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

1 Answer

Answer:

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
#color(white)("XXX")x=(D_x)/Dcolor(white)("XXX")y=(D_y)/D#

#D=(color(red)2xx(color(blue)(-3)))-(color(red)(1)xx(color(blue)(-5)))= -1#

#D_x=(color(green)(12)xxcolor(blue)((-3)))-((color(green)(-3))xx(color(blue)(-5))) = -51#

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

Therefore
#color(white)("XXX")x=51 and y=18#