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Question Number 183240 by ali009 last updated on 23/Dec/22
$${find}\:{the}\:{value}\:{of}\:{cofficent}\:\mu\:{in}\:{the}\:{following} \\ $$$${system}\:{from}\:{the}\:{determinat}: \\ $$$$\mathrm{2}{x}_{\mathrm{1}} +\mu{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mu−\mathrm{1}\right){x}_{\mathrm{1}} −{x}_{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} =\mathrm{0} \\ $$
Answered by TheSupreme last updated on 24/Dec/22
![[(2,μ,1),((μ−1),(−1),2),(4,1,4) ] [(x_1 ),(x_2 ),(x_3 ) ]= [(0),(0),(0) ] det: 2(−4−2)−μ[4(μ−1)−8]+[(μ−1)+4]=0 −12−4μ^2 +12μ+μ+3=0 −4μ^2 +13μ−9=0 μ=((−13±(√(169−144)))/(−8))=((−13±5)/(−8))=1, (9/4)](Q183259.png)
$$\begin{bmatrix}{\mathrm{2}}&{\mu}&{\mathrm{1}}\\{\mu−\mathrm{1}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{4}}\end{bmatrix}\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\\{{x}_{\mathrm{3}} }\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$${det}:\:\mathrm{2}\left(−\mathrm{4}−\mathrm{2}\right)−\mu\left[\mathrm{4}\left(\mu−\mathrm{1}\right)−\mathrm{8}\right]+\left[\left(\mu−\mathrm{1}\right)+\mathrm{4}\right]=\mathrm{0} \\ $$$$−\mathrm{12}−\mathrm{4}\mu^{\mathrm{2}} +\mathrm{12}\mu+\mu+\mathrm{3}=\mathrm{0} \\ $$$$−\mathrm{4}\mu^{\mathrm{2}} +\mathrm{13}\mu−\mathrm{9}=\mathrm{0} \\ $$$$\mu=\frac{−\mathrm{13}\pm\sqrt{\mathrm{169}−\mathrm{144}}}{−\mathrm{8}}=\frac{−\mathrm{13}\pm\mathrm{5}}{−\mathrm{8}}=\mathrm{1},\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$ \\ $$