The augmented matrix for the system:

{ 2 -2 2 | -6 }

{ 3 -5 6 | 3 }

{ 0 7 -4 | -7 }

The curly brackets extending over three lines are meant to represent the matrix enclosure.

Use the letter R to denote a matrix row and 1, 2 and 3 to define which row it is: R1, R2, R3.

Work out R1/2: { 1 -1 1 -3 } that is R1→1/2R1.

Work out 3R1-R2:

{ 0 2 -3 | -12 } that is, R2→3R1-R2.

We now have two rows beginning with 0, this row and R3, so let's reduce the problem to a 3 by 2 matrix for brevity and to avoid unnecessary clutter and repetition.

{ 2 -3 | -12 }

{ 7 -4 | -7 }

Call the rows r1 and r2 to distinguish from R1 and R2. Remember we're aiming for an identity matrix with zeroes and one 1 to the left of the vertical bar. Let's work out 3r2-4r1 in two steps:

{ 8 -12 | -48 }

{ 21 -12 | -21 }

3r2-4r1={ 13 0 | 27 } and divide this row by 13: { 1 0 | 27/13 } which can be expanded to { 0 1 0 | 27/13 } in the original matrix format, row 2.

Similarly 2r2-7r1 becomes { 0 13 | 70 }, and dividing this by 13 gives { 0 1 | 70/13 } which can be expanded to { 0 0 1 | 70/13 in the original matrix format, row 3.

This what we have:

{ 1 -1 1 | -3 }

{ 0 1 0 | 27/13 }

{ 0 0 1 | 70/13 }

Finally, we can convert R1 by performing R1→R1+R2-R3:

{ 1 0 0 | -82/13 }, so the solution to the system is x=-82/13, y=27/13 and z=70/13.