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Question Number 15232 by tawa tawa last updated on 08/Jun/17

$$\mathrm{For}\mathrm{what}\mathrm{values}\mathrm{of}\mathrm{a}\mathrm{does}\mathrm{the}\mathrm{following}\mathrm{system}\mathrm{have}\mathrm{a}\mathrm{non}\mathrm{trivial}$$$$\mathrm{solution}$$$${\mathrm{ax}}_1+{3\mathrm{x}}_2-{2\mathrm{x}}_3=0$$$$-{\mathrm{x}}_1+{4\mathrm{x}}_2+{\mathrm{ax}}_3=0$$$${5\mathrm{x}}_1-{6\mathrm{x}}_2-{7\mathrm{x}}_3=0$$

Answered by mrW1 last updated on 08/Jun/17

$$\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a}& 3& -2\\ -1& 4& \mathrm{a}\\ 5& -6& -7\end{array}\right]$$$$\mid \mathrm{A}\mid =\mathrm{a}(-28+6\mathrm{a})-3(7-5\mathrm{a})-2(6-20)$$$$={6\mathrm{a}}^2-28\mathrm{a}-21+15\mathrm{a}+28$$$$={6\mathrm{a}}^2-13\mathrm{a}+7=0$$$$\mathrm{a}=\frac{13\pm \sqrt{{13}^2-4\times 6\times 7}}{2\times 6}=\frac{13\pm 1}{12}=1,\frac{7}{6}$$

Commented by tawa tawa last updated on 08/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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