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BRANSON

You just need to rewrite your system of N equations in matrix form, AX=B where A is the N by N matrix of coefficients, X is the N by 1 column of variables, and B is the N by 1 column of constants. For the example you gave,

2x+y=5
x+3y=4

so N=2 and in matrix form (and using MATLAB syntax),

A = [2, 1; 1, 3]
X = [x; y]
B = [5; 4]

So if AX=B, then X=(A^-1)B where A^-1 is the matrix inverse of A. This is where MATLAB does the hard work for you -- once you give it A and B, just type (A^-1)B to have it multipy B by the inverse of A. The N by 1 matrix it spits out will be the solution to your system of equations, which for the example above looks like:

ans=
2.2
0.6

So the solution to your system is x=2.2, y=0.6. It works exactly the same way with any linear system of equations (no exponents, products of two variables, trig functions, etc. -- you will need a somewhat more advances method for those types of problems), and it's always a good idea to plug the numbers back in to the equations for a quick check at the end.

Does that all make sense? I can give you step-by-step instructions if you need them.

Hello!

If it is for linear equations, you can use cramer's rules, with a couple of "for" cicles.

http://en.wikipedia.org/wiki/Cramer's_rule

Yes, this is a variation on the technique that I described above.

a easier way you can use Gauss Jordan Method put your matrix and use this comand

you just order you matrix for example: for the next 2 equations system

2x+y=4

x+5y=-2

the matrix will be

2 1 4

1 5 -2

MATLAB Function Reference

rref

Reduced row echelon form

Syntax

R = rref(A)
[R,jb] = rref(A)
[R,jb] = rref(A,tol)

Description

R = rref(A) produces the reduced row echelon form of A using Gauss Jordan elimination with partial pivoting. A default tolerance of (max(size(A))*eps *norm(A,inf)) tests for negligible column elements.

[R,jb] = rref(A) also returns a vector jb such that:

• r = length(jb) is this algorithm's idea of the rank of A.
• x(jb) are the pivot variables in a linear system Ax = b.
• A(:,jb) is a basis for the range of A.
• R(1:r,jb) is the r-by-r identity matrix.

[R,jb] = rref(A,tol) uses the given tolerance in the rank tests.

Roundoff errors may cause this algorithm to compute a different value for the rank than rank, orth and null.

Examples

Use rref on a rank-deficient magic square:

A = magic(4), R = rref(A) A = 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1 R = 1 0 0 1 0 1 0 3 0 0 1 -3 0 0 0 0

See Also

inv, lu, rank

Yes, this would certainly also work. I think, though, that the technique I described above is the most straightforward and easy to understand for someone who is not familiar with MATLAB or linear algebra,

well it is just order the matrix, load the matrix or write it on matlab

A=[2 1 4;5 1 -2]

rref(A) and etc ...

sorry wrong link

http://www.engin.umich.edu/group/ctm/extras/plot.html

Hello,

If you have the symbolic toolbox you could also try the expression "solve", here is your example "solved".

Good look!

equation1='2*x+y=5'

equation2='x+3*y=4'

[solution]=solve(equation1,equation2)

x=solution.x

y=solution.y

Oh, I forgot, with solve you can also use non linearities, and for your computational purposes it uses the least square error method.

equation1='2*x^2+y=5'

equation2='x+3*y^2=4'

[solution]=solve(equation1,equation2)

x=double(solution.x)

y=double(solution.y)

use matrix to solve equation.

read its example. if you have 10 equations, there are 10 raw and 10 colume matrix. then divide. just like write on pa\per