If `|(x,-3i,1),(y,1,i),(0,2i,-i)|=6+11i`. then what are the value of x and y respectively ?

Options

A. A) `-3, 4`
B. B) 3, 4
C. C) 3, -4
D. D) `-3, -4`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) such that the determinant of the matrix

\[ \begin{pmatrix} x & -3i & 1 \\ y & 1 & i \\ 0 & 2i & -i \end{pmatrix} \]

is equal to \( 6 + 11i \).

### Step 1: Calculate the Determinant

We will expand the determinant using the first column:

\[ \text{Det} = x \cdot \begin{vmatrix} 1 & i \\ 2i & -i \end{vmatrix} - (-3i) \cdot \begin{vmatrix} y & i \\ 0 & -i \end{vmatrix} + 1 \cdot \begin{vmatrix} y & 1 \\ 0 & 2i \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

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

2. For the second determinant:

\[ \begin{vmatrix} y & i \\ 0 & -i \end{vmatrix} = (y)(-i) - (i)(0) = -yi \]

3. For the third determinant:

\[ \begin{vmatrix} y & 1 \\ 0 & 2i \end{vmatrix} = (y)(2i) - (1)(0) = 2yi \]

### Step 3: Substitute Back into the Determinant Expression

Now substituting these back into the determinant expression:

\[ \text{Det} = x(2 - i) + 3i(-yi) + 2yi \]

This simplifies to:

\[ \text{Det} = x(2 - i) + 3y + 2yi \]

### Step 4: Expand and Combine Like Terms

Expanding \( x(2 - i) \):

\[ \text{Det} = 2x - xi + 3y + 2yi \]

Now, combine the real and imaginary parts:

\[ \text{Det} = (2x + 3y) + (-x + 2y)i \]

### Step 5: Set Equal to Given Value

We know that:

\[ (2x + 3y) + (-x + 2y)i = 6 + 11i \]

This gives us two equations:

1. \( 2x + 3y = 6 \) (Real part)
2. \( -x + 2y = 11 \) (Imaginary part)

### Step 6: Solve the System of Equations

From the first equation:

\[ 2x + 3y = 6 \quad \text{(1)} \]

From the second equation:

\[ -x + 2y = 11 \quad \text{(2)} \]

We can express \( x \) from equation (2):

\[ x = 2y - 11 \]

Substituting \( x \) into equation (1):

\[ 2(2y - 11) + 3y = 6 \]

Expanding this gives:

\[ 4y - 22 + 3y = 6 \]

Combining like terms:

\[ 7y - 22 = 6 \]

Adding 22 to both sides:

\[ 7y = 28 \]

Dividing by 7:

\[ y = 4 \]

### Step 7: Substitute Back to Find \( x \)

Now substituting \( y = 4 \) back into the equation for \( x \):

\[ x = 2(4) - 11 = 8 - 11 = -3 \]

### Final Result

Thus, the values of \( x \) and \( y \) are:

\[ x = -3, \quad y = 4 \]