Thats one of my problems. I'm told to find out whether the vectors are linearly independent or not, and the textbook tells me that if the linear system has non trivial solutions, it is dependent, and if it has trivial solutions, it will be dependent. Am I supposed to find row echelon form? And if so, what do I do with it?
I think the problem is asking to define
And then determine if has non-trivial solutions.
Where and .
Since the number of equations exceede the number of variables the homogenous system will have non-trivial solutions.
If you want to find those solutions you need to use Gaussian-Jordan elimination.
Do you know how to do that?
Then you find the row reduced echelon form of the matrix.
Once you do that it should be clear if the system has only a trivial solution of not*.
*)To get only trivial solutions the row reduced echelon form must be
Because this matrix leads to the following equations:
Which immediately means .
Therefore, it has only the trivial solutions.
He wasnt saying that was the answer.. he was saying that if you get this as your answer then your vectors would be linearly independent. The row of zeros indicates that that specific vector was a linear combination of the others so it is dependent ( ie. not independent... )
With a little more work we can actually find all the solutions. Multiply the 4th row by 1/3 and add it to the 2nd. After that multiply the 4th row by -1. Finally interchange the 3rd and 4th rows. The resulting matrix is:
This tells us the following:
The fourth equation is redundant, since it is obviously true.
Solving each equation for the leading variables we get:
Let then it means .
What does this means? It means we found all solutions. Let be any number you want it to be. It follows if we set and then we found a solution. Try it for and convince these are solutions. Since varies for infinitely many values it means there are infinitely many solutions. This is an example when the number of equations is equal to the number of variables and yet the number of solutions is infinite.