How do you solve #x-6y=31# and #6x+9y=-84# using matrices?

1 Answer
Sep 30, 2016

Answer:

#(x,y)=color(green)(-5,-6)#

Explanation:

Given the equations:
#color(white)("XXX")x-6y=31#
#color(white)("XXX")6x+9y=-84#

We can write these in "augmented matrix form as:)
#color(white)("XXX")( (1,-6,31),(6,9,-84) )#

This can be solved using the normal operations we would perform on the original equations (with the variables and equal side "assumed").

#color(red)("Alternately, we can use Cramer's Rule with Matrix Determinants")#

#color(white)("XXX")color(blue)("Although I demonstrate the soloution (below) using hand calculations")#
#color(white)("XXX")color(blue)("the power of the Matrix Methods lies in their compatibility with")#
#color(white)("XXX")color(blue)("computer systems. For example, the evaluation of Determinants")#
#color(white)("XXX")color(blue)("is a build-in function for most spreadsheets.")#

If #M_(xy)=((1,-6),(6,9))color(white)("XX")M_(cy)=((31,-6),(-84,9))color(white)("XX")M_(xc)=((1,31),(6,-84))#

#x=(|M_(cy)|)/(|M_(xy)|)" and " y=(|M_xc|)/(|M_(xy)|)#

Where the Determinants:
#color(white)("XXX")|M_(xy)|= 1xx9-6xx(-6)= 9+36=45#

#color(white)("XXX")|M_(cy)|= 31xx9-(-84)xx(-6))=279-504=-255#

#color(white)("XXX")|M_(xc)|=1xx(-84)-6xx31=-84-186=-270#

Giving
#color(white)("XXX")x=-255/45=-6#
and
#color(white)("XXX")y=-270/45=-5#