## anonymous one year ago Find the general solutions of 1 0 -5 0 -8 3 0 1 4 -1 0 6 0 0 0 0 1 0 0 0 0 0 0 0 Last question I'm going to ask, I promise.

1. anonymous

Is there even a solution? The third row doesn't make a lot of sense 0 + 0 + 0 + 0 + 1 = 0?

2. ganeshie8

is that the augmented matrix in row echelon form ?

3. anonymous

I'm not sure. The problem doesn't say.

4. anonymous

Wait! okay, it is the augmented matrices.

5. ganeshie8

good, how many pivots are there ?

6. anonymous

I don't know. How do I tell if they are pivots?

7. ganeshie8

in echelon form, pivot is the first "non zero" number in a row

8. anonymous

So there would be three pivots then.

9. ganeshie8

Right, 1 0 -5 0 -8 3 0 1 4 -1 0 6 0 0 0 0 1 0 0 0 0 0 0 0 you have 3 pivot columns and 2 free columns lets find a particular solution by letting free variables = 0

10. anonymous

Okay.

11. ganeshie8

just for definiteness, call the variables : x1, x2, x3, x4, x5 so to find a particular solution you let the free variables x3 = x4 = 0 and solve for the pivot variables x1, x2, x5

12. ganeshie8

x1 x2 x3 x4 x5 b 1 0 -5 0 -8 3 0 1 4 -1 0 6 0 0 0 0 1 0 0 0 0 0 0 0 ^ ^ free columns

13. ganeshie8

notice that the corresponding system is  x5 = 0 x2 + 4x3 - x4 = 6 x1 - 5x3 - 8x5 = 3  let x3 = x4 = 0 and solve remaining variables

14. anonymous

Okay, so how do I solve for the pivot points?

15. ganeshie8

plugin x3=x4=0 in above system

16. ganeshie8

then you will have 3 equations and 3 unknowns in triangular form which is a piece of cake to solve by back substitution

17. anonymous

so.. x2 = 5 x5 = 0 x1=0

18. ganeshie8

try again

19. ganeshie8

the corresponding system is  x5 = 0 x2 + 4x3 - x4 = 6 x1 - 5x3 - 8x5 = 3  letting x3 = x4 = 0, the system becomes  x5 = 0 x2 = 6 x1 - 8x5 = 3  yes ?

20. anonymous

Yes.

21. ganeshie8

solve it

22. ganeshie8

its trivial, but you have to do it

23. anonymous

I'm trying to do it right now, give me a moment.

24. anonymous

x5 = 0 x2 + 4(0) - 0 = 6 -> x2 = 6 x1 - 5(0) - 8(0) = 3 -> x1 = 3

25. ganeshie8

so $$\begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix}$$ is a particular solution we still need to find the null solution, familiar with the process ?

26. anonymous

Nope

27. ganeshie8

null solution is the solution to following system :  x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 

28. ganeshie8

Notice that I have put 0s on the right hand side now

29. ganeshie8

again we mess with the free variables to cookup the null solution : since we have two free variables, we expect to get two independent null solutions

30. anonymous

Do I use the values for the system that we came up with earlier?

31. ganeshie8

let x3 = 0, x4 = 1 and find one null solution let x3 = 1, x4 = 0 and find the other null solution

32. ganeshie8

null solution is the solution to following system :  x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0  plugin x3 = 0, x4 = 1 and solve the other variables

33. anonymous

x2 - 1 = 0 x2 = 1

34. ganeshie8

yes, keep going find the remaining two variables

35. anonymous

x1 = 0

36. anonymous

Unless you wanted me to use "let x3 = 1, x4 = 0 and find the other null solution" to solve x1 - 5x3 - 8x5 = 0

37. anonymous

Because then x1 would equal 5

38. ganeshie8

one thing at a time

39. ganeshie8

null solution is the solution to following system :  x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0  plugin x3 = 0, x4 = 1 and solve the other variables this gives a null solution $$\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix}$$, yes ?

40. anonymous

Yeah.

41. ganeshie8

save it, find the other null solution also similarly

42. ganeshie8

null solution is the solution to following system :  x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0  plugin x3 = 1, x4 = 0 and solve the other variables

43. anonymous

Okay, so then x1 = 1 and x2 = -1

44. ganeshie8

try again

45. anonymous

Thanks for your help ganeshie8, but I have to go to sleep, it's really late here and I have to wake up for class. Have a wonderful evening, and I'm sorry for not being about to finish the problem.

46. ganeshie8

we're mostly done just wait one minute

47. ganeshie8

null solution is the solution to following system :  x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0  plugin x3 = 1, x4 = 0 and solve the other variables that gives the null solution $$\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}$$

48. ganeshie8

so the nullspace is given by $\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}$

49. ganeshie8

then the final general solutions is given by $$x_p + x_{null}$$ : $x_{\text{general }} = \begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix} +\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}$

50. anonymous

Closing now as it is no longer helpful.