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Dido525

  • 3 years ago

Solve the following system by Gauss–Jordan elimination. 2x₁+5x₂+3x₃=35 12x₁+31x₂+20x₃=216

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  1. Dido525
    • 3 years ago
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    I got x₁=5/2 x₂=6 x₃=t (free variable) The answer says I am wrong. Care to tell me what I did wrong?

  2. Dido525
    • 3 years ago
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    @jim_thompson5910

  3. Dido525
    • 3 years ago
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    @ParthKohli @UnkleRhaukus

  4. UnkleRhaukus
    • 3 years ago
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    x_3=0

  5. Dido525
    • 3 years ago
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    Right. But whats wrong with my other answers?

  6. Dido525
    • 3 years ago
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    Here is what the online page looks like:

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  7. wio
    • 3 years ago
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    Shouldn't the other variables be in terms of the free variable?

  8. wio
    • 3 years ago
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    What did were your elimination steps?

  9. Dido525
    • 3 years ago
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    It's difficult to post the elimination steps here. I checked with my calculator though. It's correct.

  10. Dido525
    • 3 years ago
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    But yeah, that's what I think @wio . I don't know how to express it though.

  11. UnkleRhaukus
    • 3 years ago
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    when we check your solution , we see t must be 0 x₁=5/2 x₂=6 x₃=t _________ 2x₁+5x₂+3x₃=35 5+30+3t = 35 35 +3t = 35 _________ 12x₁+31x₂+20x₃ = 216 60 + 186 + 20t = 216 216 + 20t = 216

  12. Dido525
    • 3 years ago
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    but t is a free variable is it not? SO it can be any value it wants.

  13. wio
    • 3 years ago
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    \[ \begin{bmatrix} 2&5&3 \\ 12&31&20 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 35 \\ 216 \\ t \end{bmatrix} \]

  14. wio
    • 3 years ago
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    Did you solve it like this?

  15. Dido525
    • 3 years ago
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    No... How did you get 0,0 and 1?

  16. wio
    • 3 years ago
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    I just added in the equation \(x_3=t\)

  17. Dido525
    • 3 years ago
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    Sorry, that doesn't make sense.

  18. wio
    • 3 years ago
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    I added in the equation: \[ 0x_1+ 0x_2 +1x_3 = t \]

  19. Dido525
    • 3 years ago
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    Ohh I see.

  20. wio
    • 3 years ago
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    All of the elimination you did before should work for the most part.

  21. Dido525
    • 3 years ago
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    So I do a back substitution?

  22. wio
    • 3 years ago
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    You first want to get the second row to have a zero in it's first colum

  23. Dido525
    • 3 years ago
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    because x3=t so solve the others n terms of t.

  24. wio
    • 3 years ago
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    then you can back substitute

  25. Dido525
    • 3 years ago
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    1 second.

  26. Dido525
    • 3 years ago
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    x₁=2.5+3.5t ?

  27. Dido525
    • 3 years ago
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    x₂=6-2t?

  28. wio
    • 3 years ago
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    yeah I'm getting that too

  29. wio
    • 3 years ago
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    did you try it?

  30. Dido525
    • 3 years ago
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    One attempt left :3 . Fingers crossed.

  31. wio
    • 3 years ago
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    you can double check it you know

  32. Dido525
    • 3 years ago
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    How?

  33. wio
    • 3 years ago
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    plug in each equation into the equations

  34. Dido525
    • 3 years ago
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    Assign a random number to t and see if it works?

  35. wio
    • 3 years ago
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    like, plug in t for x_3 and 6-2t for x_2 into the equations

  36. wio
    • 3 years ago
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    hmm actually I'm not quite sure how to double check it

  37. wio
    • 3 years ago
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    let me look at wolfram

  38. Dido525
    • 3 years ago
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    You can do this on wolfram? O_O .

  39. wio
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=2x%2B5y%2B3z%3D35%2C+12x%2B31y%2B20z%3D216%2C+z+%3D+t

  40. wio
    • 3 years ago
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    wolfram agrees... it's just a matter of the stupid answer verify thing not being a wingspan

  41. Dido525
    • 3 years ago
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    THANKS!!! It's right :) .

  42. wio
    • 3 years ago
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    No problem. Matrix is your friend.

  43. Dido525
    • 3 years ago
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    I love Linear algebra <3 .

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