Question Number 124071 by zarminaawan last updated on 30/Nov/20 | ||
$${Given}\:{the}\:{system}\:{of}\:{linear}\:{equation} \\ $$$${x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} −\mathrm{3}{x}_{\mathrm{3}} =\mathrm{4} \\ $$$$\mathrm{3}{x}_{\mathrm{1}} −{x}_{\mathrm{2}} +\mathrm{5}{x}_{\mathrm{3}} =\mathrm{2} \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +\left({a}^{\mathrm{2}} −\mathrm{14}\right){x}_{\mathrm{3}} ={a}+\mathrm{2} \\ $$$${for}\:{which}\:{value}\:{of}\:``{a}''\:{the}\:{system}\:{have}\:{one}\:{solution},{infinite}\:{sol}.{and}\:{number}\:{solution}. \\ $$$$ \\ $$ | ||
Commented by danishnaveed last updated on 30/Nov/20 | ||
$${send}\:{ans} \\ $$ | ||
Answered by mr W last updated on 30/Nov/20 | ||
$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right): \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{3}} =\mathrm{6}\:\:\:...\left(\mathrm{4}\right) \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +\left({a}^{\mathrm{2}} −\mathrm{14}\right){x}_{\mathrm{3}} ={a}+\mathrm{2}\:\:\:\:...\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{4}\right)−\left(\mathrm{3}\right): \\ $$$$\left(\mathrm{4}+{a}\right)\left(\mathrm{4}−{a}\right){x}_{\mathrm{3}} =\mathrm{4}−{a}\:\:\:...\left(\mathrm{5}\right) \\ $$$${if}\:{a}=\mathrm{4}\:\Rightarrow\:{infinite}\:{solutions} \\ $$$${if}\:{a}=−\mathrm{4}\:\Rightarrow\:{no}\:{solution} \\ $$$${if}\:{a}\neq\mathrm{4}\:\&\:{a}\neq−\mathrm{4}\:\Rightarrow\:{one}\:{solution}\:{x}_{\mathrm{3}} =\frac{\mathrm{1}}{{a}+\mathrm{4}} \\ $$ | ||