Let `Delta = |(a,a+b,a+b+c),(3a,4a+3b,5a+4b+3c),(6a,9a+6b,11a+9b+6c)|` where `a =i,b=omega and c=omega^2` , then `Delta` equals

Options

A. `omega`
B. `-omega^2`
C. `i`
D. `-i`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{vmatrix} \) where \( a = i \), \( b = \omega \), and \( c = \omega^2 \), we will follow these steps:

### Step 1: Substitute the values of \( a \), \( b \), and \( c \)
We will substitute \( a = i \), \( b = \omega \), and \( c = \omega^2 \) into the determinant.

\[ \Delta = \begin{vmatrix} i & i+\omega & i+\omega+\omega^2 \\ 3i & 4i+3\omega & 5i+4\omega+3\omega^2 \\ 6i & 9i+6\omega & 11i+9\omega+6\omega^2 \end{vmatrix} \]

### Step 2: Simplify the columns
We can perform column operations to simplify the determinant. Let's perform the operation \( C_3 = C_3 - C_2 \).

\[ C_3 = (i+\omega+\omega^2) - (i+\omega) = \omega^2 \]

Thus, the determinant becomes:

\[ \Delta = \begin{vmatrix} i & i+\omega & \omega^2 \\ 3i & 4i+3\omega & 3\omega \\ 6i & 9i+6\omega & 6\omega^2 \end{vmatrix} \]

Next, we can perform \( C_2 = C_2 - C_1 \):

\[ C_2 = (i+\omega) - i = \omega \]

Now, the determinant is:

\[ \Delta = \begin{vmatrix} i & \omega & \omega^2 \\ 3i & 3\omega & 3\omega \\ 6i & 6\omega & 6\omega^2 \end{vmatrix} \]

### Step 3: Factor out constants
Notice that the second and third rows can be factored out:

\[ \Delta = 3 \cdot 3 \cdot \begin{vmatrix} i & \omega & \omega^2 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \]

### Step 4: Calculate the determinant
The determinant of the matrix \( \begin{vmatrix} i & \omega & \omega^2 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \) can be calculated. Notice that the second and third rows are linearly dependent (they are multiples of each other), which means the determinant is zero.

Thus, we have:

\[ \Delta = 0 \]

### Step 5: Conclusion
Since the determinant simplifies to zero, we conclude:

\[ \Delta = 0 \]