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 5 of 7: Solve the three relations for x2,x3x_2,x_3
Previous stepNext step
In plain words

Three linear relations among four unknowns, once combined pairwise, pin down two of the unknowns completely and force the remaining two to be equal.

x2=x3=0,x1=x4x_2=x_3=0,\quad x_1=x_4
Detailed analysis

Subtracting the equation of step 3 from that of step 2 gives 2x2=02x_2=0, so x2=0x_2=0; subtracting the equation of step 4 from that of step 3 gives 2x3=02x_3=0, so x3=0x_3=0. Substituting into the equation of step 2 gives x1=x4x_1=x_4.