Question

If the matrix has free variables. Is it automatically linearly dependent?

Answer 1

See below, your question is unclear. A specific example might be helpful.

Explanation:

the wording here is a bit loose, i think.

let's say you have a nxn system, ie n equations and n unknown variables, so that `A mathbf x = mathbf b`.

Now, if you cannot row reduce that A to a smaller matrix m x n matrix, or if `det A ne 0`, or if you cannot show that at least 2 of its column or row vectors are linearly dependent, ..... , well these are all ways of saying that the matrix is non-singular, and a non-singular nxn matrix will show all of these features at the same time . And the original equation `A mathbf x= mathbf b` will have only one solution.

On the other hand, if you can row reduce that A to a smaller matrix m x n matrix, or if det A = 0, or if you can show that at least 2 of its column or row vectors are linearly dependent, ..... , well these are all ways of saying that the matrix is singular, and the singular matrix will display all these features at the same time . In this case, the original equation `A mathbf x = mathbf b` has either no solution or an infinite number of solutions.

I am not sure about your question, though, because you may mean you have already row-reduced A to mxn....look at the wording. If you have, so that the row vectors are no longer linearly independent, then you have already removed the dependency.