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   " 2  0  0 −1  0
     0  6  0 −5 -4 
     0  0  2  0  4
     0  0  0  0  0

From here you can get two solution vectors:

    x1 = (3,5,0,6,0)    x2 = (0,2,-6,0,3)"

How do you go from this matrix to the solution vectors?

related to an answer for: Finding S-invariant
in * Other Teaching Fields by (460 points)

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Well, it is a linear equation system with five variables and three linear independent equations so the solutions form a two-dimensional vector space that can be described as the set of linear combinations of two vectors. These vectors are not unique, and can be easily found as usual in linear algebra:

    2a-s = 0        ==> a = 1/2s
    6b-5s-4t = 0    ==> b = (5s+4t)/6
    2c+4t =0        ==> c = -2t

Hence, the solutions are

    (a,b,c,s,t) = (1/2,5/6,0,1,0)*s + (0,4/6,-2,0,1)*t

which could alternatively also represented as follows

    (a,b,c,s,t) = (3,5,0,6,0)*s' + (0,2,-6,0,3)*t'

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