How do you solve 3x+3y=15 and 2x+3y=-5 using matrices?

1 Answer
Feb 25, 2016

Either use the Gauss-Jordan method or Cramer's Rule:
color(white)("XXX")(x,y)=(20,-15)

Explanation:

In either case we will need to rewrite the given equations into matrix form:

((3,3,color(blue)(15)),(2,3,color(blue)(-5)))

Gauss-Jordan
Divide row 1 (by 3 so that the leading term is 1
color(white)("XXX")((1,1,5),(2,3,-5))
Subtract 2xx this new row 1 from row 2 to reduce its leading term to 0
color(white)("XXX")((1,1,5),( 0,1,-15 ))

Subtract this new row 2 from row 1:
color(white)("XXX")((1,0,20),(0,1,-15))
The first column is x, the second is y and the third is the value they are equal to.

Cramer's Rule
D=((3,3),(2,3))color(white)("XX")D_x=((color(blue)(15),3),(color(blue)(-5),3))color(white)("XX")D_y=((3,color(blue)(15)),(2,color(blue)(-5)))

The determinants:
color(white)("XXX")abs(D) = 3xx3-2xx3=3

color(white)("XXX")abs(D_x)= 15xx3-(-5)xx3=60

color(white)("XXX")abs(D_y)=3xx(-5)-2xx15=-45

x=abs(D_x)/abs(D) = 60/3=20

y=abs(Dy)/abs(D) = (-45)/3=-15