vera_ewing
  • vera_ewing
Math question
Mathematics
chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
here your linear system can be rewritten as follows: \[\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right)\]
Michele_Laino
  • Michele_Laino
now, we have to find the inverse matrix of the coefficients matrix: \[\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right)\]
vera_ewing
  • vera_ewing
Ok and then what?

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Michele_Laino
  • Michele_Laino
such inverse matrix, is: \[\left( {\begin{array}{*{20}{c}} {5/7}&{ - 4/7} \\ { - 2/7}&{3/7} \end{array}} \right)\] please check my computations
vera_ewing
  • vera_ewing
Ok, and then how do we find the product of the solution?
Michele_Laino
  • Michele_Laino
it is simple, since the solution of your system is: \[\left( {\begin{array}{*{20}{c}} {5/7}&{ - 4/7} \\ { - 2/7}&{3/7} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\]
vera_ewing
  • vera_ewing
Ohh so the answer is C?
Michele_Laino
  • Michele_Laino
are you sure?
vera_ewing
  • vera_ewing
I think so...can you check?
Michele_Laino
  • Michele_Laino
the solution of your system, can be rewritten as follows: \[\frac{1}{7}\left( {\begin{array}{*{20}{c}} 5&{ - 4} \\ { - 2}&3 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\]
vera_ewing
  • vera_ewing
Oh so D! Thanks Michele! :)
Michele_Laino
  • Michele_Laino
correct! :)

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