Explicit root of symbolic equations

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Ductho Le
Ductho Le 2019-10-22
评论： Ductho Le 2019-10-22
I am trying to find the explicit roots of symbolic equations like the code bellow:
syms d1 d2 d3 d4 d5 d6 h11 h12 h13 h14 h21 h22 h23 h24 h41 h42 h45 h46 h51 h52 h55 h56 h61 h62 h65 h66
eq1 = h11*d1+h12*d2+h13*d3+h14*d4 == 0;
eq2 = h21*d1+h22*d2+h23*d3+h23*d4 == 0;
eq3 = d1+d2+d3+d4-d5-d6 == 0;
eq4 = h41*d1+h42*d2-h41*d3-h42*d4-h45*d5-h46*d6 == 0;
eq5 = h51*d1+h52*d2+h51*d3+h52*d4-h55*d5-h56*d6 == 0;
eq6 = h61*d1+h62*d2-h61*d3-h62*d4-h65*d5-h66*d6 == 0;
eq = [eq1,eq2,eq3,eq4,eq5,eq6];
var = [d1 d2 d3 d4 d5 d6];
S = solve(eq, var)
but the program return me the results like that
S =
struct with fields:
d1: [1×1 sym]
d2: [1×1 sym]
d3: [1×1 sym]
d4: [1×1 sym]
d5: [1×1 sym]
d6: [1×1 sym]
Can anybody please help me to find the explicit roots? Thanks you so much!

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采纳的回答

Stephan
Stephan 2019-10-22
syms d1 d2 d3 d4 d5 d6 h11 h12 h13 h14 h21 h22 h23 h24 h41 h42...
h45 h46 h51 h52 h55 h56 h61 h62 h65 h66
eq1 = h11*d1+h12*d2+h13*d3+h14*d4 == 0;
eq2 = h21*d1+h22*d2+h23*d3+h23*d4 == 0;
eq3 = d1+d2+d3+d4-d5-d6 == 0;
eq4 = h41*d1+h42*d2-h41*d3-h42*d4-h45*d5-h46*d6 == 0;
eq5 = h51*d1+h52*d2+h51*d3+h52*d4-h55*d5-h56*d6 == 0;
eq6 = h61*d1+h62*d2-h61*d3-h62*d4-h65*d5-h66*d6 == 0;
% Solve for d1...d6
sol_d1 = isolate(eq1,d1)
sol_d2 = isolate(eq2,d2)
sol_d3 = isolate(eq3,d3)
sol_d4 = isolate(eq4,d4)
sol_d5 = isolate(eq5,d5)
sol_d6 = isolate(eq6,d6)
gives:
sol_d1 =
d1 == -(d2*h12 + d3*h13 + d4*h14)/h11
sol_d2 =
d2 == -(d1*h21 + d3*h23 + d4*h23)/h22
sol_d3 =
d3 == d5 - d2 - d4 - d1 + d6
sol_d4 =
d4 == -(d3*h41 - d2*h42 - d1*h41 + d5*h45 + d6*h46)/h42
sol_d5 =
d5 == (d1*h51 + d2*h52 + d3*h51 + d4*h52 - d6*h56)/h55
sol_d6 =
d6 == -(d3*h61 - d2*h62 - d1*h61 + d4*h62 + d5*h65)/h66
This is what you wanted to get?
  1 个评论
Ductho Le
Ductho Le 2019-10-22
actually I want more detail but thanks you i have found solution based on your help.

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