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sloppycanada

  • one year ago

Solve x + y = 5 using matrices. 2x – y = 1 I have the answer, x = 2 and y =3, how do I phrase this as a matrices? (2,3)?

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  1. rishavraj
    • one year ago
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    u know Cramer's Rule??

  2. sloppycanada
    • one year ago
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    I don't think that was covered.

  3. rishavraj
    • one year ago
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    Cramer's Rule. If the equations are \[a_1 x + b_1 y = c_1\] \[a_2 x + b_2 y = c_2\] then the solutions are are \[x = \frac{ D_1 }{ D } ~~~~ and ~~~~~ y = \frac{ D_2 }{ D }\]

  4. rishavraj
    • one year ago
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    where |dw:1435751082539:dw|

  5. sloppycanada
    • one year ago
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    Okay... so I'm phrasing mine wrong?

  6. rishavraj
    • one year ago
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    hmmm wht do u mean?? u tried it using matrices??

  7. sloppycanada
    • one year ago
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    I think so, I'm thinking it's [2 3]

  8. rishavraj
    • one year ago
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    u hav been askd to solve using matrices ...thts why u need to solve it tht way

  9. sloppycanada
    • one year ago
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    but I don't understand how I'm supposed to do that. Just plug in the equations to what you gave me?

  10. rishavraj
    • one year ago
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    yup....bt use with appropriate signs.....the answers will remain the same....kinda obvious :P

  11. sloppycanada
    • one year ago
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    |dw:1435751899086:dw|

  12. rishavraj
    • one year ago
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    like the equations are x + y = 5 and 2x - y = 1 \[a_1 = 1~~~~b_1 = 1~~~~~c_1 = 5 \]

  13. rishavraj
    • one year ago
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    \[a_2 = 2~~~~b_2 = -1~~~~c_2 = 1\]

  14. sloppycanada
    • one year ago
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    Oohhhhh then I add them? Or subtract?

  15. rishavraj
    • one year ago
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    |dw:1435752333034:dw|

  16. rishavraj
    • one year ago
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    solve D , D_1, D_2

  17. rishavraj
    • one year ago
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    after tht \[x = \frac{ D_1 }{ D } ~~~ and y = \frac{ D_2 }{ D }\]

  18. sloppycanada
    • one year ago
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    |dw:1435752595951:dw|

  19. sloppycanada
    • one year ago
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    |dw:1435752646115:dw|

  20. rishavraj
    • one year ago
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    on solving those determinants \[D = -3~~~~~D_1 = -6~~~~~~~~D_2 = -9\]

  21. sloppycanada
    • one year ago
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    But it's divided... I am so confused..

  22. rishavraj
    • one year ago
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    u need to first of all calculate D, D_1 and D_2 and then divide ..

  23. sloppycanada
    • one year ago
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    So for the first one - [2] [3]

  24. rishavraj
    • one year ago
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    u got the values of all deteminants???

  25. sloppycanada
    • one year ago
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    -3, -6, and -9

  26. rishavraj
    • one year ago
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    ooo cool....xD so u get x = 2 and y = 3 :))

  27. sloppycanada
    • one year ago
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    Okay, so I was right to begin with... I just did it a different way (elimination method)

  28. sloppycanada
    • one year ago
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    But I should phrase it [2 3]?

  29. rishavraj
    • one year ago
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    i don't think so.....

  30. mathmate
    • one year ago
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    @sloppycanada In math, if you are asked to use a specific method, you have to use the method specified to solve the problem. Otherwise the answer is considered wrong. If you have learned Gaussian Elimination (using matrices) it is considered a matrix method, so is Cramer's rule. The normal elimination method is not considered a matrix method.

  31. UsukiDoll
    • one year ago
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    Solve x + y = 5 using matrices. 2x – y = 1 So we need to have an augmented matrix in the form of a l b anything before the = sign goes in the a portion and anything after the = sign goes in the b portion

  32. UsukiDoll
    • one year ago
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    we just grab the values.. you don't have to write x or y in the matrix as that is not necessary

  33. UsukiDoll
    • one year ago
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    |dw:1435753678426:dw| you should have a matrix that looks like this and looks like we need the Reduced row echelon form

  34. UsukiDoll
    • one year ago
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    |dw:1435753725392:dw| see these triangles? this is where my main diagonal is located. Our goal for reduced row echelon form is to have all numbers below and above the main diagonal go to 0

  35. UsukiDoll
    • one year ago
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    do you know any row operations ? @sloppycanada

  36. sloppycanada
    • one year ago
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    Yes? + or - and multiplace and division... but generally you need to [] [] of those things.

  37. UsukiDoll
    • one year ago
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    ok we have three of them either we can multiply a number throughout the row notation is like this \[2r_1 \] which means multiply the whole first row by 2

  38. UsukiDoll
    • one year ago
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    \[r_1 \leftrightarrow r_2\] we can swap rows.. this means switch row 1 and row 2

  39. UsukiDoll
    • one year ago
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    \[r_1+r_2 \rightarrow r_2 \] meaning we add row one to row two and our result goes to row 2 same thing applies with subtraction

  40. UsukiDoll
    • one year ago
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    |dw:1435753941663:dw| so back to this problem what row operations do we need to get rid of that 2 on the bottom diagonal?

  41. sloppycanada
    • one year ago
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    The adding thing. r1 + r2?

  42. UsukiDoll
    • one year ago
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    ok good start but there's a row operation that we need to do first and it has something to do with the first row

  43. UsukiDoll
    • one year ago
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    so since I see a 2 on the bottom of the main diagonal I have to multiply what number to the first row

  44. sloppycanada
    • one year ago
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    2?

  45. UsukiDoll
    • one year ago
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    yes... close... but switch signs

  46. UsukiDoll
    • one year ago
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    what is the opposite of positive?

  47. UsukiDoll
    • one year ago
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    instead of 2 it has to be - ?

  48. UsukiDoll
    • one year ago
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    what's (-1)x2?

  49. sloppycanada
    • one year ago
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    -2

  50. UsukiDoll
    • one year ago
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    yes so we multiply -2 throughout the entire first row -2[ 1 1 l 5] and that row becomes -2 -2 l -10 now we add this to row 2 |dw:1435754420055:dw|

  51. UsukiDoll
    • one year ago
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    so what's -2+2 -2-1 -10+1

  52. sloppycanada
    • one year ago
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    0 -3 -9

  53. UsukiDoll
    • one year ago
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    yes now we have 0 -3 -9 should we place this in the first row or in the second row? One of the rules for reduce row echelon form is that we need a 1 on the top left.

  54. UsukiDoll
    • one year ago
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    err I meant 0 -3 l -9

  55. UsukiDoll
    • one year ago
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    and we need a 1 on the top left ( meaning the entry of the first row and first column should always be 1)

  56. UsukiDoll
    • one year ago
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    so does 0 -3 l -9 belong in the first row or second row?

  57. sloppycanada
    • one year ago
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    Okay.. I think I'm starting to get it, kinda.

  58. sloppycanada
    • one year ago
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    basically once you find the determinant of the first A you put it into the original matrix and then divide

  59. UsukiDoll
    • one year ago
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    well determinants are for square matrices only like 2 x 2 (an n x n matrix) we have an m x n which is 2 rows 3 columns

  60. UsukiDoll
    • one year ago
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    determinant for 3 x 3 is possible but nasty!

  61. sloppycanada
    • one year ago
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    I have one for practice that I've been doing as I've been listening or watching your explanation.

  62. UsukiDoll
    • one year ago
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    |dw:1435754875634:dw|

  63. UsukiDoll
    • one year ago
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    now we have to get rid of that 1 in the first row second column... let's repeat what we did the last time.. let's multiply the entire first row by ???

  64. UsukiDoll
    • one year ago
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    keep in mind that the numbers in the main diagonal can't be 0! What we are aiming for is using row operations to have a reduce row echelon form. In the end we should have something similar to an Identity Matrix

  65. UsukiDoll
    • one year ago
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    |dw:1435755082603:dw|

  66. UsukiDoll
    • one year ago
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    |dw:1435755112509:dw|

  67. UsukiDoll
    • one year ago
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    first column is x second column is y eventually third column is z followed by w or something similar like that

  68. sloppycanada
    • one year ago
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    Something like this - Solve x – 3y + z = -9 using matrices. x + y – 3z = -5 3x – y + z = 11

  69. UsukiDoll
    • one year ago
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    oh lord... well weeee we need a 3 x 4 augmented matrix.

  70. UsukiDoll
    • one year ago
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    |dw:1435755254310:dw|

  71. UsukiDoll
    • one year ago
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    the main diagonal is in entries \[a_{11},a_{22},a_{33} \]

  72. sloppycanada
    • one year ago
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    I got this - x = 4 y = 6 z = 5

  73. UsukiDoll
    • one year ago
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    ok so it's x =4, y =6 and z=5 now we just plug them back into the equations to see if they are true.

  74. UsukiDoll
    • one year ago
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    3x-y+z = 11 3(4)-6+5=11 12-6+5=11 6+5=11 11=11

  75. UsukiDoll
    • one year ago
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    x+y-3z = -5 4+6-3(5)=-5 4+6-15=-5 10-15=-5 -5=-5

  76. UsukiDoll
    • one year ago
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    x-3y+z = -9 4-3(6)+5=-9 4-18+5=-9 -14+5=-9 -9=-9

  77. UsukiDoll
    • one year ago
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    yup everything checks out

  78. sloppycanada
    • one year ago
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    Thanks so much I just have five more of these stupid review problems then I can do fun things!

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