + 3i)(5 - 3i) = -21 - (25 - 9) = -21 - 16 = -37 \]

2. For \( \begin{vmatrix} 1 - 2i & 5 + 3i \\ 5i & 7 \end{vmatrix} \):
\[ = (1 - 2i)(7) - (5 + 3i)(5i) = 7 - 14i - (25i + 15) = 7 - 14i - 25i - 15 = -8 - 39i \]

3. For \( \begin{vmatrix} 1 - 2i & -3 \\ 5i & 5 - 3i \end{vmatrix} \):
\[ = (1 - 2i)(5 - 3i) - (-3)(5i) = (5 - 3i - 10i + 6) + 15i = 11 - 8i + 15i = 11 + 7i \]

### Step 4: Substitute Back into the Determinant
Now substituting back into the determinant expression:

\[ z = 1 \cdot (-37) - (1 + 2i)(-8 - 39i) - 5i(11 + 7i) \]

### Step 5: Simplify Each Term
1. First term:
\[ -37 \]

2. Second term:
\[ -(-8 - 39i) - 2i(-8 - 39i) = 8 + 39i + 16i + 78 = 8 + 55i + 78 = 86 + 55i \]

3. Third term:
\[ -5i(11 + 7i) = -55i - 35 = -35 - 55i \]

### Step 6: Combine All Terms
Combining all the terms:

\[ z = -37 + (86 - 35) + (55i - 55i) = -37 + 51 + 0i = 14 \]

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

\[ z = 14 \]