How do you solve #x + 2y + 2z = 0# and #3x - 2y + 3z = -4# and #2x - y + z = -1# using matrices?

1 Answer
Feb 17, 2016

Answer:

Use Cramer's Rule (see below)
to get #x=2/3, y=1, z=-4/3#

Explanation:

Given:
#( (x, y, z), (color(red)(1),color(blue)(2),color(green)(2)),(color(red)(3),color(blue)(-2),color(green)(3)),(color(red)(2),color(blue)(-1),color(green)(1)))=((c),(color(brown)(0)),(color(brown)(-4)),(color(brown)(-1)))#

The determinant:
#| (color(red)(1),color(blue)(2),color(green)(2)),(color(red)(3),color(blue)(-2),color(green)(3)),(color(red)(2),color(blue)(-1),color(green)(1)) |#

#color(white)("XXX")=color(red)(1) * | (color(blue)(-2), color(green)(3)), (color(blue)(-1),color(green)(1)) |-color(blue)(2) * | (color(red)(3), color(green)(3)),(color(red)(2),color(green)(1)) | + color(green)(2) * | (color(red)(3),color(blue)(-2)),(color(red)(2), color(blue)(-1)) |#

#color(white)("XXX")=(-2+3) - 2(3-6) + 2(-3+4)#

#color(white)("XXX")= 9#

Other determinants can be evaluated in a similar method.

Cramer's Rule tells us that for the given values:
#x = ( | (color(brown)(0),color(blue)(2),color(green)(2)),(color(brown)(-4),color(blue)(-2),color(green)(3)),(color(brown)(-1),color(blue)(-1),color(green)(1)) | ) / (| (color(red)(1),color(blue)(2),color(green)(2)),(color(red)(3),color(blue)(-2),color(green)(3)),(color(red)(2),color(blue)(-1),color(green)(1)) | )=6/9 = 2/3#

#y = (| (color(red)(1),color(brown)(0),color(green)(2)),(color(red)(3),color(brown)(-4),color(green)(3)),(color(red)(2),color(brown)(-1),color(green)(1)) |) / (| (color(red)(1),color(blue)(2),color(green)(2)),(color(red)(3),color(blue)(-2),color(green)(3)),(color(red)(2),color(blue)(-1),color(green)(1)) | ) = 9/9 =1#

#z = ( | (color(red)(1),color(blue)(2),color(brown)(0)),(color(red)(3),color(blue)(-2),color(brown)(-4)),(color(red)(2),color(blue)(-1),color(brown)(-1)) | ) / (| (color(red)(1),color(blue)(2),color(green)(2)),(color(red)(3),color(blue)(-2),color(green)(3)),(color(red)(2),color(blue)(-1),color(green)(1)) | )=-12/9 = -4/3#