Put the following matrix into row reduced echelon form: (1, 2, -1, 0), (-1, 1, 2 ,1), (0, 3, 1, 1) Does this have a unique solution, or an infinite number?

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Put the matrix `[(1,2,-1,0),(-1,1,2,1),(0,3,1,1)]` in reduced row echelon form:

Labeling the top row I, the second row II and the third row III we use permissable transformations; we can add a multiple of one row to another and we can switch any rows, and we can multiply any row by a nonzero constant.

`[(1,2,-1,0),(-1,1,2,1),(0,3,1,1)]`    II=I+II

`=[(1,2,-1,0),(0,3,1,1),(0,3,1,1)]`   III=II-III;II=1/3(II)

`=[(1,2,-1,0),(0,1,1/3,1/3),(0,0,0,0)]` I=-2(II)+I

`=[(1,0,-5/3,-1/3),(0,1,1/3,1/3),(0,0,0,0)]`

This is in reduced row-echelon form.

There are an infinite number of solutions; they are of the form:

`(x,-1/5x+1/5,3/5x+2/5)` (You can get this by solving `x-5/3z=-2/3` for z and substituting into `y+1/3z=1/3` getting an expression for y in terms of x. Now you have expressions for y and z in terms of x. There are other possibilities.