# How do you solve 2x - 5y = 12 and x - 3y = -3 using matrices?

Mar 17, 2016

You could write the equations as matrices and apply standard row operations as if they were normal linear equations
or you could use Cramer's Rule

#### Explanation:

{: (color(red)(2)x,color(blue)(-5)y," = ",color(green)(12)), (color(red)(1)x,color(blue)(-3)y," = ",color(green)(-3)) :}hArr ((color(red)(2),color(blue)(-5),color(green)12),(color(red)(1),color(blue)(-3),color(green)(-3)))

Defining the determinants:
D=|(color(red)(2),color(blue)(-5)),(color(red)(1),color(blue)(-3))| color(white)("XX") Dx=|(color(green)(12),color(blue)(-5)),(color(green)(-3),color(blue)(-3))| color(white)("XX") Dy=|(color(red)(2),color(green)(12)),(color(red)(1),color(green)(-3))|

Cramer's Rule tells us that
$\textcolor{w h i t e}{\text{XXX")x=(D_x)/Dcolor(white)("XXX}} y = \frac{{D}_{y}}{D}$

$D = \left(\textcolor{red}{2} \times \left(\textcolor{b l u e}{- 3}\right)\right) - \left(\textcolor{red}{1} \times \left(\textcolor{b l u e}{- 5}\right)\right) = - 1$

${D}_{x} = \left(\textcolor{g r e e n}{12} \times \textcolor{b l u e}{\left(- 3\right)}\right) - \left(\left(\textcolor{g r e e n}{- 3}\right) \times \left(\textcolor{b l u e}{- 5}\right)\right) = - 51$

D_y=((color(red)(2)xx(color(green)(-3)))-(color(red)(1xxcolor(green)(12))) = -18

Therefore
$\textcolor{w h i t e}{\text{XXX}} x = 51 \mathmr{and} y = 18$