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vera_ewing
 one year ago
Math question
vera_ewing
 one year ago
Math question

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.1now, we have to find the inverse matrix of the coefficients matrix: \[\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1such 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok, and then how do we find the product of the solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it 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
 one year ago
Best ResponseYou've already chosen the best response.0Ohh so the answer is C?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0I think so...can you check?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh so D! Thanks Michele! :)
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