Let `z = |(1,1-2i,3+5i),(1+2i,-5,10i),(3-5i,-10i,11)|`, then

Options

A. z is purely imaginary
B. z is purely real
C. z = 0
D. None of these

Answer

To solve the problem, we need to calculate the determinant of the given 3x3 matrix:

\[ z = \begin{vmatrix} 1 & 1-2i & 3+5i \\ 1+2i & -5 & 10i \\ 3-5i & -10i & 11 \end{vmatrix} \]

### Step 1: Calculate the determinant using cofactor expansion

We can expand the determinant along the first row:

\[ z = 1 \cdot \begin{vmatrix} -5 & 10i \\ -10i & 11 \end{vmatrix} - (1-2i) \cdot \begin{vmatrix} 1+2i & 10i \\ 3-5i & 11 \end{vmatrix} + (3+5i) \cdot \begin{vmatrix} 1+2i & -5 \\ 3-5i & -10i \end{vmatrix} \]

### Step 2: Calculate the first 2x2 determinant

\[ \begin{vmatrix} -5 & 10i \\ -10i & 11 \end{vmatrix} = (-5)(11) - (10i)(-10i) = -55 + 100 = 45 \]

### Step 3: Calculate the second 2x2 determinant

\[ \begin{vmatrix} 1+2i & 10i \\ 3-5i & 11 \end{vmatrix} = (1+2i)(11) - (10i)(3-5i) = 11 + 22i - (30i - 50) = 11 + 22i - 30i + 50 = 61 - 8i \]

### Step 4: Calculate the third 2x2 determinant

\[ \begin{vmatrix} 1+2i & -5 \\ 3-5i & -10i \end{vmatrix} = (1+2i)(-10i) - (-5)(3-5i) = -10i - 20i^2 + 15 - 25i = -10i + 20 + 15 - 25i = 35 - 35i \]

### Step 5: Substitute the determinants back into the expression for z

Now substituting back into the expression for \( z \):

\[ z = 1 \cdot 45 - (1-2i)(61-8i) + (3+5i)(35-35i) \]

### Step 6: Simplify the expression

Calculating each term:

1. The first term is \( 45 \).
2. The second term:
\[ (1-2i)(61-8i) = 61 - 8i - 122i + 16i^2 = 61 - 130i - 16 = 45 - 130i \]
So, \( - (1-2i)(61-8i) = -45 + 130i \).
3. The third term:
\[ (3+5i)(35-35i) = 105 - 105i + 175i - 175i^2 = 105 + 70i + 175 = 280 + 70i \]

### Step 7: Combine all the terms

Now we combine all the terms:
\[ z = 45 - 45 + 280 + (130i + 70i) = 280 + 200i \]

### Step 8: Analyze the result

The final result is \( z = 280 + 200i \).

### Conclusion

Since the imaginary part is not zero, \( z \) is not purely real. However, it has both real and imaginary parts.