## anonymous one year ago how do i find a solution to a system of equations represented by a matrix?

1. anonymous

this specifically

2. anonymous

Well first you need to know how to read a matrix. Do you know how to read a matrix?

3. anonymous

no....

4. anonymous

im kind of clueless

5. anonymous

6. anonymous

|dw:1438709526900:dw|

7. anonymous

Okay I'm just writing the problem here so I can see it more easier. And it's okay. I'll write how to solve the problem step by step. Just tell me when you are lost.

8. anonymous

aw, thank you

9. anonymous

You're welcome! :)

10. anonymous

itll help alot if i understand this

11. anonymous

Alright I'll try the best I can in getting you to understand this. |dw:1438709667710:dw|

12. anonymous

First you need to know how to read the matrix. As you can see there is a column name called x y and z. Those numbers below the variables are going to be part of the equation.

13. anonymous

For example in the first row you see the numbers 1 2 and -2. When you put this into an equation you will use the variables above the first row to combine it as shown below:$x+2y-2z$The 11 will be after the equals sign since that line separating -2 and 11 technically means equals. So now the equation will look like this:$x+2y-2z=11$

14. anonymous

Did you get that so far?

15. anonymous

uuhm..

16. anonymous

oh, yeah okay i get it.

17. anonymous

since one would be just x, sorry takes me awhile to work things out.

18. anonymous

It's okay! Take your time! :)

19. anonymous

Do you know what the other two equations will be if you used my method that I showed you?

20. anonymous

5x+4y+7z=81?

21. anonymous

Yes!

22. anonymous

-4x+y-6z=-37

23. anonymous

Great job! Now those three equations are needed when you are trying to solve equations within a matrix.

24. anonymous

$x+2y-2z=11$$5x+4y+7z=81$$-4x+y-6z=-37$

25. anonymous

Now do you know how to solve for x's and y's?

26. anonymous

replace them with the numbers or..

27. anonymous

Not in this case since they didn't give us any numbers to put in for x y or z. So the first step is to pick any two sets of equations out of the three.

28. anonymous

29. anonymous

sorry, ah :(

30. anonymous

|dw:1438710552563:dw|

31. anonymous

And it's okay. That was a rhetorical question.

32. anonymous

Oh wait let me rewrite what I wrote on the side.

33. anonymous

Pick which variable to make common. For example the y's can be made common by multiplying the 2y by 2 and leaving the 4y. When you multiply 2y by 2 you can get 4y. 4y and 4y equal each other so now you can subtract the whole equation to get rid of y.

34. anonymous

Tell me if you get that part.

35. anonymous

hm..

36. anonymous

i do, but at the same time i guess i need to know why we are doing that.

37. anonymous

what is the reason for finding commons, is that like a big part of the equation?

38. anonymous

what is the end result supposed to be? like, is the Y now 1? bleh...

39. anonymous

We are doing that so that we can get rid of one term and just concentrate on finding x and z. We will find y at the end.

40. anonymous

why are we doing that?

41. anonymous

In order to find the solution to this matrix we need to find what x y and z equals. So we need to break it down by finding one variable at a time.

42. anonymous

Oh alright. so whenever im doing this i have to find a way to get rid of a variable?

43. anonymous

Yes.

44. anonymous

okay, so, like in this equation for example: |dw:1438711521979:dw|

45. anonymous

just as an example

46. anonymous

they have to be in the same row to be able to get rid of them right? the 9 and -9 cant cancel each other out

47. anonymous

I'm sorry but where did 31 -25 and 28 come from? It's the last column. And no you can't cancel out the 9 and -9 because one is a y and the other is a z. Remember the 2nd column are y's and the 3rd column are z's.

48. anonymous

|dw:1438711724574:dw|

49. anonymous

its an example problem of one i did and failed.

50. anonymous

totally different problem sorry just trying to see where this applies to other problems too

51. anonymous

You understood more than before at least :) And oh ok. I'll help you out with those if you don't understand still.

52. anonymous

thank you, youre a saint ;-;

53. anonymous

okay, so lets finish this problem first

54. anonymous

Lol you're welcome! And okay. We have one equation: -3x-11z=-59 after canceling the y. Now when you look at the 3 equations from earlier, you can see that we need an equation with just an x and z and no y but we don't have that. So we need to repeat the step in finding another 2 equations to cancel out y.

55. anonymous

did you subtract?

56. anonymous

im lost again

57. anonymous

-3x-11z=-59 came from the drawing I drew earlier. Take a look at that and tell me if you get that.

58. anonymous

where did the other two come from?

59. anonymous

It came from the three equations we wrote from the matrix.

60. anonymous

|dw:1438712602958:dw| gosh im so sorry, im not sure where the two came from. i wrote down the equation again, to see if i can do what youre doing

61. anonymous

Oh lol okay.

62. anonymous

https://join.me/592-430-542 im sharing my screen with you, is that alright??

63. anonymous

I'm pretty freaked out by that screen. Lol sorry but I don't feel like it's safe. No offense to you or anything. I just feel like it's a hack since that kind of screen popped up when my uncle's computer got hacked.

64. anonymous

pfft, im not trying to hack anyone and learn math at the same time but thats understandable. i usually use joinme with my friends when im drawin and stuff

65. anonymous

Oh okay. I'm really sorry about that! Are you able to draw here?

66. anonymous

We're almost done anyways.

67. anonymous

|dw:1438713360885:dw|

68. anonymous

alright, sorry i scared you haha

69. anonymous

the brushes on here kind of suck there for i kind of suck

70. anonymous

It's okay you didn't know! :) And at least you draw better than me XD

71. anonymous

$x+2y-2z=11$$5x+4y+7z=81$$-4x+y-6z=-37$*Just rewriting this so I don't have to keep scrolling up.*

72. anonymous

i changed the background so its less menacing lol

73. anonymous

but anyways, MATH.

74. anonymous

Lol okay. Now let's use the 1st and 3rd equations so we can get rid of y again.

75. anonymous

alright

76. anonymous

|dw:1438713734611:dw|

77. anonymous

|dw:1438713838855:dw|

78. anonymous

|dw:1438713940961:dw|

79. anonymous

Did you get that?

80. anonymous

noooooooooooooooooo

81. anonymous

What part did you get lost on?

82. anonymous

|dw:1438714161470:dw| i dont understand any of it, how do we get to the answers that it wants, why am i so BAD at UNDERSTANDING THINGS

83. anonymous

Oh we didn't get to finding the answers yet. We needed to find two equations to find x and z. We did those last two steps to get rid of y.

84. anonymous

why did we multiply that row by 2 though?

85. anonymous

To get rid of the y. You need to multiply the equation by a certain number to cancel out the y. In that case the y in the 1st equation is 2 while the 3rd equation's y is 1. So if you were to just look at$2y$&$y$How would you be able to get that y to equal the 2y?

86. anonymous

multiply it by two but...

87. anonymous

what happened to the row in the middle

88. anonymous

89. anonymous

Oh it was cancelled because we need to cancel out variables to find x or z. In order to find what x y and z equals we need to have two variables in two equations in order to solve it. That's why we got rid of y.

90. anonymous

Did you get that?

91. anonymous

so...that whole row was canceled out?

92. anonymous

No just y was cancelled out. We subtracted the two equations to get rid of y and have only x and z left in a new equation.

93. anonymous

i dont get iiiit

94. anonymous

Which part?

95. anonymous

im a kinesthetic learner, i figure things out differently im sorry this must be frustrating.

96. anonymous

everything

97. anonymous

|dw:1438715563780:dw|

98. anonymous

And it's okay.

99. anonymous

Did you understand this picture?

100. anonymous

no

101. anonymous

Do you know where I got the 1st and 3rd equations from?

102. anonymous

lets solve this whole thing and then break it down because i dont know where we are headed

103. anonymous

no

104. anonymous

Oh okay then hold on.

105. anonymous

its really cool of you to help me

106. anonymous

thank you

107. anonymous

$x+2y-2z=11$$5x+4y+7z=81$$-4x+y-6z=-37$----------------------- 1st step:$x+2y-2z=11$$5x+4y+7z=81$----------------------- $2(x+2y-2z=11)$$5x+4y+7z=81$----------------------- $2x+4y-4z=22$$5x+4y+7z=81$----------------------- $(2x+4y-4z=22)$minus $(5x+4y+7z=81)$equals$(-3x+0y-11z=-59)$OR$-3x-11z=-59$(1st equation with no y's) ----------------------- 2nd step: (Pick another two sets of equations to cancel out y.)$x+2y-2z=11$$-4x+y-6z=-37$-----------------------$x+2y-2z=11$$2(-4x+y-6z=-37)$-----------------------$x+2y-2z=11$$-8x+2y-12z=-74$-----------------------$x+2y-2z=11$minus $-8x+2y-12z=-74$equals$9x+0y+10z=85$OR$9x+10z=85$(2nd equation with no y's) -----------------------

108. anonymous

And you're welcome! :) This doesn't give you the answers yet but I'm just going to show you this part to see if you understand it.

109. anonymous

oooooh

110. anonymous

You understand it now?

111. anonymous

yes

112. anonymous

Okay great! I should have just showed you the steps earlier like that lol. XD

113. anonymous

yeah omg

114. anonymous

1st equation with no y's:$-3x-11z=-59$2nd equation with no y's:$9x+10z=85$Now we need to get rid of x's in this step since that's the easiest to multiply with. We need to get -3x equal to 9x. We have to multiply the 1st equation with no y's by 3.$3(-3x-11z=-59)$$9x+10z=85$-----------------------$-9x-33z=-177$$9x+10z=85$-----------------------$-9x-33z=-177$plus $9x+10z=85$equals$0x-23z=-92$OR$-23z=-92$----------------------- Now that we have one variable left, we can finally find what z equals. Divide -92 by -23.$z=\frac{ -92 }{ -23 }$$z=4$----------------------- After finding one variable, you can now substitute the z in the equation with no y's. You can do either the 1st or 2nd equation with no y's. It doesn't matter.$9x+10z=85$$9x+10(4)=85$$9x+40=85$$9x=85-40$$9x=45$$x=5$----------------------- Lastly we can solve for y by using the equations givin to us in the beginning of the question (from the matrices). Either 3 of the equations can work. Make sure to substitute the x and z.$x+2y-2z=11$$(5)+2y-2(4)=11$$5+2y-8=11$$2y-3=11$$2y=11+3$$2y=14$$y=7$

115. anonymous

oh wow

116. anonymous

Tell me if you understand that. This is the final step and it gives the answer.

117. anonymous

that is a process

118. anonymous

Ya it might look really long but don't let that scare you. Once you keep practicing it, it will be really easy! :)

119. anonymous

omg i actually understand

120. anonymous

Yay that's great! :D

121. anonymous

see, im taking a summer course and its online, the section imtaking is all about that so im stuck

122. anonymous

Oh okay. Well hopefully this was helpful in figuring out how to solve it. :)

123. anonymous

124. anonymous

125. anonymous

i might need help going through another problem because i know there are exceptions

126. anonymous

|dw:1438718495088:dw|

127. anonymous

Okay. If it's another problem can you write it in the Ask a question box? That way I can scroll through the steps that belong to that problem?

128. anonymous

oh sure

129. anonymous

Okay thanks!

130. anonymous

so do i get rid of the x?

131. anonymous

because 4 can go into the 8 or no

132. anonymous

|dw:1438718635151:dw|

133. anonymous

Ya you can get rid of the x or y since you can just multiply the 3rd equation with 4 so it can be the same x and y in the 2nd equation.

134. anonymous

|dw:1438718776581:dw|

135. anonymous

|dw:1438718875081:dw|