This is a 3*5 matrix and there are 2 free variables. I don't know if this is the best way to solve this in MATLAB. But it doesn't work and output Empty sym: 0-by-1
clear x_1 x_2 x_3 x_4 x_5
syms x_1 x_2 x_3 x_4 x_5
eqn1 = x_1+0*x_2+3*x_3+2*x_4-4*x_5==4 ;
eqn2 = 2*x_1+x_2+6*x_3+5*x_4+0*x_5==7 ;
eqn3 = -x_1+x_2-3*x_3-x_4+x_5==-5 ;
tic ;
res = solve([eqn1,eqn2,eqn3]) ;
toc ;
You don't need the symbolic math toolbox for a simple system of linear equations. You're better off using mldivide, which is common enough that it has the shorthand \.
For a system Ax = b, where x is a vector of x values, A is a matrix of coefficients, and b is their product (right hand side of your system), you can solve it by
x = A\b;
So
A = [1 0 3 2 -4;
2 1 6 5 0;
-1 1 -3 -1 1];
b = [4; 7; -5];
x = A\b
>> ans = [0; 0; 2; -1; 0];
You can manually check that this result (x3=2, x4=-1, x1=x2=x5=0) works.
Use a QR or LU decomposition
>> [l,u] = lu(A)
l =
0.5 -0.33333 1
1 0 0
-0.5 1 0
u =
2 1 6 5 0
0 1.5 0 1.5 1
0 0 0 0 -3.6667
>> c = l\b
c =
7
-1.5
0
Then switch to Uy = L\b = c problem. Now you have,
[2 1 6 5 0][x1] [ 7 ]
[0 1.5 0 1.5 1][x2] = [-1.5]
[0 0 0 0 -3.6667][x3] [ 0 ]
[x4]
[x5]
Last one removes x5 and hence row 3/column 5. And from the rest you can check the solution space.
Let's say that A is an m-by-n matrix, and that b is a n-dimensional vector. You want to know all the solution to A x = b in the case of m<n, and it is expressed as following:
x = x_0 + K y, .. (1)
where x_0 is an arbitrary solution to the original equation, A x = b, K is a n-by-(n-m) matrix whose columns constitute a basis of the null space (also called kernel) of A, and y is an arbitrary (n-m)-dimensional vector. This y corresponds to what you call "free parameter".
Let me explain about the matrix K more. The null space of the matrix A is a set of vectors v such that A v = 0. This set of v is actually a vector space of (n-m)-dimension, and any v can be expressed as a linear combination of (n-m) vectors, k_1, .., k_{n-m}, which are linearly independent with each other and satisfy A k_i = 0 (i=1,..,n-m). The choice of k_1,..,k_{n-m} is not unique.
You can substitute eq. (1) to A x =b to verify that it's indeed the solution because A K = 0. The term K y expresses all vectors v such that A v=0.
One method to find the matrix K is the singular value decomposition,
A V = U S,
where V and U are n-by-n and m-by-m orthogonal matrices, respectively, and S is a m-by-n diagonal matrix like,
S = [s_1 0 0 0 0 ]
[ 0 s_2 0 0 0 ]
[ 0 0 s_3 0 0 ]
Since the last (n-m) columns of S are zero vectors, the last (n-m) columns of the product U S are also zero vectors. This means that the last (n-m) columns of V will yield zero vectors if we multiply A on them from the left. Therefore, these columns of V belongs to the null space of A. As they are linearly independent (as V is an orthogonal matrix), they constitute the basis of the null space. Therefore, we can set K as the last (n-m) columns of V found in the singular value decomposition.
I don't know how to do the singular value decomposition in Matlab, but there must be a function for it.
