## anonymous one year ago How would I go abouts solving a 3x4 matrix system?

1. anonymous

$\left[\begin{matrix}5 & 3 & -9 & 2 \\ -4 & -4 & -8 & 5 \\ 21 & 11 & -53 & 15\end{matrix}\right]$

2. anonymous

Specifically one such as above, except with a line down the middle before the last column?

3. ganeshie8

$\left[\begin{matrix}5 & 3 & -9 &|& 2 \\ -4 & -4 & -8 &|& 5 \\ 21 & 11 & -53 &|& 15\end{matrix}\right]$ like this ?

4. anonymous

Yes, how did you manage to do that on here by the way?

5. ganeshie8

you may right click and see latex commands : |dw:1440239723612:dw|

6. anonymous

Oh, okay then.

7. ganeshie8

btw, we call it a 3x3 system the size of matrix $$A$$ is $$3\times 3$$ in the system of equations : $Ax=b$

8. ganeshie8

you're just attaching that "b" also as the last column to get that 3x4 matrix that 3x4 matrix with "b" attached is called "augmented matrix"

9. ganeshie8

familiar with elimination/row opearations ?

10. anonymous

Familiar but not well enough to solve the problem at hand unfortunately.

11. ganeshie8

first see that the solutions are not affected by row operations because you're just adding/subtracting a multiple of one row from the other... this doesn't change the solutions

12. ganeshie8

if you have x + y = 2 x+2y = 3 then, adding one equation to other won't change the solutions because you're adding the same thing to both sides of a particular equation

13. ganeshie8

first convince yourself that below two systems will have same solutions : system1  x + y = 2 x+2y = 3  system2  x + y = 2 2x+3y = 5 

14. anonymous

Okay..

15. ganeshie8

your goal in elimination is to get a triangular matrix

16. anonymous

So, not 4x + 6y = 10?

17. ganeshie8

that is a valid row operation too, you're allowed to multiply entire row by a constant

18. ganeshie8

|dw:1440240774738:dw|

19. ganeshie8

you want to change the starting matrix to above form

20. ganeshie8

start by working the element at 2,1 $\left[\begin{matrix}5 & 3 & -9 &|& 2 \\ -4 & -4 & -8 &|& 5 \\ 21 & 11 & -53 &|& 15\end{matrix}\right]$ adding 4/5 times first row to second row does the job, right ?

21. anonymous

Yes, at least I think so anyway

22. ganeshie8

do it

23. anonymous

So now the second and third row match?

24. ganeshie8

no, what do you get after doing that first step

25. anonymous

$\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]$

26. ganeshie8

how ?

27. anonymous

5 times the 1st row added into the 2nd row?

28. ganeshie8

yeah you could do that!

29. anonymous

Was I not supposed to do that?

30. ganeshie8

that is not a standard method, but what you did is a better one..

31. ganeshie8

so lets keep going

32. ganeshie8

for second step, maybe subtract second row from third row

33. ganeshie8

$\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]$ R3-R2 gives : $\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 0& 0& 0& |0\end{matrix}\right]$

34. anonymous

Okay then, what's the next step?

35. ganeshie8

next look at the element at 2,1 position you want to make it 0 so that you get a triangular matrix

36. anonymous

So 21 needs to be 0?

37. ganeshie8

yes how to do that using row operations ?

38. anonymous

Multiply the 2nd row by 5 and then add 21x1st row to the second row?

39. ganeshie8

Multiply the 2nd row by 5 and then "subtract" 21x1st row to the second row?

40. ganeshie8

you mean that right ?

41. anonymous

Yes, sorry I was thinking in terms of -21

42. ganeshie8

that works perfectly! do it

43. anonymous

$\left[\begin{matrix}5 & 3 & -9 & |2 \\ 0 & -8 & -76 & |33 \\ 0 & 0 & 0 & |0\end{matrix}\right]$

44. ganeshie8

Looks good! so above reduced system is equivalent to :  5x + 3y - 9z = 2 - 8y - 76z = 33 

45. ganeshie8

let $$z=0$$ and find a particular solution

46. anonymous

x-y=7?

47. ganeshie8

how ? plugin $$z=0$$ in the bottom equation first

48. anonymous

Oh, I just plugged 0 in for z for both equations, my bad.

49. ganeshie8

the advantage of triangular matrix is just that solving is trivial from bottom

50. anonymous

So y = -33/8 for the bottom solution?

51. ganeshie8

yes, find x also

52. ganeshie8

plugin z=0, y=-33/8 in first equation

53. anonymous

x = 23/8

54. ganeshie8

right, so our particular solution is (23/8, -33/8, 0)

55. ganeshie8

save that, we need to find nullspace

56. ganeshie8

for nullspace, you want to solve :  5x + 3y - 9z = 0 - 8y - 76z = 0 

57. ganeshie8

plugin $$z=1$$ and solve y, x values

58. anonymous

x = 15/2 & y = -19/2?

59. ganeshie8

Yes, so the nullspace is t(15/2, -19/2, 1)

60. ganeshie8

the complete solution is given by particular + nullspace

61. anonymous

So the one with z being 0 + what we just got?

62. ganeshie8

Complete solution :  (23/8, -33/8, 0) + t(15/2, -19/2, 1) 

63. anonymous

Oh I see, that makes sense!

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