Let $\alpha,\beta \in \mathbf{R}$ be such that the system of linear equations$$x + 2y + z = 5 $$$$2x

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Let $\alpha,\beta \in \mathbf{R}$ be such that the system of linear equations

$$x + 2y + z = 5 $$

$$2x + y + \alpha z = 5 $$

$$8x + 4y + \beta z = 18 $$has no solution. Then $\frac{\beta}{\alpha}$ is equal to :

Accepted Answer

Sol. [Key Concept: A system has no solution when its determinant $\Delta = 0$; this gives the relation between $\alpha$ and $\beta$.]$$x + 2y + z = 5 $$$$2x + y + \alpha z = 5 $$$$8x + 4y + \beta z = 18 $$for no solution$$\begin{matrix} & \Delta = \left| \begin{matrix} 1 & 2 & 1 \ 2 & 1 & \alpha \ 8 & 4 & \beta \end{matrix} ight| = 0 \ & \Delta = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(0) = 0 \ & \ = \beta - 4\alpha - 4\beta + 16\alpha = 0 \ & \ \Rightarrow 4\alpha = \beta \end{matrix}$$

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