solve with many variables

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I had to start an answer, since a comment would not fit in the given space. The present "answer" is only showing that the resulting equation of degree six -- even in some particular cases -- is "too generic" to offer a possibility of expressing solutions by "simple means", e.g. radicals (combined with simple algebra.) First of all, let us rewrite the code, so that it can be copy-pasted.

x,a,b,c,s1,s2,s3,s4,s5,s6 = var( 'x a b c s1 s2 s3 s4 s5 s6' )

k = matrix( [
    [-3*a-b+c*s1,a,a,0,a,0],
    [a,-2*a-b+c*s2,a,0,0,0],
    [a,a,-4*a-b+c*s3,a,a,0],
    [0,0,a,-2*a-b+c*s4,a,0],
    [a,0,a,a,-4*a-b+c*s5,a],
    [0,0,0,0,a,  -a-b+c*s6]
] )

B = matrix( k - x*matrix.identity(6) )
print k

And the matrix k is:

[c*s1 - 3*a - b              a              a              0              a              0]
[             a c*s2 - 2*a - b              a              0              0              0]
[             a              a c*s3 - 4*a - b              a              a              0]
[             0              0              a c*s4 - 2*a - b              a              0]
[             a              0              a              a c*s5 - 4*a - b              a]
[             0              0              0              0              a   c*s6 - a - b]

A first question is: Is this the matrix to be studied?

Now we want to subtract $x$ on the diagonal, apply the determinant, and solve with respect to $x$. Then we can forget about $b$. (It has only a shifting contribution of the eigenvalues.) We can set without loss of generality $b=0$. The variable $c$ is the next one that we should get rid of. Since $s_1,s_2,\dots,s_6$ are general enough, we can also set either $c=0$, the degenerated case which removes the dependency on the $s$-variables, or $c=1$ else. Than, replacing $s_1-3a$ by "an other" $s_1$ we have to understand the eigenvalues of a matrix $M$ of the following "too general" shape:

M = matrix( [
    [s1,a,a,0,a,0],
    [a,s2,a,0,0,0],
    [a,a,s3,a,a,0],
    [0,0,a,s4,a,0],
    [a,0,a,a,s5,a],
    [0,0,0,0,a,s6]
] )
M

We get:

[s1  a  a  0  a  0]
[ a s2  a  0  0  0]
[ a  a s3  a  a  0]
[ 0  0  a s4  a  0]
[ a  0  a  a s5  a]
[ 0  0  0  0  a s6]

If $a=0$, a very special case, we have of course no problems. Else, we can multiply the whole matrix by $1/a$ and get an equivalent problem, namely to find the eigenvalues of the matrix:

M = matrix( [
    [s1,1,1,0,1,0],
    [1,s2,1,0,0,0],
    [1,1,s3,1,1,0],
    [0,0,1,s4,1,0],
    [1,0,1,1,s5,1],
    [0,0,0,0,1,s6]
] )
M

Explicity:

[s1  1  1  0  1  0]
[ 1 s2  1  0  0  0]
[ 1  1 s3  1  1  0]
[ 0  0  1 s4  1  0]
[ 1  0  1  1 s5  1]
[ 0  0  0  0  1 s6]

Some rows / columns are simpler than the other ones. For instance, the row / column through the entry $s_6$ has only one non-diagonal $1$. This matrix is still too complicated to allow a general formula for its eigenvalues. For instance, let us plug in some explicit values for $s_1,s_2,\dots,s_6$:

def MM( s1, s2, s3, s4, s5, s6 ):
    return matrix( [
        [s1,1,1,0,1,0],
        [1,s2,1,0,0,0],
        [1,1,s3,1,1,0],
        [0,0,1,s4,1,0],
        [1,0,1,1,s5,1],
        [0,0,0,0,1,s6]
    ] )

MM( 0,0,0,0,0,0 ).charpoly()

This gives:

x^6 - 8*x^4 - 6*x^3 + 7*x^2 + 4*x - 1

Well, the above polynomial has no "obvious roots", that can be expressed by radicals. So the more general equation also has no "obvious roots". Which is the origin of the question?