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vera_ewing

  • one year ago

Math question

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  1. Michele_Laino
    • one year ago
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    here your linear system can be rewritten as follows: \[\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right)\]

  2. Michele_Laino
    • one year ago
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    now, we have to find the inverse matrix of the coefficients matrix: \[\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right)\]

  3. vera_ewing
    • one year ago
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    Ok and then what?

  4. Michele_Laino
    • one year ago
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    such inverse matrix, is: \[\left( {\begin{array}{*{20}{c}} {5/7}&{ - 4/7} \\ { - 2/7}&{3/7} \end{array}} \right)\] please check my computations

  5. vera_ewing
    • one year ago
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    Ok, and then how do we find the product of the solution?

  6. Michele_Laino
    • one year ago
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    it is simple, since the solution of your system is: \[\left( {\begin{array}{*{20}{c}} {5/7}&{ - 4/7} \\ { - 2/7}&{3/7} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\]

  7. vera_ewing
    • one year ago
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    Ohh so the answer is C?

  8. Michele_Laino
    • one year ago
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    are you sure?

  9. vera_ewing
    • one year ago
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    I think so...can you check?

  10. Michele_Laino
    • one year ago
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    the solution of your system, can be rewritten as follows: \[\frac{1}{7}\left( {\begin{array}{*{20}{c}} 5&{ - 4} \\ { - 2}&3 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\]

  11. vera_ewing
    • one year ago
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    Oh so D! Thanks Michele! :)

  12. Michele_Laino
    • one year ago
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    correct! :)

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