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While calibrating a three-phase system's voltage measurement, I stuck with the following system of equations.

$$a_1=b_1x-b_3z\\a_2=b_2y-b_1x\\a_3=b_3z-b_2y$$

I need solution for $x$, $y$, and $z$. If it is helpful, we can use the following facts about this system.

$$a_1+a_2+a_3=0\\b_1+b_2+b_3=0$$

I used Wolfram alpha with the following code with no success.

Solve[{Subscript[a, 1] == Subscript[b, 1] x - Subscript[b, 3] z, Subscript[a, 2] == Subscript[b, 2] y - Subscript[b, 1] x, Subscript[a, 3] == Subscript[b, 3] z - Subscript[b, 2] y, Subscript[a, 1] + Subscript[a, 2] + Subscript[a, 3] == 0, Subscript[b, 1] + Subscript[b, 2] + Subscript[b, 3] == 0}, {x, y, z}, MaxExtraConditions -> Automatic]

Please help/guide me for the solution. Thank you.

SKGadi
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    This is a linear system, but the $,3^{rd},$ equation is the sum of the first two, so the equations are not independent, so the system does not have a unique solution. – dxiv Jun 28 '22 at 22:44
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    @AidenChow You left out the other side of the equal sign. Adding the first $2$ equations and using that $,a_1+a_2+a_3=0,$ gives $,-b_3 z + b_2 y = a_1 + a_2 = -a_3,$ which is the $3^{rd}$ equation with the signs changed. – dxiv Jun 28 '22 at 23:08
  • I understand that they are not independent. – SKGadi Jun 29 '22 at 12:50

1 Answers1

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As pointed out in the comments, the third equation is a combination of the first two. Since $a_3=-a_1-a_2$ and $b_3=-b_1-b_2$, we have the equivalent system

$$a_1=b_1x+(b_1+b_2)z\\a_2=b_2y-b_1x$$

or

$$x=\frac{a_1-(b_1+b_2)z}{b_1}\\y=\frac{a_2+a_1-(b_1+b_2)z}{b_2}$$

where $z$ can be chosen arbitrarily.

(If $b_1$ and $b_2$ are both non-zero, then we can solve for $x$ and $y$ as shown. If $b_1$ and $b_3$ are non-zero, we can solve for $x$ and $z$. If $b_2$ and $b_3$ are non-zero, we can solve for $y$ and $z$. If only one of them is non-zero, that contradicts $b_1+b_2+b_3=0$. If all three are zero, the system is trivial.)

mr_e_man
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