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Question Number 158724 by HongKing last updated on 08/Nov/21

$$\mathrm{let}\mathrm{a}>\mathrm{b}>\mathrm{c}>0\mathrm{solve}\mathrm{in}\mathbb{R}$$ $$\{\begin{array}{l}\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{a}\\ \mathrm{bx}+\mathrm{cy}+\mathrm{az}=\mathrm{b}\\ \mathrm{cx}+\mathrm{ay}+\mathrm{bz}=\mathrm{c}\end{array}$$ $$$$

Answered by ajfour last updated on 08/Nov/21

$$x+y+z=1$$ $$ax+by+c(1-x-y)=a$$ $$\Rightarrow (a-c)x+(b-c)y=a-c$$ $$x+\left(\frac{b-c}{a-c}\right)y=1$$ $$(a-c)x+(b-a)y$$ $$+(c-b)(1-x-y)=a-c$$ $$\Rightarrow (a+b-2c)x+(2b-a-c)y$$ $$=a+b-2c$$ $$\Rightarrow x-y=1$$ $$\Rightarrow y+1+\left(\frac{b-c}{a-c}\right)y=1$$ $$\Rightarrow y=0,x=1,z=0\cdot$$ $$$$

Commented byHongKing last updated on 08/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}\mathrm{cool}$$

Answered by som(math1967) last updated on 08/Nov/21

$$D=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=a(bc-a^2)-b(b^2-ca)+c(ab-c^2)$$ $$=3abc-a^3-b^3-c^3$$ $$D_1=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=abc-a^3-b^3+abc+abc-c^3$$ $$=3abc-a^3-b^3-c^3$$ $$D_2=\left|\begin{array}{ccc}a& a& c\\ b& b& a\\ c& c& b\end{array}\right|=0\left[C_{1,}{C}_{2}identical\right]$$ $$D_3=\left|\begin{array}{ccc}a& b& a\\ b& c& b\\ c& a& c\end{array}\right|=0[C_1,C_{3}identical]$$ $$x=\frac{D_1}{D}=1,y=\frac{D_2}{D}=0,z=\frac{D_3}{D}=0$$ $$a>b>c>0\therefore (3abc-a^3-b^3-c^3)\ne 0$$

Commented byHongKing last updated on 08/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}\mathrm{cool}$$

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