Question

Math question

Answer 1

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

Answer 2

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

Answer 3

Ok and then what?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Ohh so the answer is C?

Answer 8

are you sure?

Answer 9

I think so...can you check?

Answer 10

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

Answer 11

Oh so D! Thanks Michele! :)

Answer 12

correct! :)