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anonymous

  • one year ago

Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5 x - y + 3z = -1 4x + y + z = -2

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  1. anonymous
    • one year ago
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    @UnkleRhaukus

  2. UnkleRhaukus
    • one year ago
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    [ 1 1 1 ] [ x ] [-5] [ 1 -1 3 ] [ y ] = [-1] [ 4 1 1 ] [ z ] [-2]

  3. anonymous
    • one year ago
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    yesss ok so what do i do here

  4. UnkleRhaukus
    • one year ago
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    [ 1 1 1 | -5] [ 1 -1 3 |-1] [ 4 1 1 | -2]

  5. UnkleRhaukus
    • one year ago
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    we want make this [ .........|..] [ x...... |..] [ ........ |..] element into a zero

  6. UnkleRhaukus
    • one year ago
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    take away the first row from the second row

  7. anonymous
    • one year ago
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    how

  8. UnkleRhaukus
    • one year ago
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    [ 1 1 1 | -5] (R1) [ 1 -1 3 | -1] (R2) [ 4 1 1 | -2] (R3) (R2)-(R1) = [(1-1) (-1-1) (3-1) | (-1-5)]

  9. anonymous
    • one year ago
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    ohhh

  10. anonymous
    • one year ago
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    whats next or is that it

  11. UnkleRhaukus
    • one year ago
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    we get [ 1 1 1 | -5] (R1) [ 0 -2 2 | -6] (R2) [ 4 1 1 | -2] (R3)

  12. UnkleRhaukus
    • one year ago
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    now we want make this [ 1.......|..] [ 0...... |..] [ x........ |..] element zero

  13. UnkleRhaukus
    • one year ago
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    so take away line row 1 from row 3, four times

  14. anonymous
    • one year ago
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    soo 111 i think

  15. UnkleRhaukus
    • one year ago
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    [ 4 1 1 | -2] - 4*[ 1 1 1 | -5] =

  16. anonymous
    • one year ago
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    umm idk im not good with these i just started this

  17. anonymous
    • one year ago
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    @loser66

  18. anonymous
    • one year ago
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    @ganeshie8

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