Q.3. Let be such that the system of linear equations , , has no solution. Then is equal to :

Q: What is the correct answer for JEE Main Mathematics Q3?

The correct answer is B. Sol. For a system of linear equations to have no solution, we must have $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. $\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 0$ $1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$ $\beta - 4\alpha - 4\beta + 16\alpha = 0$ $-3\beta + 12\alpha = 0 \Rightarrow \beta = 4\alpha$ Now, con

Q: Q.3. Let be such that the system of linear equations , , has no solution. Then is equal to :

The correct answer is B. Sol. For a system of linear equations to have no solution, we must have $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. $\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 0$ $1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$ $\beta - 4\alpha - 4\beta + 16\alpha = 0$ $-3\beta + 12\alpha = 0 \Rightarrow \beta = 4\alpha$ Now, con

Question

Q.3. Let  be such that the system of linear equations , ,  has no solution. Then  is equal to :

FirstInTest · Accepted Answer

Sol. For a system of linear equations to have no solution, we must have $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
$\Delta = \begin{vmatrix} 1 & 2 & 1 \ 2 & 1 & \alpha \ 8 & 4 & \beta \end{vmatrix} = 0$
$1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$
$\beta - 4\alpha - 4\beta + 16\alpha = 0$
$-3\beta + 12\alpha = 0 \Rightarrow \beta = 4\alpha$
Now, consider $\Delta_z = \begin{vmatrix} 1 & 2 & 5 \ 2 & 1 & 5 \ 8 & 4 & 18 \end{vmatrix}$
$1(18 - 20) - 2(36 - 40) + 5(8 - 8)$
$-2 - 2(-4) + 0 = -2 + 8 = 6 
e 0$
Since $\Delta = 0$ and $\Delta_z 
e 0$, the system has no solution. This condition is satisfied when $\beta = 4\alpha$.
Therefore, $\frac{\beta}{\alpha} = 4$.

Answer

-4

Answer

4

Answer

8

Answer

-8

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