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anonymous

  • one year ago

matrix equation help? {{12,7},{5,3}}x={{2,-1},{3,2}}

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  1. amistre64
    • one year ago
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    an augment might be beneficial

  2. amistre64
    • one year ago
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    unless you know how to work an inverse? Ax = b A'Ax = A'b Ix = A'b x = A'b

  3. anonymous
    • one year ago
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    how can i plug in matrices?

  4. amistre64
    • one year ago
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    what are you working with? ti83?

  5. anonymous
    • one year ago
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    No. I meant how can I plug in matrices on Openstudy

  6. anonymous
    • one year ago
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    So I can show the equation correctly

  7. amistre64
    • one year ago
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    its labor intensive, but i did make an app for it, if you are using a laptop or desktop that is

  8. amistre64
    • one year ago
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    \[\left[ \begin{array}c 12 & 7\\5 & 3\\\end{array} \right] x= \left[ \begin{array}c 2 & -1\\3 & 2\\\end{array} \right]\]

  9. anonymous
    • one year ago
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    \[\[\left[ \begin{array}c 12 & 7\\5 & 3\\\end{array} \right]\] X = \[\left[ \begin{array}c 2 & -1\\3 & 2\\\end{array} \right]\]\]

  10. anonymous
    • one year ago
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    hows that?

  11. anonymous
    • one year ago
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    yeah

  12. amistre64
    • one year ago
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    it was designed to post one matrix hence the oddity of trying multiples, but knowing the latex sorts it out

  13. amistre64
    • one year ago
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    so, how do we approach this? inverse are augment? either way is fine. inverse may be simpler in this case tho

  14. amistre64
    • one year ago
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    we can augment it as this, and row reduce. \[\left[ \begin{array}c 12 & 7 & | & 2 & -1\\5 & 3 & | & 3 & 2\\\end{array} \right]\]

  15. anonymous
    • one year ago
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    how do you reduce it

  16. amistre64
    • one year ago
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    elementary row operations for course, get the left side worked into the identity matrix. and the right side works itself into x

  17. amistre64
    • one year ago
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    divide row1 by 12 and multiply by -5 and add to row2 12,7,2,-1 1,7/12,2/12,-1/12 -5,-35/12,-10/12,5/12 5, 3, 3, 2 -5,-35/12,-10/12, 5/12 5, 36/12, 36/12,24/12 ---------------------- 0, 1/12, 26/12, 29/12 multiply by 12 0, 1, 26, 29 new setup 1, 7/12, 2/12, -1/12 0, 1 , 26 , 29 multiply row 2 by -7/12 and add it to row1 1, 7/12, 2/12, -1/12 0, -7/12 , -26(7)/12 , -29(7)/12 -------------------------------- 1, 0 , -90/6 , -204/12 1, 0 , -15 , -17 1, 0, -15, -17 0, 1, 26, 29 I x

  18. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=rref%7B%7B12%2C7%2C2%2C-1%7D%2C%7B5%2C3%2C3%2C2%7D%7D

  19. amistre64
    • one year ago
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    did you try for an inverse?

  20. amistre64
    • one year ago
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    -35 12 7 5 3 36 det = 1 by some basic shortcut, A' has a pattern swap corners and negate the other 2 A' = 3 -7 -5 12

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