In a matrix row-echelon form means the main diagonal is all 1's and the elements below the main diagonal are all 0, in other words a (upper) triangular matrix with all 1's on the main diagonal.
In this form it is a relatively easy job to "solve" the simultaneous set of equations represented by the matrix equation.
Getting the matrix into this form can be done in the way you may have previously called "elimination".
Start out by dividing each row (and the constant) by first element of each row as long as it is not 0 (if it is 0 leave it alone).
Subtract the first row from each of the other 2 rows so that the first element in each row except the first is 0.
The proceed in a similar fashion with the rows: divide by the first element and subtract the 2nd row from the third. This should give you row-echelon form.
I cannot help you with GeoGebra...sorry!

Paul M.
07/28/19
Wilcox A.
Using (x = 3, y = -2 and z = 1). I arrived at, (x + y + z = 2, (3) + (−2) + (1) = 2), (3x + 2y − 2z = 3, 3(3) + 2(−2) − 2(1) = 3, 9 − 4 − 2 = 3) and (2x + 3y + 4z = 4, 2(3) + 3(−2) + 4(1) = 4, 6 − 6 + 4 = 4). Changing these equations below to augmented matrix form, (1 1 1 | 2), (3 2 −2 | 3), ( 2 3 4 | 4). I need help to perform row operations on the system to obtain a row-echelon form and the solution. To answer question 3 above.07/29/19
Wilcox A.
Please can you help me with the calculations.07/28/19