@Alejandro Jakubi  Your wonderful programmatic way  works, for example, in Standard Maple 12 on Windows 8.1:

cat(`&#`, convert("25AA", decimal, hex), `;`);

@Markiyan Hirnyk   Without a doubt, your plot out of competition!

@Rouben Rostamian   The parentheses are necessary.

Test the solution:

P:=Matrix([[ 0 , .5 , .5 , 0 , 0 , 0 ], [ 1/3 , 0 , 0 , 1/3 , 1/3 , 0 ], [ 1/3 , 0 , 0 , 0 , 1/3 , 1/3 ], [ 0 , 1 , 0 , 0 , 0 , 0 ], [ 0 , .5 , .5 , 0 , 0 , 0 ], [ 0 , 0 , 1 , 0 , 0 , 0 ]]):

pii:=Vector[row]([ a , b , c , d , e , f ]):

solve({seq((pii.P)[i]=pii[i], i=1..6)});

eval([seq((pii.P)[i]=pii[i], i=1..6)], %);

                                                       {a = 2.*f, b = 3.*f, c = 3.*f, d = f, e = 2.*f, f = f}

             [2.000000000*f = 2.*f, 3.0*f = 3.*f, 3.0*f = 3.*f, 1.000000000*f = f, 2.000000000*f = 2.*f, 1.000000000*f = f]

Statement of the problem is not clear. Your  beta  depends on  l  rather than on  n .  What does  beta[n]  mean?

Addition: The same question remains. The fact that in the equation  beta  is replaced by  beta[n] does not change anything. The roots of the equation does not depend on their designations.

@rit  If you have the matrix of the size  n  by  n , the number of its entries that lie below (or above) of the main diagonal equal to the sum of the arithmetic progression 1 + 2 + 3 + .. + (n-1) = n * (n-1) / 2.  If each element is  0  or  1 , the number of all possible ordered combinations of  n * (n-1) / 2  such elements obviously equal to the number of placements of  [1$(n*(n-1)/2), 0$(n*(n-1)/2)]  by  n*(n-1)/2 . This number equals  2^(n*(n-1)/2) .

For example, if  n=5  then we take of the set  {1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0}  all possible ordered combinations of  10  numbers .  The total number of such combinations equals  2^10=1024

@acer   Thank you very much! A brilliant solution.

@one_man  Thank you for your suggestion! But I want the fractions to be displayed in 2D mode in beautiful style. Such output is important to me in one application.  See my solution in this thread.

@Carl Love  For example, should be  

rather than

or

@beidouxing   The quote "So  x-y  is a solution for the  eqn1=0 , so the eqn1 can be divided by the  x-y".  This is true only for rational expressions.

For example,  eval(1 - exp(x-y), x=y) = 0 , but you can not divide  1 - exp(x-y)  by  x-y

You should submit a complete version of your problem, ie, all the equations of the system and etc,  rather than only one equation.

@Markiyan Hirnyk   Yes, you're right. A higher degree is needed.

V := <t, piecewise(t >= 0, t^5, 0), piecewise(t >= 0, 0, -t^5)>;

Student[VectorCalculus][Torsion](V, t);

Student[VectorCalculus][SpaceCurve](V, t = -1 .. 1);

@Markiyan Hirnyk  For example, if you need to find the derivative at a point, then you can find  separately left and right derivatives and make sure they match.

@Markiyan Hirnyk 

V1:=<t, t^3, 0>;

V2:=<t, 0, -t^3>;

Student[VectorCalculus][Torsion](V1, t);

Student[VectorCalculus][Torsion](V2, t);

eval(Student[VectorCalculus][Curvature](V1, t), t=0);

Obviously  piecewise(t>=0, V1, t<0, V2)  is not a plane curve.

@Markiyan Hirnyk  If you are able to supplement my answer, no one is stopping you from doing so.

@taro  Compare two examples:

restart;

assign(a, c);

a := d;

a, c, d;

and

restart;

a := c;

assign(a, d);

a, c, d;

@taro  Quote "what is assing(a,c)? Does it mean both a:=c and c:=a?"

No. This means that in the future, writing  a  we mean  с  and  if after this we write assign(a,d)  then it means that с is d  and  a  is  d .