MathLabs

Problem 5

Solve the following system of equations for real numbers x1x_1, x2x_2, x3x_3, x4x_4: ∣a1−a2∣x2+∣a1−a3∣x3+∣a1−a4∣x4=1∣a2−a1∣x1+∣a2−a3∣x3+∣a2−a4∣x4=1∣a3−a1∣x1+∣a3−a2∣x2+∣a3−a4∣x4=1∣a4−a1∣x1+∣a4−a2∣x2+∣a4−a3∣x3=1\begin{aligned} |a_1-a_2|x_2 + |a_1-a_3|x_3 + |a_1-a_4|x_4 &= 1 \\ |a_2-a_1|x_1 + |a_2-a_3|x_3 + |a_2-a_4|x_4 &= 1 \\ |a_3-a_1|x_1 + |a_3-a_2|x_2 + |a_3-a_4|x_4 &= 1 \\ |a_4-a_1|x_1 + |a_4-a_2|x_2 + |a_4-a_3|x_3 &= 1 \end{aligned} where a1a_1, a2a_2, a3a_3, a4a_4 are mutually distinct real numbers.
Step 1 of 7: Fix an ordering of the aia_i
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In plain words

Fixing an order is free (it is just renaming), but it turns every ∣ai−aj∣|a_i-a_j| into an ordinary difference, which is what makes linear elimination possible.

a1>a2>a3>a4a_1>a_2>a_3>a_4
Detailed analysis

The system is symmetric under simultaneously relabeling every index, so we may relabel 1,2,3,41,2,3,4 so that a1>a2>a3>a4a_1>a_2>a_3>a_4. Under this ordering, ∣ai−aj∣=amax⁡(i,j)−amin⁡(i,j)|a_i-a_j|=a_{\max(i,j)}-a_{\min(i,j)} for every pair, which removes all absolute values from the system.