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 4 of 7: Subtract equation 3 minus equation 4
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In plain words

Three such differences (steps 2-4) turn the original nonlinear-looking system into an ordinary linear system in x1,x2,x3,x4x_1,x_2,x_3,x_4.

−x1−x2−x3+x4=0-x_1-x_2-x_3+x_4=0
Detailed analysis

Equation 4 is (a1−a4)x1+(a2−a4)x2+(a3−a4)x3=1(a_1-a_4)x_1+(a_2-a_4)x_2+(a_3-a_4)x_3=1. Subtracting it from equation 3 and dividing by a3−a4≠0a_3-a_4\ne0 gives −x1−x2−x3+x4=0-x_1-x_2-x_3+x_4=0.