Solve [{2 + 2 a*d + 2 a*e == 0, 1 - 2 b + 2 d*b + 2 e*b == 0, 1 + 2 d*c + 2 e*c == 0, -2 + a^2 - 2 b + b^2 + c^2 == 0, -2 + a^2 + b^2 + c^2 == 0}, {a, b, c, d, e}]My rty: 2 + 2 λ 1 x 1 + 2 λ 2 x 1 = 0 1 + 2 λ 1 x 2 + 2 λ 2 x 2 − 2 λ 2 = 0 1 + 2 λ 1 x 3 + 2 λ 2 x 3 = 0 x 1 2 + x 2 2 − 2 x 2 + x 3 2 − 2 = 0 x 1 2 + x 2 2 + x 3 2 − 2 = 0

Question

Solve [{2 + 2 a*d + 2 a*e == 0, 1 - 2 b + 2 d*b + 2 e*b == 0, 1 + 2 d*c + 2 e*c == 0, -2 + a^2 - 2 b + b^2 + c^2 == 0, -2 + a^2 + b^2 + c^2 == 0}, {a, b, c, d, e}]My rty:  2  +  2      λ    1        x    1    +  2      λ    2        x    1    =  0  1  +  2      λ    1        x    2    +  2      λ    2        x    2    −  2      λ    2    =  0  1  +  2      λ    1        x    3    +  2      λ    2        x    3    =  0      x    1    2    +      x    2    2    −  2      x    2    +      x    3    2    −  2  =  0      x    1    2    +      x    2    2    +      x    3    2    −  2  =  0

Clarence Mills · Accepted Answer

Using the notations of your Mathematica command, you should notice that variables   a  ,  b  ,  c can be eliminated using the first three equations. This leads to  a  =            −      1              d      +      e        b  =  −      1          2      (      d      +      e      −      1      )        c  =      1          2      (      d      +      e      )      Now, replacing these in the next equations, the fourth becomes      5          4      (      d      +      e              )        2              +      1          d      +      e      −      1        +      1          4      (      d      +      e      −      1              )        2              −  2  =  0and the fifth becomes      5          (      d      +      e              )        2              +      1          (      d      +      e      −      1              )        2              −  8  =  0So, only   (  d  +  e  ) terms everywhere and then the problem.

Solve [{2 + 2 ad + 2 ae == 0, 1 - 2 b + 2 db + 2 eb == 0, 1 + 2 dc + 2 ec == 0, -2 + a^2 - 2 b + b^2 + c^2 == 0, -2 + a^2 + b^2 + c^2 == 0}, {a, b, c, d, e}]

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