The value of q for which the following set of linear equation 2x + 3y = 0; 6x + qy = 0 can have non-trivial solution is A. 2 B. 7 C. 9 D. 11

The correct answer is $\boxed{9}$.

For a system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. In this case, the coefficient matrix is $$\begin{bmatrix}2 & 3 \\ 6 & q\end{bmatrix}$$ The determinant of this matrix is $-6q+18$. For the system to have a non-trivial solution, $-6q+18=0$. Solving for $q$, we get $q=\boxed{9}$.

Here is a brief explanation of each option:

• Option A: $q=2$. In this case, the determinant of the coefficient matrix is $-12$, which is not zero. Therefore, the system of equations does not have a non-trivial solution.
• Option B: $q=7$. In this case, the determinant of the coefficient matrix is $-3$, which is not zero. Therefore, the system of equations does not have a non-trivial solution.
• Option C: $q=9$. In this case, the determinant of the coefficient matrix is zero. Therefore, the system of equations has a non-trivial solution.
• Option D: $q=11$. In this case, the determinant of the coefficient matrix is $-54$, which is not zero. Therefore, the system of equations does not have a non-trivial solution.