How do you solve #x + 2y = 4# and #2x - y = 3# using matrices?

1 Answer
Feb 15, 2016

Answer:

#x=2color(white)("XXXX")y=1#

Explanation:

#[(color(red)(x),color(blue)(y),,color(brown)(c)),(color(red)(1),color(blue)(2),"|",color(brown)(4)),(color(red)(2),color(blue)(-1),"|",color(brown)(3))]#

Using Cramer's Rule
#x=(|(color(brown)(4),color(blue)(2)),(color(brown)(3),color(blue)(-1))|) /(|(color(red)(1),color(blue)(2)),(color(red)(2),color(blue)(-1))|)=(-4-6)/(-1-4)=(-10)/(-5)=2#

#y=(|(color(red)(1),color(brown)(4)),(color(red)(2),color(brown)(3))|) /(|(color(red)(1),color(blue)(2)),(color(red)(2),color(blue)(-1))|)=(3-8)/(-1-4)=(-5)/(-5)=1#