Math Help Forum: Augmented Matrix in Calculator, TI-83

I'm confused here, I have a basic system:

x+3y+2z=5
-x+z=1
x-y-2z=-3

So I plut in the augmented matix into my calculator TI-83

1st row [1, 3, 2, 5]
2nd row [-1,0,1,1]
3rd row [1,-1,-2,-3]

When I put it in RREF, I end up with many solutions, I even did it online here to double check: Linear Algebra Toolkit

Yet my book is giving soultions of 2, -1, and 3 respectively. The solutions do work, but I don't see what's going on with the RREF in my calculator. I've used this like a hundred times to do problems like this, it's not working this time, I can't figure out my error...

I am not fimilar with the row operation functions on the TI-83, but I know you can find the inverse of the 3x3 coeffecient matrix and multiply it by the 3x1 RHS of your system. That will give you the solution.

I can get a solution on my calculator by just puting the augmented matrix in row-reduced echelon form. When you reduce it to that form, you end up with many solutions. The solution in the book is unique, x=2, y=-1, and z=3. Just plug it into any calculator and put it in row reduced echelon form and see what you get. I even tried it here: Linear Algebra Toolkit

I derived this system to find coordinate vectors for a certain polynomial, the coordinatee vector should be <2, -1, 3>, which are the solutions to that system. But I'm getting many solutions on my calculator, and the online tool I linked above.

The determinant of the coefficient matrix is zero. Therefore, if there is at least one solution, there are infinitely many solutions.

the solution i come up with is (t, 1-t, t+1). without some way of choosing one polynomial over another, i see no reason to choose t = 2 as THE solution.