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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I have the following system of equations where the m&#039;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&#039;t apply and I don&#039;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

aniizl · Accepted Answer

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&#039;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.

codosse2e · Accepted Answer

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

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