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I have a system of equations and I would like your help to show that it has a unique solution with respect to $\lambda_1,\lambda_2,\lambda_3$.

More precisely, let the system be $$ \begin{cases} c\lambda_1=\lambda_3a\\ b\lambda_1=\lambda_2a\\ \lambda_1+\lambda_2+\lambda_3=1\\ \end{cases} $$ where it is assumed that $a,b,c,\lambda_1,\lambda_2,\lambda_3$ are strictly positive and $a+b+c=1$.

Question: I want to show that $\lambda_1=a, \lambda_2=b, \lambda_3=c$ is the unique solution of the system. I tried various substitution routes but couldn't come up with any clean steps. Could you help?

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  • $\begingroup$ The solution is not unique. If we take your solution, then the system is reduced to a single equation $\lambda_1+\lambda_2+\lambda_3=1$, where we have infinitely many solutions "with respect to" $\lambda_i$. $\endgroup$ – Dietrich Burde Mar 6 '19 at 17:51
  • $\begingroup$ I'm not sure I got your comment. Could you clarify more? Also in relation to the answer below. Thanks $\endgroup$ – TEX Mar 6 '19 at 18:08
  • $\begingroup$ $4$ equations ??? $\endgroup$ – user65203 Mar 6 '19 at 18:26
  • $\begingroup$ You have changed the system now. It used to be four equations in the six variables $a,b,c,\lambda_1,\lambda_2,\lambda_3$, which did not have a unique solution. $\endgroup$ – Dietrich Burde Mar 6 '19 at 19:06
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We get $$\lambda_2=\frac{b}{a}\lambda_1$$ $$\lambda_3=\frac{c}{a}\lambda_1$$ so $$\lambda_1+\frac{b}{a}\lambda_1+\frac{c}{a}\lambda_1=1$$ Can you proceed? From this equation (using that $$a+b+c=1$$) we get $$\lambda_1=a$$

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  • $\begingroup$ Thanks, got it. I'm confused with the comment I received above about the fact that the solution provided is not unique. Even if it is not your comment, could you clarify? Thanks $\endgroup$ – TEX Mar 6 '19 at 17:57
  • $\begingroup$ With $$ \lambda_1=a$$ you will get $$\lambda_2= \frac{b}{a}\cdot a=b$$ and $$\lambda_3=\frac{c}{a}\cdot a=c$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 6 '19 at 18:07
  • $\begingroup$ And this was and is not my comment! $\endgroup$ – Dr. Sonnhard Graubner Mar 6 '19 at 18:08
  • $\begingroup$ Can you read? $$a,b,c,\lambda_1,\lambda_2,\lambda_3$$ are strictly POSITIVE! $\endgroup$ – Dr. Sonnhard Graubner Mar 6 '19 at 18:39
  • $\begingroup$ Ok all things are ok! $\endgroup$ – Dr. Sonnhard Graubner Mar 6 '19 at 18:43
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Hint Multiplying both sides of the last equation gives $$a = a \lambda_1 + a \lambda_2 + a \lambda_3,$$ and substituting using the first two equations gives $$a = a \lambda_1 + (b \lambda_1) + (c \lambda_1) .$$

Factoring gives $a = (a + b + c) \lambda_1 = \lambda_1$. Now, the first two equations tell us that $\lambda_1$ determines $\lambda_2$ and $\lambda_3$.

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The determinant of the system is

$$\begin{vmatrix}c&0&-a\\b&-a&0\\1&1&1\end{vmatrix}=-a(a+b+c)\ne0,$$

hence the solution is unique.

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Rescaling the variables by the factors $a,b,c$ respectively,

$$\mu_1=\mu_2=\mu_3,\\a\mu_1+b\mu_2+c\mu_3=1$$ is trivial to solve.

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