Show that a system of equations has a unique solution and indicate the solution

linear algebrasystems of equations

Consider the following system of equations with strictly positive unknowns $\lambda_1,\lambda_2,\lambda_3,\lambda_4$

$$
\begin{cases}
\lambda_1 d=\lambda_4 a\\
\lambda_2 d=\lambda_4 b\\
\lambda_1 c=\lambda_3 a\\
\lambda_2 c=\lambda_3 b\\
\lambda_3 d=\lambda_4 c\\
\lambda_1+\lambda_2+\lambda_3+\lambda_4=1
\end{cases}
$$

and parameters $a>0,b>0,c>0,d>0$ with $a+b+c+d=1$. Show that the unique solution of the system is

$$
\lambda_1=a\\
\lambda_2=b\\
\lambda_3=c\\
\lambda_4=d\\
$$

I tried by taking several routes but I couldn't find any clear pattern.

Best Answer

Using SymPy, we create the augmented matrix:

>>> from sympy import *
>>> a, b, c, d = symbols('a b c d', real=True, positive=True)
>>> 
>>> M = Matrix([[ d, 0, 0,-a, 0],
                [ 0, d, 0,-b, 0],
                [ c, 0,-a, 0, 0],
                [ 0, c,-b, 0, 0],
                [ 0, 0, d,-c, 0],
                [ 1, 1, 1, 1, 1]])

Imposing the constraint $a + b + c + d = 1$:

>>> M.subs(d,1-a-b-c)
Matrix([
[-a - b - c + 1,              0,              0, -a, 0],
[             0, -a - b - c + 1,              0, -b, 0],
[             c,              0,             -a,  0, 0],
[             0,              c,             -b,  0, 0],
[             0,              0, -a - b - c + 1, -c, 0],
[             1,              1,              1,  1, 1]])

Using Gaussian elimination to compute the RREF:

>>> _.rref()
(Matrix([
[1, 0, 0, 0, a/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 1, 0, 0, b/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 0, 1, 0, c/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 0, 0, 1,                    1/(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1)],
[0, 0, 0, 0,                                                                                       0],
[0, 0, 0, 0,                                                                                       0]]), [0, 1, 2, 3])

Simplifying:

>>> simplify(_)
(Matrix([
[1, 0, 0, 0,              a],
[0, 1, 0, 0,              b],
[0, 0, 1, 0,              c],
[0, 0, 0, 1, -a - b - c + 1],
[0, 0, 0, 0,              0],
[0, 0, 0, 0,              0]]), [0, 1, 2, 3])
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