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Question Number 56422 by gunawan last updated on 16/Mar/19

$$\mathrm{If}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}$$$$ax+by+(a\lambda +b)z=0$$$$bx+cy+(b\lambda +c)z=0$$$$(a\lambda +b)x+(b\lambda +c)y=0$$$$\mathrm{has}\mathrm{a}\mathrm{non}-\mathrm{trivial}\mathrm{solution},\mathrm{then}$$

Answered by MJS last updated on 17/Mar/19

$$\mathrm{solving}\mathrm{we}\mathrm{get}$$$$x=\frac{(a\lambda +b)(b\lambda +c)}{b^2-ac}z$$$$y=-\frac{{(a\lambda +b)}^2}{b^2-ac}z$$$$(a{\lambda }^2+2b\lambda +c)z=0$$$$z=0\Rightarrow x=y=0$$$$a{\lambda }^2+2b\lambda +c=0\Rightarrow \lambda =-\frac{b}{a}\pm \frac{\sqrt{b^2-ac}}{a}$$$$\Rightarrow \left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}(\frac{b}{a}\pm \frac{\sqrt{b^2-ac}}{a})p\\ -p\\ p\end{array}\right)$$$$\mathrm{with}p\in \mathbb{R}\wedge a\ne 0\wedge b^2⩾ac$$

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