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For-what-values-of-a-does-the-following-system-have-a-non-trivial-solution-ax-1-3x-2-2x-3-0-x-1-4x-2-ax-3-0-5x-1-6x-2-7x-3-0-




Question Number 15232 by tawa tawa last updated on 08/Jun/17
For what values of  a  does the following system have a non trivial   solution   ax_1  + 3x_2  − 2x_3  = 0  −x_1  + 4x_2  + ax_3  = 0  5x_1  − 6x_2  − 7x_3  = 0
$$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\:\mathrm{a}\:\:\mathrm{does}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{have}\:\mathrm{a}\:\mathrm{non}\:\mathrm{trivial}\: \\ $$$$\mathrm{solution}\: \\ $$$$\mathrm{ax}_{\mathrm{1}} \:+\:\mathrm{3x}_{\mathrm{2}} \:−\:\mathrm{2x}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$−\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{4x}_{\mathrm{2}} \:+\:\mathrm{ax}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{5x}_{\mathrm{1}} \:−\:\mathrm{6x}_{\mathrm{2}} \:−\:\mathrm{7x}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$
Answered by mrW1 last updated on 08/Jun/17
A= [(a,3,(−2)),((−1),4,a),(5,(−6),(−7)) ]  ∣A∣=a(−28+6a)−3(7−5a)−2(6−20)  =6a^2 −28a−21+15a+28  =6a^2 −13a+7=0  a=((13±(√(13^2 −4×6×7)))/(2×6))=((13±1)/(12))=1,(7/6)
$$\mathrm{A}=\begin{bmatrix}{\mathrm{a}}&{\mathrm{3}}&{−\mathrm{2}}\\{−\mathrm{1}}&{\mathrm{4}}&{\mathrm{a}}\\{\mathrm{5}}&{−\mathrm{6}}&{−\mathrm{7}}\end{bmatrix} \\ $$$$\mid\mathrm{A}\mid=\mathrm{a}\left(−\mathrm{28}+\mathrm{6a}\right)−\mathrm{3}\left(\mathrm{7}−\mathrm{5a}\right)−\mathrm{2}\left(\mathrm{6}−\mathrm{20}\right) \\ $$$$=\mathrm{6a}^{\mathrm{2}} −\mathrm{28a}−\mathrm{21}+\mathrm{15a}+\mathrm{28} \\ $$$$=\mathrm{6a}^{\mathrm{2}} −\mathrm{13a}+\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{a}=\frac{\mathrm{13}\pm\sqrt{\mathrm{13}^{\mathrm{2}} −\mathrm{4}×\mathrm{6}×\mathrm{7}}}{\mathrm{2}×\mathrm{6}}=\frac{\mathrm{13}\pm\mathrm{1}}{\mathrm{12}}=\mathrm{1},\frac{\mathrm{7}}{\mathrm{6}} \\ $$
Commented by tawa tawa last updated on 08/Jun/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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