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boomerang285

  • 2 years ago

Solve the matrix : |1 2 1| 0 | |0 1 0|-2| |0 0 1| 3 |

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  1. technopanda13
    • 2 years ago
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    open study is for help not answers :)

  2. boomerang285
    • 2 years ago
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    i know that, I am looking for help in solving this matrix

  3. nubeer
    • 2 years ago
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    i think the person has just stated the question. maybe he needs help solving it

  4. nubeer
    • 2 years ago
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    hmm @boomerang285 just want to know.. familiar with Guass jordan? is it related to it?

  5. boomerang285
    • 2 years ago
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    yes I believe so, I am looking in my textbook, it is just very confusing

  6. nubeer
    • 2 years ago
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    ok and you have to find x, y, and z right?

  7. boomerang285
    • 2 years ago
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    yes

  8. nubeer
    • 2 years ago
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    ok see your matrix.. all the elements below the peincipal diagnol are 0.. so you can apply Guass method.

  9. boomerang285
    • 2 years ago
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    ok

  10. nubeer
    • 2 years ago
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    ok so in guass u make equations like this. just ignore the last column in you question and replace it with x,y ,z|dw:1352480536058:dw|

  11. nubeer
    • 2 years ago
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    1x+2y+z = 0 y = -2 0x+0y+1z = 3 just now solve these equations

  12. boomerang285
    • 2 years ago
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    so the answer is then (1, -2, 3) ? thats what i get

  13. nubeer
    • 2 years ago
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    yes .. that should be the answer.. you have to remmeber you can do like this just when you have all elements under principal diagonal are 0

  14. boomerang285
    • 2 years ago
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    thank you very much, i have a couple questions i need checked, are you willing?

  15. nubeer
    • 2 years ago
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    yeah sure..

  16. boomerang285
    • 2 years ago
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    these ones i have to solve using cramers rule, If you are familiar. x-2y=5 5x-y=-2 The answer for this one that I come up with is (-1, -3)

  17. boomerang285
    • 2 years ago
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    solve with cramers rule 2x-9y=5 3x-3y=11 i come up with (4, 1/3)

  18. nubeer
    • 2 years ago
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    second one is right.

  19. nubeer
    • 2 years ago
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    yup.. both of them are right :)

  20. boomerang285
    • 2 years ago
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    thank you so much!

  21. nubeer
    • 2 years ago
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    you are most welcome

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