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Question-198862

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Question Number 198862 by sonukgindia last updated on 25/Oct/23
Answered by Frix last updated on 25/Oct/23
determinant ((a,b,c),(b,c,a),(c,a,b))=3abc−(a^3 +b^3 +c^3 )= =(a+b+c)(ab+ac+bc−(a^2 +b^2 +c^2 ))= =(a+b+c)(3(ab+ac+bc)−(a+b+c)^2 )= =1×(3×0−1^2 )=−1 (x−a)(x−b)(x−c)= =x^3 −(a+b+c)_(=1) x^2 +(ab+ac+bc)_(=0) x−abc_(=−2)
$$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{vmatrix}=\mathrm{3}{abc}−\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)= \\ $$$$=\left({a}+{b}+{c}\right)\left({ab}+{ac}+{bc}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right)= \\ $$$$=\left({a}+{b}+{c}\right)\left(\mathrm{3}\left({ab}+{ac}+{bc}\right)−\left({a}+{b}+{c}\right)^{\mathrm{2}} \right)= \\ $$$$=\mathrm{1}×\left(\mathrm{3}×\mathrm{0}−\mathrm{1}^{\mathrm{2}} \right)=−\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)= \\ $$$$\:\:\:\:\:={x}^{\mathrm{3}} −\underset{=\mathrm{1}} {\left({a}+{b}+{c}\right)}{x}^{\mathrm{2}} +\underset{=\mathrm{0}} {\left({ab}+{ac}+{bc}\right)}{x}−\underset{=−\mathrm{2}} {{abc}} \\ $$
Commented by Frix last updated on 25/Oct/23
This means x^3 +px^2 +qx+r=0 determinant ((x_1 ,x_2 ,x_3 ),(x_2 ,x_3 ,x_1 ),(x_3 ,x_1 ,x_2 ))=p^3 −3pq
$$\mathrm{This}\:\mathrm{means} \\ $$$${x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\begin{vmatrix}{{x}_{\mathrm{1}} }&{{x}_{\mathrm{2}} }&{{x}_{\mathrm{3}} }\\{{x}_{\mathrm{2}} }&{{x}_{\mathrm{3}} }&{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{3}} }&{{x}_{\mathrm{1}} }&{{x}_{\mathrm{2}} }\end{vmatrix}={p}^{\mathrm{3}} −\mathrm{3}{pq} \\ $$
Answered by cortano12 last updated on 25/Oct/23
x^3 −x^2 +2 = 0 abc = −2 ; a+b+c = 1 ab+ac+bc = 0 det(A) = 3abc−(a^3 +b^3 +c^3 ) = −6−{(a+b+c)^2 −2(ab+ac+bc)−6} = −6−{1−0−6 }= −1
$$\:\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\mathrm{abc}\:=\:−\mathrm{2}\:;\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{1} \\ $$$$\:\:\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\:=\:\mathrm{0} \\ $$$$\:\mathrm{det}\left(\mathrm{A}\right)\:=\:\mathrm{3abc}−\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \right) \\ $$$$\:=\:−\mathrm{6}−\left\{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right)−\mathrm{6}\right\} \\ $$$$\:=\:−\mathrm{6}−\left\{\mathrm{1}−\mathrm{0}−\mathrm{6}\:\right\}=\:−\mathrm{1}\: \\ $$

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