## Imtiaz7 one year ago Help Please...!!!

1. Imtiaz7

Convert following matrix into Reduced Echlon Form:

2. Imtiaz7

|dw:1438863140426:dw|

3. UsukiDoll

just reduce echelon form right?

4. UsukiDoll

so what we need is an upper triangular matrix which means that below the main diagonal we need all 0's we can achieve this by using row operations by swapping rows, multiplying a row by a scalar number, or add/subtract rows and replace the old row with the new result after that computation

5. UsukiDoll

keep in mind that we need a 1 always at the top left the 1 also has to go towards the right and the bottom... we also need a row of zeros

6. Imtiaz7

wait I've solve this, can you please tell me is this right?

7. UsukiDoll

|dw:1438863440670:dw| so the places I've circled is the main diagonal so we need to get rid of the 3 , -1, and 4 using row operations... yeah sure.

8. UsukiDoll

anyway... -3r1 + r2 -> r2 r1+ r3 - > r3

9. UsukiDoll

|dw:1438863691395:dw||dw:1438863716259:dw|

10. UsukiDoll

|dw:1438863747952:dw|

11. UsukiDoll

hmmm gonna make life a bit easier by 1/2r3->r3

12. UsukiDoll

|dw:1438863802231:dw|

13. UsukiDoll

-r2+11r3 ->r3

14. UsukiDoll

|dw:1438863836507:dw|

15. UsukiDoll

1/10 r3 - >r3 1/11 r2 - > r2

16. UsukiDoll

|dw:1438863882447:dw|

17. UsukiDoll

do you have a matrix that looks like this? ^

18. Imtiaz7

19. UsukiDoll

how did you get 20/11 ?

20. Imtiaz7

by multiplying Row 2 by -2 and add it in row 3(-2*1/11 + 2)

21. UsukiDoll

hmmm dealing with fractions that early is killer though

22. UsukiDoll

because what I've done to the third row was divide by 1/2 all over row 3 so I would have just 0 1 1

23. UsukiDoll

so when I had 0 11 1 0 1 1 I distributed the negative sign on row 2 and then multiply row 3 with 11 -r2+11r^3 -> r^3

24. Imtiaz7

Uhm so i i did wrong?

25. UsukiDoll

0 -11 -1 0 11 11 adding them together would make it 0 0 10 and then 1/10r3 ->r3 to get rid of that 10 to have 0 0 1

26. Imtiaz7

but in last it becomes in reduced echlon form; isnt?

27. UsukiDoll

and then to have the 1 in the middle I had to divide the whole second row by 1/11

28. UsukiDoll

somehow our first rows are different but the second and third is the same..

29. Imtiaz7

30. UsukiDoll

it should be as long as it followed the reduce echelon rules.. and it does. we do have an upper triangular matrix and our 1 is going down and shifted to the right

31. Imtiaz7

Yes! So our answers are different, but rules are same!

32. UsukiDoll

hmm I'm assuming that an extra step was taken to get rid of the -2 on the first row. unless we are required to get RREF form, that's not required. What's required is the numbers until the main diagonal go to 0.

33. UsukiDoll

check this one out... it also did something different but ended up with the same thing before taking RREF form oh that 1 in the third row got blocked out, but noticed how this calculator did something I disagree with, but still ends up following the rules?

34. Imtiaz7

wait let me check it

35. UsukiDoll

that was done by a calculator bot... I really hated that 0 0 54/11 but by multiplying 11/54r3 -> r3 0 0 1

36. Imtiaz7

@UsukiDoll the question is different there, that is there is 1 instead of 1 in first row

37. Imtiaz7

-1*

38. UsukiDoll

ugh a typo.. yeah this is way late at night... on my end.

39. UsukiDoll

one sec.

40. UsukiDoll

41. UsukiDoll

still has that big fraction 20/11 on row 3 but 11/20r3 - >r3 makes it 0 0 1

42. Imtiaz7

43. UsukiDoll

mhm

44. Imtiaz7

:) lol Thank you so much @UsukiDoll to help me..

45. UsukiDoll

XD you're welcome... these questions always come up very late at night, so typos are bound to happen... typed too fast XD

46. Imtiaz7

Its okay, when i post i dont run xD

47. UsukiDoll

yay I got it right too XD |dw:1438865533703:dw| row operation #5 on the correct image has the same matrix I had... only in fewer steps.

48. UsukiDoll

but as long as the rules aren't broken for row echelon form, then everything will be fine ^_^

49. Imtiaz7

Yup @UsukiDoll rules are for obey uhm i think so lol

50. Imtiaz7

@UsukiDoll ?

51. UsukiDoll

hmmm?

52. Imtiaz7

There is still a confusion uhm the answer you showed me are correct one but in my answer in fist row there is -9/11, still is it right?

53. UsukiDoll
54. UsukiDoll

as long as you have an upper triangular matrix which means all 0's below the main diagonal and have a 1 on the top left that goes to the bottom and right... I don't see a problem.

55. Imtiaz7

Okay :)

56. Imtiaz7

Once again thanks a lot :)

57. UsukiDoll

we could RREF that last matrix with that -9/11 if you want

58. UsukiDoll

to show that we can get an identity matrix out of some more row operations..

59. UsukiDoll

|dw:1438866425233:dw| this was the matrix you had in REF form right?

60. Imtiaz7

Yes

61. UsukiDoll

I would use row 3 to get rid of the remaining numbers above the main diagonal, but I hate dealing with fractions so 11r2 - >r2 11r1 -> r1

62. UsukiDoll

|dw:1438866522476:dw|

63. UsukiDoll

-r3+r2 - > r2 -9r3 + r1 - > r1

64. UsukiDoll

|dw:1438866568376:dw||dw:1438866589363:dw|

65. Imtiaz7

@UsukiDoll can i give you another medal bc your math is too good

66. UsukiDoll

1/11r2 - >r2 1/11 r1 -> r1... I wish we can give more than 1 medal XD

67. UsukiDoll

|dw:1438866634762:dw| a-ha identity matrix! so we got RREF XD

68. Imtiaz7

lol

69. UsukiDoll

which means your row operations from earlier were correct... otherwise we wouldn't get RREF

70. Imtiaz7

yes...