Let ,R be such that the system of linear equations [x+2y+z=5, 2x+y+ z=5, 8x+4y+ z=18] has …

Question

Let $\alpha,\beta\in\mathbb{R}$ be such that the system of linear equations
$$x+2y+z=5,\qquad 2x+y+\alpha z=5,\qquad 8x+4y+\beta z=18$$
has no solution. Then $\dfrac{\beta}{\alpha}$ is equal to:

Accepted Answer

For the system to have no solution, the coefficient determinant must vanish:
$$\Delta=\begin{vmatrix}1&2&1\2&1&\alpha\8&4&\beta\end{vmatrix}=\beta-4\alpha-4\beta+16\alpha=0 \Rightarrow \beta=4\alpha.$$

Substituting $\beta=4\alpha$, the third equation becomes $8x+4y+4\alpha z=18$, which is exactly $4$ times the second equation $2x+y+\alpha z=5$ on the left side --- giving $4\times5=20$ on the left, but the right side is $18\ne20$. So the system is indeed inconsistent (no solution) for any such $\alpha\ne0$, confirming $\beta=4\alpha$.

So $\dfrac{\beta}{\alpha}=4$.

Solutions · 1