CODEWITHSUNDEEP
matlab
TYL
TYL

Reputation: 1637

Substitute a vector of variables for syms when using linsolve for linear equations

I am trying to solve linear equations with many variables. So i used this to create multiple variables.

xvariables = sym('x', [n 1])

where n is the number of variables and it created a vector of variables x1 x2 x3...xn

So how do I use this vector beside syms when using linsolve instead of listing out all the variables?

% syms xvariables (something like that) instead of:

syms x2 x3 x4 x5 x6 x7 x8

eqn1 = 0.5*x2 + 0.12*x3 == 21.8;
eqn2 = 0.12*x2 + 0.5*x3 + 0.12*x4 == 21.9;
eqn3 = 0.12*x3 + 0.5*x4 + 0.12*x5 == 47.8;
eqn4 = 0.12*x4 + 0.5*x5 + 0.12*x6 == 37.6;
eqn5 = 0.12*x5 + 0.5*x6 + 0.12*x7 == 27.5;
eqn6 = 0.12*x6 + 0.5*x7 + 0.12*x8 == 52.5;
eqn7 = 0.12*x7 + 0.5*x8 == 59;

[A,B] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5, eqn6, eqn7], [x2, x3, x4, x5, x6, x7, x8])
X = linsolve(A,B)

I hope that makes sense. Thanks!

Upvotes: 0

Views: 187

Answers (2)

obchardon
obchardon

Reputation: 10792

You have better to use the mldivideoperator. (symbol \) (in my opinion!)

You have your matrix P that contains the equation:

p =

   [0.50000   0.12000   0.00000   0.00000   0.00000   0.00000   0.00000;
   0.12000   0.50000   0.12000   0.00000   0.00000   0.00000   0.00000;
   0.00000   0.12000   0.50000   0.12000   0.00000   0.00000   0.00000;
   0.00000   0.00000   0.12000   0.50000   0.12000   0.00000   0.00000;
   0.00000   0.00000   0.00000   0.12000   0.50000   0.12000   0.00000;
   0.00000   0.00000   0.00000   0.00000   0.12000   0.50000   0.12000;
   0.00000   0.00000   0.00000   0.00000   0.00000   0.12000   0.50000]

or if you don't want to write the 0:

p = zeros(7)

p(1,1) = 0.5;
p(1,2) = 0.12;
p(2,1) = 0.12;
...

Your vector s that contains the solution

s = [21.8,21.9,47.8,37.6,27.5,52.5,59]'

And you can resolve your system with:

>> res = p\s

res =

   40.0000 %x2
   15.0000 %x3
   80.0000 %x...
   50.0000
   25.0000
   75.0000
  100.0000

Upvotes: 0

Matt
Matt

Reputation: 13933

When using sym('x',[n 1]) you are creating a symbolic vector with automatically generated elements [x1; x2; ...; xn]. The elements do not directly appear in the MATLAB workspace but they are accessible by using parentheses on the created variable (xvariables in your question, x in my answer). Therefore we use the same syntax as with a 'normal' vector/matrix. Thus xn can be accessed with x(n).

For the equations we can use a vector as well and store them in a single variable eqn. This additionally simplifies the call to equationsToMatrix.

This yields the following code:

x = sym('x', [7 1]);

eqn(1) = 0.50*x(1) + 0.12*x(2) == 21.8;
eqn(2) = 0.12*x(1) + 0.50*x(2) + 0.12*x(3) == 21.9;
eqn(3) = 0.12*x(2) + 0.50*x(3) + 0.12*x(4) == 47.8;
eqn(4) = 0.12*x(3) + 0.50*x(4) + 0.12*x(5) == 37.6;
eqn(5) = 0.12*x(4) + 0.50*x(5) + 0.12*x(6) == 27.5;
eqn(6) = 0.12*x(5) + 0.50*x(6) + 0.12*x(7) == 52.5;
eqn(7) = 0.12*x(6) + 0.50*x(7) == 59;

[A,B] = equationsToMatrix(eqn, x)
X = linsolve(A,B)

Upvotes: 1

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