# Thread: Find the inhomogeneous solution for set of equations, check please :)

1. ## Find the inhomogeneous solution for set of equations, check please :)

2x1 + 4x2 -4x3 + 6x4 = 4
2x1 + 4x2 -3x3 + 4x4 = 5
5x1 + 10x2 -8x3 + 11x4 = 6

2 4 -4 6 ¦ 4
2 4 -3 4 ¦ 5
5 10 -8 11 ¦ 6

1 2 -2 3 ¦ 2
2 4 -3 4 ¦ 5
5 10 -8 11 ¦ 6

1 2 -2 3 ¦ 2
0 0 1 -2 ¦ 1
0 0 2 -4 ¦ -4

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

So it doesnt work, is this right?

$\displaystyle 2x1 + 4x2 -4x3 + 6x4 = 4 2x1 + 4x2 -3x3 + 4x4 = 5 5x1 + 10x2 -8x3 + 11x4 = 6$

2 4 -4 6 ¦ 4
2 4 -3 4 ¦ 5
5 10 -8 11 ¦ 6

1 2 -2 3 ¦ 2
2 4 -3 4 ¦ 5
5 10 -8 11 ¦ 6

1 2 -2 3 ¦ 2
0 0 1 -2 ¦ 1
0 0 2 -4 ¦ -4

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

So it doesnt work, is this right?

I'm not sure what you mean by "doesn't work", but what you proved is that the linear system is incongruent, i.e. it has no solutions.

Tonio

3. Originally Posted by tonio
I'm not sure what you mean by "doesn't work", but what you proved is that the linear system is incongruent, i.e. it has no solutions.

Tonio
Yeah no solutions is what i meant, have i done the workings out right as well?

4. My friend is getting an answer for this with many solutions. Can someone check if i've done it right please. Thank you.

He didn't make the top row lead with a one, can you do this?

5. Your friend is wrong; you're right. There are no solutions.

You don't have to make the top row lead with a one. Computer solutions, in fact, rarely do that because for a computer, it doesn't make the solution come any quicker. The ERO of multiplying an entire row by a constant is not done in computers.

6. Originally Posted by Ackbeet
Your friend is wrong; you're right. There are no solutions.

You don't have to make the top row lead with a one. Computer solutions, in fact, rarely do that because for a computer, it doesn't make the solution come any quicker. The ERO of multiplying an entire row by a constant is not done in computers.
Yeah we went threw his workings out and he missed out a minus half way through his reduction which changed his results. Cheers.

7. Cool. All set on this one, then?

8. Originally Posted by Ackbeet
Cool. All set on this one, then?
Yep, ta.

9. All right. Have a good one!