Respuesta :

merlynthewhizz
We have the linear system:

[tex] \left \{ {{-2x-6y=-26} \atop {5x+2y=13}} \right. [/tex]

which in Matrix format is

[tex] \left[\begin{array}{ccc}a_1&b_1\\a_2&b_2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}c_1\\c_2\end{array}\right] [/tex]

[tex] \left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}-26\\13\end{array}\right] [/tex]

We then find the value of x and y use the Cranmer's Rule:

[tex]x= \frac{ \left[\begin{array}{ccc}c_1&b_1\\c_2&b_2\end{array}\right] }{ \left[\begin{array}{ccc}a_1&a_2\\b_1&b_2\end{array}\right] } = \frac{c_1b_2-b_1c_2}{a_1b_2-a_2b_1} [/tex]

[tex]x= \frac{ \left[\begin{array}{ccc}-26&-6\\13&2\end{array}\right] }{ \left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] } = \frac{(-26)(2)-(-6)(13)}{-2)(2)-(-6)(5)} = \frac{26}{26}=1 [/tex]

[tex]y= \frac{ \left[\begin{array}{ccc}a_1&c_1\\a_2&c_2\end{array}\right] }{ \left[\begin{array}{ccc}a_1&b_1\\a_2&b_2\end{array}\right] } = \frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1} [/tex]

[tex]y= \frac{ \left[\begin{array}{ccc}-2&-26\\5&13\end{array}\right] }{ \left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] }= \frac{(-2)(13)-(5)(-26)}{(-2)(2)-(-6)(5)}= \frac{104}{26}=4 [/tex]

So we have the answers:
x = 1 and y = 4

Answer: Option A


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