I have the following system of equations where the m's are known but a , b , c ,

Simone Werner 2022-05-22 Answered
I have the following system of equations where the m's are known but a , b , c , x , y , z are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply and I don't want to do it through tedious substitutions.
a + b + c = m 0 a x + b y + c z = m 1 a x 2 + b y 2 + c z 2 = m 2 a x 3 + b y 3 + c z 3 = m 3 a x 4 + b y 4 + c z 4 = m 4 a x 5 + b y 5 + c z 5 = m 5
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Answers (2)

aniizl
Answered 2022-05-23 Author has 12 answers
Let x , y , z be roots of t 3 + p t 2 + q t + r = 0 .
i.e.
x 3 + p x 2 + q x + r = 0 ----- (1)
y 3 + p y 2 + q y + r = 0 ----- (2)
z 3 + p z 2 + q z + r = 0 ----- (3)
Multiply the (1) by a , (2) by b and (3) by c and adding gives
m 3 + p m 2 + q m 1 + r m 0 = 0 ----- (4)
Multiply the (1) by a x , (2) by b y and (3) by c z and adding gives
m 4 + p m 3 + q m 2 + r m 1 = 0 ----- (5)
Multiply the (1) by a x 2 , (2) by b y 2 and (3) by c z 2 and adding gives
m 5 + p m 4 + q m 3 + r m 2 = 0 ----- (6)
Now (4), (5), (6) is a set of 3 linear equations in 3 variables ( p , q , r ) and can be solved easily.
This give us the cubic which x , y , z satisfy (because we can find out that p , q , r are) which can be solved using Cardano's Method, or more simply by the Trigonometric and Hyperbolic Method.
Once you know x , y , z you can solve for a , b , c , as those become just linear equations.
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codosse2e
Answered 2022-05-24 Author has 2 answers

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