## UsukiDoll 3 years ago solve the system of linear equations using the Gauss–Jordan elimination method. 2x-3y=8 4x+y=-2

1. UsukiDoll

|dw:1361067093589:dw|

2. UsukiDoll

|dw:1361067118822:dw|

3. UsukiDoll

ugh I missed that one on my paper! that subtraction because -16-(-2) is - 14

4. UsukiDoll

anyway|dw:1361067211535:dw|

5. UsukiDoll

|dw:1361067247174:dw|

6. UsukiDoll

|dw:1361067302376:dw|

7. Callisto

|dw:1361067336310:dw|

8. UsukiDoll

|dw:1361067340708:dw|

9. Callisto

|dw:1361067371987:dw|

10. UsukiDoll

I'm using elementary row operations.

11. UsukiDoll

oh crud

12. UsukiDoll

spotted anotehr error. not my day today

13. UsukiDoll

|dw:1361067436003:dw|

14. UsukiDoll

arghhh...

15. UsukiDoll

ok really...I put this in reduced row echleon form. got my solution, but geez doesn't even equal back into the equation

16. Callisto

Hmm... Try to do it again...

17. Callisto

Since you've made a mistake in the first operation, it's hard to get the right answer :|

18. UsukiDoll

something is wrong with the y

19. UsukiDoll

ok let's redo this

20. UsukiDoll

|dw:1361067604121:dw|

21. UsukiDoll

augmented matrix [a b]

22. UsukiDoll

|dw:1361067630686:dw| triangles indicate main diagonal

23. UsukiDoll

so I have to get rid of the 4 and the -3.

24. UsukiDoll

2row 1 - row 2 --> row2

25. Callisto

Hmm.. My usual practice is -2R1 + R2 -> R2

26. UnkleRhaukus

do you mean $R_2\to R_2-2R_1$ @UsukiDoll

27. UsukiDoll

yes

28. UsukiDoll

|dw:1361067762579:dw|

29. UsukiDoll

|dw:1361067803078:dw|

30. UsukiDoll

so now... to get rid of the -3... 7row1 -3row2 --->row1

31. Callisto

Why not do this: -1/7 R2 -> R2

32. UsukiDoll

achhhhhhhhhhhh

33. UsukiDoll

I don't wanna deal with fractions

34. Callisto

But the answers are fractions :P

35. UnkleRhaukus

|dw:1361067930957:dw|

36. UsukiDoll

arghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I feel like Rarity Belle now

37. Callisto

|dw:1361067978329:dw|

38. UsukiDoll

oh wait. sign errors everywhere. Is that why I am getting frustrateD?

39. UsukiDoll

wait a sec something is wrong already with row 2

40. UsukiDoll

|dw:1361068119755:dw|

41. UsukiDoll

|dw:1361068152547:dw|

42. Callisto

Relax~~~ |dw:1361068054013:dw|

43. UsukiDoll

|dw:1361068165204:dw|

44. UsukiDoll

argh there could be so many combinations with this thing!

45. Callisto

|dw:1361068179917:dw|

46. UsukiDoll

|dw:1361068217096:dw|

47. UsukiDoll

|dw:1361068228254:dw|

48. Callisto

Just stop for a while, if you don't mind?

49. UsukiDoll

arghhhhhhhhhhhhh doing that got me x = 7

50. Callisto

This is basically what you do for the first row operation -2R1 + R2 -> R2. I split it into two steps so that you can catch up the the arithmetic easier (I hope..) |dw:1361068290374:dw|

51. UsukiDoll

big numbers -_-

52. UsukiDoll

that means to get rid of the 6 I have to do 7r1 - 6r2

53. Callisto

No?!

54. UsukiDoll

or -7r1+6r2

55. UsukiDoll

-.-

56. UsukiDoll

|dw:1361068464895:dw|

57. UsukiDoll

|dw:1361068472375:dw| main diagonal in triangles

58. UsukiDoll

|dw:1361068495592:dw|

59. UsukiDoll

|dw:1361068548208:dw|

60. Callisto

Yup..Do you want to deal with big numbers or fractions?

61. UsukiDoll

|dw:1361068571208:dw|

62. UsukiDoll

I don't want to deal with fractions

63. UsukiDoll

|dw:1361068639980:dw|

64. Callisto

Yes!!!

65. UsukiDoll

|dw:1361068650303:dw|

66. UsukiDoll

|dw:1361068693152:dw|

67. UsukiDoll

ok why the heck am I getting different answers despite the fact that I got it into reduced row echleon form. that's just...grrrrrrrrrrrrrrrrrr

68. UsukiDoll

just earlier I got x = -1 and y = -2

69. Callisto

Hmm.. Careless mistakes perhaps?!

70. UsukiDoll

now x = 1/7 y = -18/7

71. UsukiDoll

what!

72. Callisto

The one x=1/7; y=-18/7 is right :\

73. UsukiDoll

-.- is there more than one answer to this? There could be many many things that I could have done here.

74. Callisto

No.

75. UnkleRhaukus

x = 1/7 y = -18/7 $$\LARGE\checkmark$$

76. Callisto

There are only 3 cases for system of linear equations.. 1. No solution 2. Exactly one solution 3. Infinitely many solution.

77. UsukiDoll

78. UnkleRhaukus

A quick string of row operations $R_2\to R_2-2R_1\\R_2\to R_2/7\\ R_1\to R_1+3R_2\\R_1\to R_1/2$, if you have gotten a different solution you must have a made a mistake somewhere, it is very easy to make mistakes using Gauss–Jordan elimination .

79. Callisto

1. No solution with a row that has all zero entries on the left, but non-zero entry on right => system is inconsistent 2. Exactly one solution After elementary row operations, no. of rows = no. of unknowns (/columns) => system is consistent 3. Infinitely many solution. After elementary row operations, no. of rows < no. of unknowns => system is consistent

80. UsukiDoll

but...yeah I suppose because I have seen some matrices that had a row of zeros on the left but with a nonzero on the right that is an inconsistent system .. How is it possible to make a mistake when it's so hard to spot one? I mean...I could've used different operations on the matrix. The combinations are everywhere.

81. UsukiDoll

Gauss Jordan is worse than Gaussian

82. UsukiDoll

ohhhhh what happens if I did Gaussian and then Gauss Jordon

83. UsukiDoll

because the answers to the Gaussian are supposed to be the same as the Gauss Jordon

84. Callisto

Gaussian Jordan = Gaussian then Jordan lol!!!

85. UsukiDoll

but sometimes when I put the matrix into echleon form I can see the steps to making it row reduced

86. UsukiDoll

|dw:1361069849783:dw|

87. Callisto

Guassian told you to make the left an upper triangular matrix Jordan asked you to further make the left an identity matrix

88. Callisto

*Gaussian

89. UsukiDoll

|dw:1361069869054:dw|

90. UsukiDoll

|dw:1361069916982:dw|

91. UsukiDoll

-____________________- this is just Gaussian... WHAT THEEEEE!!!!!!! I see y = -18/7 oh I haven't solved yet oops

92. UsukiDoll

|dw:1361070002077:dw|

93. Callisto

Just a minute..

94. UsukiDoll

|dw:1361070023727:dw|

95. UsukiDoll

uighhhhhhhhhhhhhhhh mind blown

96. UsukiDoll

Gaussian and then Gaussian Jordan yeahhhhh

97. UsukiDoll

I found the x = 1/7 just doing the Gaussian

98. Callisto

|dw:1361070037568:dw|

99. UsukiDoll

yeahhhh...now I really should've done Gaussian first and then the Jordan version

100. Callisto

Gaussian -> Make the left an upper triangular matrix Jordan -> Make the left an identity matrix Gaussian-Jordan => Make it an upper triangular matrix, then further make it an identity matrix. That's what Gaussan-Jordan is! Jordon helps you to get the answer by matching the left and the right That is for the row [1 0 | 1/7], you can tell immediately that x = 1/7 It's just the same as doing Gaussian elimination then back substitution.

101. UsukiDoll

dang that means I have to redo some practice problems...gawd not again wasting paper here.

102. UsukiDoll

oh yeah now I got it

103. UsukiDoll

yup works. I went way too ahead... x.x

104. UsukiDoll

so now I gotta correct all of these problems and write a proof...nice X___X

105. UsukiDoll

Callisto do you know how to write proofs? I got a partial first draft and I was wondering if you could critique it?

106. Callisto

Are you doing high school maths or college maths?

107. UsukiDoll

college math

108. Callisto

Proofs of??

109. UsukiDoll

like the one I am currently writing is a contradiction proof. contradicts a theorem at least 2-3 times

110. UsukiDoll

umm should I type it?

111. Callisto

I.. didn't know what was a proof :S Please type it, if you don't mind!

112. Callisto

113. UsukiDoll

okkk... Let A and B be n x n matrices. Show that if AB is nonsingular then A and B must be nonsingular.(Hint Use Theorem 2.9) Theorem 2.9 states that the homogeneous system of n linear equations in n unknowns Ax = 0 has a nontrivial solution if and only if A is singular.

114. UsukiDoll

There's a contradiction....namely because if Matrix A is singular, then the inverse of Matrix A does NOT exist which means that AB doesn't exist as well.

115. UsukiDoll

It also has something to do with the homogenous system as well.

116. UsukiDoll

because basically a homogeneous system is always consistent. However, a homogeneous system's solution is always 0 and it's a trivial solution. Nontrivial solution in a homogenous system means that the solution IS NOT 0 at all!

117. UsukiDoll

so, what is given before proving is that A and B are n x n matrices A and B must be nonsingular AB is also nonsingular Theorem 2.9

118. UsukiDoll

If a matrix is nonsingular, an inverse exists...so Theorem 2.9 doesn't work at all

119. Callisto

Forgive me, I don't understand the part "... which means that AB doesn't exist as well" I understand if A is singular, A^(-1) doesn't exist, but not the which means part..

120. UsukiDoll

oh yeah this is just a rough draft of it

121. UsukiDoll

hmm if A is nonsingular, an inverse exists. I remember back from Theorem 1.5 that it was AB=BA=In. B is the inverse of A. hmmm..if A is singular, yes the inverse doesn't exist. Let's have B = the inverse of A. Matrix A is singular Inverse doesn't exist so B doesn't exist AB is impossible to achieve unless A is nonsingular

122. Callisto

Hmm.. $AA^{-1} = A^{-1}A=I$provided that $$A^{-1}$$ exists... ------------- But what about $$B \ne A^{-1}$$? And we also have to show that B is also non-singular?!

123. UsukiDoll

I know. What if we let A be the inverse of B?

124. UsukiDoll

that would be Matrix B A = B^-1 BB^-1 = identity matrix

125. Callisto

What if $$A \ne B^{-1}$$ and $$B \ne A^{-1}$$ ?

126. UsukiDoll

I think that the only way that occurs is if A and B are singular

127. Callisto

Hmm.. I'm sorry that I don't know how to do it :( @hartnn Would you mind giving a hand here? The problem we are discussing now is: Let A and B be n x n matrices. Show that if AB is nonsingular then A and B must be nonsingular.(Hint Use Theorem 2.9) Theorem 2.9 states that the homogeneous system of n linear equations in n unknowns Ax = 0 has a nontrivial solution if and only if A is singular.

128. hartnn

is it necessary to use that theorem ? can't we go like this way ? if AB is non-sing., $$|AB| \ne 0 \implies |A||B|\ne 0$$ for that both |A|and |B| must not =0, so...

129. hartnn

if need to use contradiction, then let any1 of the A or B be singular, say A , then |A| =0, then |A||B|=0 , then |AB|=0 which is a contardiction....hence, neither A nor B can be 0

130. hartnn