# Show that a system of 4 equations has a unique solution

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?

• 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" $\lambda_i$. – Dietrich Burde Mar 6 '19 at 17:51
• I'm not sure I got your comment. Could you clarify more? Also in relation to the answer below. Thanks – TEX Mar 6 '19 at 18:08
• $4$ equations ??? – user65203 Mar 6 '19 at 18:26
• 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. – Dietrich Burde Mar 6 '19 at 19:06

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 $$\lambda_1=a$$

• 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 – TEX Mar 6 '19 at 17:57
• With $$\lambda_1=a$$ you will get $$\lambda_2= \frac{b}{a}\cdot a=b$$ and $$\lambda_3=\frac{c}{a}\cdot a=c$$ – Dr. Sonnhard Graubner Mar 6 '19 at 18:07
• And this was and is not my comment! – Dr. Sonnhard Graubner Mar 6 '19 at 18:08
• Can you read? $$a,b,c,\lambda_1,\lambda_2,\lambda_3$$ are strictly POSITIVE! – Dr. Sonnhard Graubner Mar 6 '19 at 18:39
• Ok all things are ok! – Dr. Sonnhard Graubner Mar 6 '19 at 18:43

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 $$\lambda_3$$.

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.

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.