Solving a system of equations using row operations

1
Solving a system of equations using row operations
2
3x 4y 4z 142x y z 2 x 2y
5z -1
3x 4y 4z 14
2x y z 2
x 2y 5z -1
3 4 -4 14 2 -1 1 2
1 2 5 -1
3
Move row three up to the top row, where you want
to have 1 in the top left corner.
3 4 -4 14 2 -1 1 2
1 2 5 -1
4
1 2 5 -13 4 -4 142 -1 1 2
Multiply row 1 by -3 and add to row 2
3 4 4 14(-1) 2 -1 1 2
3 4 -4 14 - 3 -6 -15 3
- 3 -6 -15 3
0 -2 -19 17
Now put this row where row 2 was
5
1 2 5 -1 0 -2 -19 17
2 -1 1 2
Multiply row 1 by -2 and add to row 3
3 4 4 14(-1) 2 -1 1 2
2 -1 1 2 - 2 -4 -10 2
- 2 -4 -10 2
0 -5 -9 4
Now put this row where row 3 was
6
1 2 5 -1 0 -2 -19 17
0 -5 -9 4
Multiply row 2 by -3 and add to row 3
3 4 4 14(-1) 2 -1 1 2
0 -5 -9 4 0 6 57 -51
0 6 57 -51
0 1 48 -47
Now put this row where row 3 was
7
1 2 5 -1 0 -2 -19
17 0 1 48 -47
Switch row 2 and row 3
1 2 5 -1 0 1 48 -47 0 -2 -19
17
8
1 2 5 -1 0 1 48
-47 0 -2 -19 17
Multiply row 2 by 2 and add to row 3
3 4 4 14(-1) 2 -1 1 2
0 -2 -19 17 0 2 96 -94
0 2 96 -94
0 0 77 -77
Now put this row where row 3 was
9
1 2 5 -1 0 1 48
-47 0 0 77 -77
Divide row 3 by 77
0 0 77 -77
0 0 1 -1
10
1 2 5 -1 0 1 48
-47 0 0 1 -1
So z -1
Now rewrite row 2 as an equation
0 1 48 -47
0x 1y 48z -47
Substitute -1 in for the z
y 48(-1) -47
Now solve for y
y - 48 -47
y 1
11
1 2 5 -1 0 1 48
-47 0 0 1 -1
Now rewrite row 1 as an equation
Substitute in the values for y and z
1x 2y 5z -1

1x 2(1) 5(-1) -1
x 2 - 5 -1
x - 3 -1
x 2
12
x 2y 1z -1(2, 1, -1)