1. ## Linear Algebra: Question.

In this eqn, x(sub1), x(sub2), x(sub3) all end up equaling to 0. x(sub4) is free, shouldn't this be a dependent set then? For a homogenous soln. to have a nontrivial soln. there needs to be a free variable, there is a free variable here.

This answer says it is Independent, however, it should be dependent? What am I missing here?

Or is it because since there are pviot positions for every all columns besides column 4, that implies system is trivial? Therefore, linear ind.

2. ## Re: Linear Algebra: Question.

Strictly speaking you have shown only a matrix which does not necessarily imply any question. I can guess that you are actually given a problem like $\begin{bmatrix}1 & 0 & 3 & 0 \\ 0 & -1 & 4 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.

That is equivalent to the four equations w+ 3y= 0, -x+ 4y= 0, 7y= 0, and 0= 0. From the third equation, y= 0. The first and second equations become w= 0 and -x= 0 so that we must have w= x= y= 0. But z can be anything. The solution set consists of all multiples of (0, 0, 0, 1). Now, what do you mean by "dependent set"? The solution set of such an equation is a subspace of $R^4$ and the set of all vectors in subspace is never "independent".