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mariomintchev

  • one year ago

Can someone clearly explain to me what i need to do and why I am doing this?

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  1. mariomintchev
    • one year ago
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  2. mariomintchev
    • one year ago
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    @Callisto @Mertsj @NotTim @phi @satellite73 @UnkleRhaukus

  3. mariomintchev
    • one year ago
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    It's a matrix problem.

  4. UnkleRhaukus
    • one year ago
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    \[\left[\begin{array}{ccc}-3&6&9\\0&-3&7\\0&-1&-1 \end{array}\right]\left[\begin{array}{c}1\\1\\-7 \end{array}\right]\] \[\sim\left[\begin{array}{ccc}-3&6&9\\0&-3&7\\(0-0)&(-3--3)&(-1-7)\end{array}\right]\left[\begin{array}{c}1\\1\\(-7-1) \end{array}\right]\qquad R_3\to R_3-R_2\] \[\sim\left[\begin{array}{ccc}-3&6&9\\0&-3&7\\0&0&-8\end{array}\right]\left[\begin{array}{c}1\\1\\-8 \end{array}\right]\] \[\sim\left[\begin{array}{ccc}-3&6&9\\0&-3&7\\0&0&-8/-8\end{array}\right]\left[\begin{array}{c}1\\1\\-8/-8 \end{array}\right]\qquad R_3\to R_3/-8\] \[\sim\left[\begin{array}{ccc}-3&6&9\\0&-3&7\\0&0&1\end{array}\right]\left[\begin{array}{c}1\\1\\1 \end{array}\right]\]

  5. mariomintchev
    • one year ago
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    ok so our goal is to make the second box or matrix or whatever you wanna call it, be the same?

  6. mariomintchev
    • one year ago
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    and what does REF and RREF mean??

  7. amistre64
    • one year ago
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    reduced echelon form and row reduced echelon form echelon just means "stairs"

  8. amistre64
    • one year ago
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    Unkles 3rd line is REF RREF has leading 1s

  9. mariomintchev
    • one year ago
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    btw that doesn't seem to be the right anwer

  10. mariomintchev
    • one year ago
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    its cause @UnkleRhaukus has 7 instead of -7

  11. amistre64
    • one year ago
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    multiply the 2nd row by -1 and add it to the 3rd row to get a new thrid row and a matrix in ref

  12. amistre64
    • one year ago
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    the object is to get rid of that -3 in the 2nd column of the last row

  13. amistre64
    • one year ago
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    (0 -3 -7 1) *-1 0 3 7 -1 0 -3 -1 -7 ---------- 0 0 6 -8 new 3rd

  14. mariomintchev
    • one year ago
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    why is that the goal? we want everything under the bridge we draw to equal zero?

  15. mariomintchev
    • one year ago
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    |dw:1359151416530:dw| this is what i mean by bridge ^^^^

  16. TimSmit
    • one year ago
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    If your matrix gets that form then you can easily solve for for the three unknowns. When you look at the last row, you see that there is a direct relation for the third unknown (lets call it x3): 6*x3 = -8. Now that x3 is known, there is a direct relation for the second unknown and after that for the first.

  17. mariomintchev
    • one year ago
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    ok i still dont understand the goal of this....

  18. mariomintchev
    • one year ago
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    like the purpose of changing the last row

  19. TimSmit
    • one year ago
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    If you don't change the last row, then you need to calculate x2 and x3 from the two last equations (2 equations, two unknowns). If you do change it, then one of the unknowns is directly given (the last row does not depend on x2 and x1 as the first two values in the matrix are zero).

  20. UnkleRhaukus
    • one year ago
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    i see i did copy the question incorrectly, that seven on the second row should be a negative like you say @mariomintchev. so the working/solution will be a bit different , However the method will be the same, The basic idea is that a matrix is a simpler way of representing an system of equations, performing row operations does not change the solution to the system. We choose row operations that reduce the complicated system by merging information in different rows, row echelon form is when we have zeroes below the main diagonal( or bridge) and ones as the first terms in each row. from REF we usually continue to Reduced row echelon form because then the solution can be easily be read off say \[\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1 \end{array}\right]\left[\begin{array}{c}8\tfrac16\\12\tfrac12\\-5\tfrac12 \end{array}\right]\] which is kinda shorthand for\[\left[\begin{array}{ccc|c}1&0&0&x_1\\0&1&0&x_2\\0&0&1&x_3 \end{array}\right]=\left[\begin{array}{c}8\tfrac16\\12\tfrac12\\-5\tfrac12 \end{array}\right]\] which equivalent to\[x_1+0+0=8\tfrac16\]\[0+x_2+0=12\tfrac12\]\[0+0+x_3=-5\tfrac12\] which is the solution to the system

  21. amistre64
    • one year ago
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    matrix setups are just another way of organizing the data. The elementary row operations on a matrix are the same technique/construct as working an system of equation using "elimination" method.

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