If `x + iy=|{:(6i,-3i,1),(4,3i,-1),(20,3,i):}|`, then what is x - iy equal to ?

Options

A. 3 + i
B. 1 + 3i
C. 3i
D. 0

Explanation

To solve the problem, we need to find the value of \( x - iy \) given that \( x + iy = |(6i, -3i, 1), (4, 3i, -1), (20, 3, i)| \).

### Step-by-Step Solution:

1. **Set up the Determinant**:
We need to calculate the determinant of the given 3x3 matrix:
\[ \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

2. **Calculate the Determinant**:
We can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
In our case:
- \( a = 6i, b = -3i, c = 1 \)
- \( d = 4, e = 3i, f = -1 \)
- \( g = 20, h = 3, i = i \)

Plugging in the values:
\[ \text{det}(A) = 6i \left( (3i)(i) - (-1)(3) \right) - (-3i) \left( (4)(i) - (-1)(20) \right) + 1 \left( (4)(3) - (3i)(20) \right) \]

3. **Simplify Each Term**:
- First term:
\[ 6i \left( 3i^2 + 3 \right) = 6i \left( 3(-1) + 3 \right) = 6i(0) = 0 \]
- Second term:
\[ -(-3i) \left( 4i + 20 \right) = 3i(4i + 20) = 3i(4i) + 3i(20) = 12i^2 + 60i = 12(-1) + 60i = -12 + 60i \]
- Third term:
\[ 1 \left( 12 - 60i \right) = 12 - 60i \]

4. **Combine the Results**:
Now we combine all the terms:
\[ 0 + (-12 + 60i) + (12 - 60i) = 0 \]

5. **Conclusion**:
Thus, we have:
\[ x + iy = 0 \]
This implies:
\[ x = 0 \quad \text{and} \quad y = 0 \]

6. **Find \( x - iy \)**:
Now we need to find \( x - iy \):
\[ x - iy = 0 - i(0) = 0 \]

### Final Answer:
\[ x - iy = 0 \]