`|(6i,-3i,1),(4,3i,-1),(20,3,i)|` = x + iy then (x,y) is

Options

A. (3,1)
B. (1,3)
C. (0,3)
D. (0,0)

Answer

To solve the determinant \( |(6i, -3i, 1), (4, 3i, -1), (20, 3, i)| \) and express it in the form \( x + iy \), we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant:

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

### Step 2: Apply row operations
We will perform a row operation to simplify the determinant. Let's add the second row to the first row:

\[ R_1 \rightarrow R_1 + R_2 \]

This gives us:

\[ D = \begin{vmatrix} (6i + 4) & (-3i + 3i) & (1 - 1) \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} (6i + 4) & 0 & 0 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

### Step 3: Calculate the determinant
Since the first row has two zeros, we can expand the determinant along the first row:

\[ D = (6i + 4) \cdot \begin{vmatrix} 3i & -1 \\ 3 & i \end{vmatrix} \]

Now we need to calculate the 2x2 determinant:

\[ \begin{vmatrix} 3i & -1 \\ 3 & i \end{vmatrix} = (3i)(i) - (-1)(3) = 3i^2 + 3 = 3(-1) + 3 = -3 + 3 = 0 \]

### Step 4: Substitute back into the determinant
Now substituting back, we have:

\[ D = (6i + 4) \cdot 0 = 0 \]

### Step 5: Express in the form \( x + iy \)
Since \( D = 0 \), we can express it as:

\[ 0 = 0 + 0i \]

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

### Final Answer
The values of \( (x, y) \) are:

\[ (x, y) = (0, 0) \]