This will be a little lengthy.....you also need to be very careful....one mistake dooms this!!!!
The idea is to end up with the form
1 0 0 a
0 1 0 b
0 0 1 c by the use of elementary row operations
So the original matrix form of the equations is
1 3 -3 - 27
2 1 1 -2
1 -1 3 17 multiply the first row by - 2 add to the second row
1 3 - 3 -27
0 -5 7 52
1 -1 3 17 multiply the first row by -1 add this to the third row
1 3 -3 -27
0 -5 7 52
0 -4 6 44 multiply the second row by 3/5 add this to the first row
1 0 6/5 21/5
0 -5 7 52
0 -4 6 44 multiply the second row by - 4/5 add this to the third row
1 0 6/5 21/5
0 -5 7 52
0 0 2/5 12/5 multiply the first and third rows by 5......this makes the next step easier
5 0 6 21
0 -5 7 52
0 0 2 12 divide the third row by 2
5 0 6 21
0 -5 7 52
0 0 1 6 multply the third row by -6 add this to the first row
5 0 0 -15
0 -5 7 52
0 0 1 6 multiply the third row by -7 add to the second row
5 0 0 -15
0 -5 0 10
0 0 1 6 divide the first row by 5 and the second by -5
1 0 0 -3
0 1 0 -2
0 0 1 6 This is the final matrix in Reduced-Row Echelon Form
Reading across.....the solution is { x, y, z} = ( -3, - 2, 6}
