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 7 of 7: State the answer for the original labeling
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In plain words

The relabeling in step 1 only reordered names; translating the clean answer back to the original indices shows the solution depends solely on which aia_i are extreme, not on any particular labeling.

xmax⁡=xmin⁡=1amax⁡−amin⁡,  the other two xi=0x_{\max}=x_{\min}=\frac{1}{a_{\max}-a_{\min}},\ \text{ the other two } x_i=0
Detailed analysis

Undoing the relabeling of step 1: if aia_i and aja_j are respectively the largest and smallest of the four numbers, the unique solution is xi=xj=1ai−ajx_i=x_j=\frac{1}{a_i-a_j} and the remaining two unknowns equal 00; substituting back into all four original equations confirms this satisfies the system.