If `|(6i,-3i,1),(4,3i,-1),(40,3,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 problem, we need to find the determinant of the given matrix:

\[ \begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 40 & 3 & i \end{vmatrix} \]

### Step 1: Set up the determinant

The determinant of a 3x3 matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

- \( a = 6i, b = -3i, c = 1 \)
- \( d = 4, e = 3i, f = -1 \)
- \( g = 40, h = 3, i = i \)

### Step 2: Calculate the determinant

Now, we will calculate each part of the determinant:

1. **Calculate \( ei - fh \)**:
\[ ei = (3i)(i) = 3i^2 = 3(-1) = -3 \]
\[ fh = (-1)(3) = -3 \]
\[ ei - fh = -3 - (-3) = -3 + 3 = 0 \]

2. **Calculate \( di - fg \)**:
\[ di = (4)(i) = 4i \]
\[ fg = (-1)(40) = -40 \]
\[ di - fg = 4i - (-40) = 4i + 40 \]

3. **Calculate \( dh - eg \)**:
\[ dh = (4)(3) = 12 \]
\[ eg = (3i)(40) = 120i \]
\[ dh - eg = 12 - 120i \]

### Step 3: Substitute back into the determinant formula

Now substituting back into the determinant formula:

\[ \text{det}(A) = 6i(0) - (-3i)(4i + 40) + 1(12 - 120i) \]
\[ = 0 + 3i(4i + 40) + (12 - 120i) \]
\[ = 3i(4i) + 3i(40) + 12 - 120i \]
\[ = 12i^2 + 120i + 12 - 120i \]
\[ = 12(-1) + 12 = -12 + 12 = 0 \]

### Step 4: Conclusion

The determinant is \( 0 \). Therefore, we can express this in the form \( x + iy \):

\[ 0 = 0 + 0i \]

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

### Final Answer

The values of \( x \) and \( y \) are:
- \( x = 0 \)
- \( y = 0 \)