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 6 of 7: Solve for the common value x1=x4x_1=x_4
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In plain words

Once two of the four unknowns vanish, any single original equation is enough to determine the last two.

x1=x4=1a1−a4x_1=x_4=\frac{1}{a_1-a_4}
Detailed analysis

Substitute x2=x3=0x_2=x_3=0 into the original equation for i=1i=1: (a1−a3)⋅0+(a1−a4)x4=1(a_1-a_3)\cdot0+(a_1-a_4)x_4=1, so x4=1a1−a4x_4=\frac{1}{a_1-a_4}, and hence x1=x4=1a1−a4x_1=x_4=\frac{1}{a_1-a_4}.