Solve Systems of Equations using Gauss-Jordan Elimination

Gauss-Jordan Refinement (or extension)...

Example:
2 x + 4 y + 6 z = 4
1 x + 5 y + 9 z = 2
2 x + 1 y + 3 z = 7

Solution:
in augmented matrix form as:

2	4	6	4
1	5	9	2
2	1	3	7

make a₁₁ = 1
                                      <--- (row 1) / 2

2	4	6	4
1	5	9	2
2	1	3	7

make a₂₁ = 0 & a₃₁ = 0

1	2	3	2
1	5	9	2
2	1	3	7

<--- (row 2) - 1 (row 1)
<--- (row 3) - 2 (row 1)


make a₂₂ = 0

1	2	3	2
0	3	6	0
0	-3	-3	3

<--- (row 2) / 3


make a₃₂ = 0

1	2	3	2
0	1	2	0
0	-3	-3	3

 <--- (row 3) + 3 (row 2)


make a₃₃ = 1

1	2	3	2
0	1	2	0
0	0	3	3

<--- (row 3) / 3



now in Gauss form

1	2	3	2
0	1	2	0
0	0	1	1

After Gauss Elimination we obtained...

1 x + 2 y + 3 z = 2
0 x + 1 y + 2 z = 0
0 x + 0 y + 1 z = 1

eliminate "z", make a₁₃ = 0 and a₂₃ = 0

1 x + 2 y + 3 z = 2  <--- (row 1) - 3 (row 3)
0 x + 1 y + 2 z = 0  <--- (row 2) - 2 (row 3)
0 x + 0 y + 1 z = 1



make a₁₂ = 0 

1 x + 2 y + 0 z = -1  <--- (row 1) - 2 (row 2)
0 x + 1 y + 0 z = -2
0 x + 0 y + 1 z = 1


Now in Gauss-Jordan Form

1 x + 0 y + 0 z = -3
0 x + 1 y + 0 z = -2
0 x + 0 y + 1 z = 1

--> 1 x =  3 --> x = 3
--> 1 y = -2 --> y = -2
--> 1 z =  1 -->  z = 1
the solution is:
x = 3 , y = -2 , z = 1

Gauss Elimination Tibdets:

• Q? Can we use Gauss on a system like:

1 x + 3 z = 4
3 x + 8 y + 1 z = 10
4 x  - 2 y = 5
Answer:
==> Yes!
==> Rearange / Standardize: ==>

1 x + 0 y + 3 z = 4
3 x + 8 y + 1 z = 10
4 x  - 2 y + 0 z = 5

• Q? Can we use Gauss on a system like:

2 x - 3 x + 18 y - 2 = 0
-3 x + 4 y - 26 + 2 y = 0
4 x - 2 y = 5

Answer:
==> Yes!
==> Simplify / Rearange: ==>

-1 x + 18 y = 2
-3 x + 6 y = 26