Question Number 56422 by gunawan last updated on 16/Mar/19
$$\mathrm{If}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$${ax}\:+\:{by}\:+\:\left({a}\lambda+{b}\right){z}\:=\:\mathrm{0} \\ $$$${bx}\:+\:{cy}\:+\:\left({b}\lambda+{c}\right){z}\:=\:\mathrm{0} \\ $$$$\left({a}\lambda\:+\:{b}\right){x}\:+\:\left({b}\lambda+{c}\right){y}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{non}−\mathrm{trivial}\:\mathrm{solution},\:\mathrm{then} \\ $$
Answered by MJS last updated on 17/Mar/19
$$\mathrm{solving}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\frac{\left({a}\lambda+{b}\right)\left({b}\lambda+{c}\right)}{{b}^{\mathrm{2}} −{ac}}{z} \\ $$$${y}=−\frac{\left({a}\lambda+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} −{ac}}{z} \\ $$$$\left({a}\lambda^{\mathrm{2}} +\mathrm{2}{b}\lambda+{c}\right){z}=\mathrm{0} \\ $$$${z}=\mathrm{0}\:\Rightarrow\:{x}={y}=\mathrm{0} \\ $$$${a}\lambda^{\mathrm{2}} +\mathrm{2}{b}\lambda+{c}=\mathrm{0}\:\Rightarrow\:\lambda=−\frac{{b}}{{a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −{ac}}}{{a}} \\ $$$$\Rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\left(\frac{{b}}{{a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −{ac}}}{{a}}\right){p}}\\{−{p}}\\{{p}}\end{pmatrix} \\ $$$$\mathrm{with}\:{p}\in\mathbb{R}\wedge{a}\neq\mathrm{0}\wedge{b}^{\mathrm{2}} \geqslant{ac} \\ $$