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

2
7
9
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 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube
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.
  • Exit mobile version