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

Options

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

Answer

To solve the determinant \( D = \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \) and express it in the form \( x + iy \), we will follow these steps:

### Step 1: Write down the determinant
The determinant we need to evaluate is:
\[ D = \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

### Step 2: Apply row operations
We can simplify the determinant by performing row operations. Let's perform the operation \( R_1 \rightarrow R_1 + R_2 \):
\[ R_1 = (6i + 4, -3i + 3i, 1 - 1) = (6i + 4, 0, 0) \]
So the determinant becomes:
\[ D = \begin{vmatrix} 6i + 4 & 0 & 0 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} \]

### Step 3: Expand the determinant
Since the first row has two zeros, we can expand the determinant along the first row:
\[ D = (6i + 4) \begin{vmatrix} 3i & -1 \\ 3 & i \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinant
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 5: Substitute back into the determinant
Substituting back, we have:
\[ D = (6i + 4) \cdot 0 = 0 \]

### Step 6: Express in the form \( x + iy \)
Since \( D = 0 \), we can express this as:
\[ 0 = x + iy \]
This implies:
\[ x = 0 \quad \text{and} \quad y = 0 \]

### Final Result
Thus, the values of \( x \) and \( y \) are:
\[ x = 0, \quad y = 0 \]