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cutie.patootie Group Title

Check my answer Solve with matrices. 5x -3y = 18 and 2x + 7y = -1 I got .... x=31/18 y=-3/5 Am I right?

  • 2 years ago
  • 2 years ago

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  1. dumbcow Group Title
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    no...but remember you can check for yourself, plug in those values

    • 2 years ago
  2. amistre64 Group Title
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    .... that was my line panlac lol

    • 2 years ago
  3. panlac01 Group Title
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    my bad... I can delete it

    • 2 years ago
  4. amistre64 Group Title
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    no need to delete :)

    • 2 years ago
  5. cutie.patootie Group Title
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    Oh okay, well I did & it wasn't.. Crud. Well I am confused on part of it, so when I have to do one row of the matrices that is 5 -13 18 to get it to 1 0 c1 How do I do that?

    • 2 years ago
  6. amistre64 Group Title
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    is this cramer rule? or row operations?

    • 2 years ago
  7. RyanL. Group Title
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    I think row of operations is the way to go.

    • 2 years ago
  8. cutie.patootie Group Title
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    Uhh. I don't know.. But I have to get the row to 1 0 c1, like 5 has to change to 1 and -13 has to change to 0 somehow.

    • 2 years ago
  9. RyanL. Group Title
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    yeah you have to divide the whole row by 5 then

    • 2 years ago
  10. hartnn Group Title
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    hey cutie,have u been taught matrix inversion?which makes the solution much simpler.....

    • 2 years ago
  11. cutie.patootie Group Title
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    @RyanL. Well that would give me 1 -13/5 18/5. How would I get -13/5 to zero with out changing the 1? @hartnn Maybe, math isn't my strong point to I have no idea the fancy names... Sorry.

    • 2 years ago
  12. RyanL. Group Title
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    Well when we row reduce we want every number bellow that 1 to be a zero

    • 2 years ago
  13. RyanL. Group Title
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    |dw:1344704018436:dw| well you need to subtract the second row by the first row but before that you need to multiply the first row by 2

    • 2 years ago
  14. cutie.patootie Group Title
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    Why would you multiply the first row by two?

    • 2 years ago
  15. cutie.patootie Group Title
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    Oh wait, nevermind! I gotcha now!

    • 2 years ago
  16. RyanL. Group Title
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    |dw:1344704116600:dw|

    • 2 years ago
  17. RyanL. Group Title
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    |dw:1344704166215:dw|

    • 2 years ago
  18. RyanL. Group Title
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    then divide the second row by 8.20 and then subtract the the first row by the second.

    • 2 years ago
  19. cutie.patootie Group Title
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    so the bottom row would be 0 1 -1 ? And would that give you.. 1 2/5 13/5 ?

    • 2 years ago
  20. RyanL. Group Title
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    |dw:1344704298500:dw| I like to make them 1s so if I need to divide to multiply it it's going to be easier to do it later.

    • 2 years ago
  21. RyanL. Group Title
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    Well no I see that I want to make the first row second column zero right. I'm talking about this number|dw:1344704459156:dw|

    • 2 years ago
  22. RyanL. Group Title
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    so I then multiply the second row by\[\frac{3}{5}\]and then add it to the first row

    • 2 years ago
  23. RyanL. Group Title
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    |dw:1344704521394:dw|

    • 2 years ago
  24. RyanL. Group Title
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    |dw:1344704554208:dw|

    • 2 years ago
  25. RyanL. Group Title
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    |dw:1344704592083:dw|

    • 2 years ago
  26. RyanL. Group Title
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    multiply the second row by \[\frac{5}{3}\]now to get them back to 1

    • 2 years ago
  27. RyanL. Group Title
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    |dw:1344704647344:dw|

    • 2 years ago
  28. dumbcow Group Title
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    tough to teach row operations online...there are a lot of steps use matrix algebra \[Ax = b\] \[x=A^{-1}b\] where A is coefficient matrix and b is right side constants \[\left(\begin{matrix}5 & -3 \\ 2 & 7\end{matrix}\right)\left(\begin{matrix}x \\ y\end{matrix}\right)= \left(\begin{matrix}18 \\ -1\end{matrix}\right)\] \[\left(\begin{matrix}x \\ y\end{matrix}\right)= \left(\begin{matrix}5 & -3 \\ 2 & 7\end{matrix}\right)^{-1}\left(\begin{matrix}18 \\ -1\end{matrix}\right)\] det(A) = 5*7 - 2*-3 = 41 \[\left(\begin{matrix}5 & -3 \\ 2 & 7\end{matrix}\right)^{-1} = \left(\begin{matrix}7/41 & 3/41\\ -2/41 & 5/41\end{matrix}\right)\] \[\rightarrow \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \\ -1\end{matrix}\right)\]

    • 2 years ago
  29. RyanL. Group Title
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    Yeah but row operations is essential to everything that you will learn ahead so you need to be able to do them at ease. After a while it takes seconds.

    • 2 years ago
  30. RyanL. Group Title
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    And I made a mistake I left out a negative sign here|dw:1344704797494:dw|

    • 2 years ago
  31. RyanL. Group Title
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    Thanks @dumbcow It's always good to show that there is more than one way to solve a problem.

    • 2 years ago
  32. cutie.patootie Group Title
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    Oh, so when you multiply and add and what not, does it not apply to the first number in the row?

    • 2 years ago
  33. RyanL. Group Title
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    Well I think you mean when we add the second row to the first. The thing is that the first number is a zero so if you multiply it a zero by a number or divide it by another number it will remain zero so when we add it to the first row we will be adding or subtracting a zero.

    • 2 years ago
  34. RyanL. Group Title
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    And when you add or subtract a zero from a number it stays the same.

    • 2 years ago
  35. cutie.patootie Group Title
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    Okay, I gotcha. Thanks for the explanation. :) It really helped!

    • 2 years ago
  36. RyanL. Group Title
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    No problem sorry for the messy work hope you could understand it .

    • 2 years ago
  37. cutie.patootie Group Title
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    I did :)

    • 2 years ago
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