# How do you solve 2x + 7y = - 8 and - 2x + 3y = - 12 using matrices?

Jun 23, 2017

$\left(x , y\right) = \left(3 , - 2\right)$
(see below for use of matrices)

#### Explanation:

Define the matrices:
M_(xyc)=({: (x,y,=c), (2,7,-8), (-2,3,-12) :})

${M}_{x y} = \left(\left.\begin{matrix}2 & 7 \\ - 2 & 3\end{matrix}\right.\right) \textcolor{w h i t e}{\text{XX")M_(cy)=({:(-8,7),(-12,3):})color(white)("XX}} {M}_{x c} = \left(\left.\begin{matrix}2 & - 8 \\ - 2 & - 12\end{matrix}\right.\right)$

and their determinants:
${D}_{x y} = | \left.\begin{matrix}2 & 7 \\ - 2 & 3\end{matrix}\right. | = 2 \times 3 - \left(- 2\right) \times 7 = 20$

${D}_{c y} = | \left.\begin{matrix}- 8 & 7 \\ - 12 & 3\end{matrix}\right. | = \left(- 8\right) \times 3 - \left(- 12\right) \times 7 = 60$

${D}_{x c} = | \left.\begin{matrix}2 & - 8 \\ - 2 & - 12\end{matrix}\right. | = 2 \times \left(- 12\right) - \left(- 2\right) \times \left(- 8\right) = - 40$

By Cramer's Rule
$\textcolor{w h i t e}{\text{XXX}} x = \frac{{D}_{c y}}{{D}_{x y}} = \frac{60}{20} = 3$
and
$\textcolor{w h i t e}{\text{XXX}} y = \frac{{D}_{x c}}{{C}_{x y}} = \frac{- 40}{20} = - 2$