## vera_ewing one year ago Math question

1. Michele_Laino

here your linear system can be rewritten as follows: $\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right)$

2. Michele_Laino

now, we have to find the inverse matrix of the coefficients matrix: $\left( {\begin{array}{*{20}{c}} 3&4 \\ 2&5 \end{array}} \right)$

3. vera_ewing

Ok and then what?

4. 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

5. vera_ewing

Ok, and then how do we find the product of the solution?

6. Michele_Laino

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

7. vera_ewing

Ohh so the answer is C?

8. Michele_Laino

are you sure?

9. vera_ewing

I think so...can you check?

10. Michele_Laino

the solution of your system, can be rewritten as follows: $\frac{1}{7}\left( {\begin{array}{*{20}{c}} 5&{ - 4} \\ { - 2}&3 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 2} \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)$

11. vera_ewing

Oh so D! Thanks Michele! :)

12. Michele_Laino

correct! :)

Find more explanations on OpenStudy