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anonymous

  • one year ago

Medal and Fan! Which product represents the solution to the system?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    Do you know how to multiple matrices?

  3. anonymous
    • one year ago
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    yes, this is my answer

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  4. anonymous
    • one year ago
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    yes that is correct, however, thats a lil complicated how you've done it.

  5. anonymous
    • one year ago
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    oh wait, you have just used the inverse formula, i see.

  6. anonymous
    • one year ago
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    i wouldn't leave it in this form , i would simplify this further

  7. anonymous
    • one year ago
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    because a solution to the system involves values for x and y, here you have just simply solved the inverse.

  8. anonymous
    • one year ago
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    \[AX=B\] \[A ^{-1}AX=A ^{-1} B\] \[IX=A ^{-1}B\] \[X=A ^{1}B\]

  9. anonymous
    • one year ago
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    that should be A^(-1) in the last part sorry.

  10. anonymous
    • one year ago
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    you have solved A^(-1)|dw:1443662042771:dw|

  11. anonymous
    • one year ago
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    my answer is correct?

  12. anonymous
    • one year ago
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    i would simplify it further \[X=\left[\begin{matrix}2 & -1/2 \\ -1 & 1/2\end{matrix}\right]\left(\begin{matrix}3 \\ 2\end{matrix}\right)=\left(\begin{matrix}5 \\ -2\end{matrix}\right)\]

  13. anonymous
    • one year ago
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    here, this simplification says that the solution to this system is x=5, y=-2

  14. anonymous
    • one year ago
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    so you don't get confused, \[X=\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}5 \\ 2\end{matrix}\right)\]

  15. anonymous
    • one year ago
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    im not sure if you could see the first part of it but theres a 1/2 in fornt of it

  16. anonymous
    • one year ago
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    @chris00

  17. anonymous
    • one year ago
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    i did, i just simplified it into the matrix

  18. anonymous
    • one year ago
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    \[\left[\begin{matrix}2 & -1/2 \\ -1& 1/2\end{matrix}\right]=\frac{ 1 }{ 2 }\left[\begin{matrix}4 & -1 \\ -2 & 1\end{matrix}\right]\]

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