If `|{:( 6i, - 3i, 1),( 4, 3i, -1), (20, 3, i):}| \= x + iy` then

Options

A. `x = 3, y =1`
B. `x =1, y =3`
C. `x =0, y =3`
D. `x =0, y =0`

Answer

To solve the determinant given in the question, we will follow these steps:

### Step 1: Write the Determinant
We have the determinant of the following matrix:

\[ \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

### Step 2: Calculate the Determinant
To calculate the determinant, we will 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:

\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:

- \( a = 6i \), \( b = -3i \), \( c = 1 \)
- \( d = 4 \), \( e = 3i \), \( f = -1 \)
- \( g = 20 \), \( h = 3 \), \( i = i \)

Now we can substitute these values into the determinant formula:

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

### Step 3: Simplify Each Term
1. Calculate \( (3i)(i) - (-1)(3) \):
\[ (3i^2 + 3) = (3(-1) + 3) = -3 + 3 = 0 \]

2. Calculate \( (4)(i) - (-1)(20) \):
\[ 4i + 20 \]

3. Calculate \( (4)(3) - (3i)(20) \):
\[ 12 - 60i \]

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

\[ \text{det}(A) = 6i(0) - (-3i)(4i + 20) + 1(12 - 60i) \]

This simplifies to:

\[ 0 + 3i(4i + 20) + 12 - 60i \]

### Step 5: Expand and Combine Like Terms
1. Expand \( 3i(4i + 20) \):
\[ 3i(4i) + 3i(20) = 12i^2 + 60i = 12(-1) + 60i = -12 + 60i \]

2. Now combine all terms:
\[ -12 + 60i + 12 - 60i = 0 \]

### Final Result
The determinant evaluates to \( 0 + 0i \). Therefore, we have:

\[ x + iy = 0 + 0i \]

Thus, \( x = 0 \) and \( y = 0 \).