How do you solve #x-2y=2# and #3x-5y=9# using matrices?

1 Answer
Feb 3, 2017

Answer:

#x=8, y=3#

Explanation:

Put the eqns in matrix form

#x-2y=2#
#3x-5y=9#

become

#((1,-2),(3,-5))##((x),(y))=((2),(9))#

we therefore have

#M((x),(y))=((2),(9))#

by premultiplying both sides by #M^(-1)#

#M^(-1)M((x),(y))=M^(-1)((2),(9))#

#I((x),(y))=M^(-1)((2),(9))#

where #I=((1,0),(0,1))#

#:.((x),(y))=M^(-1)((2),(9))#

so we need to find the inverse of the matrix.

#M=((1,-2),(3,-5))#

for any # 2xx2 # mx

#M=((a,b),(c,d))#

#M^(-1)=1/Delta((d,-b),(-c,a))" providing "DeltaM!=0#

find its determinant

#Delta M=|(1,-2),(3,-5)|#

#DeltaM=1xx-5-(3xx-2)=-5--6=1#

#because Delta M!=0 " the inverse exists"#

#:.M^(-1)=1/1((-5,2),(-3,1))=((-5,2),(-3,1))#

so#((x),(y))=M^(-1)((2),(9))=((-5,2),(-3,1))((2),(9))#

#((x),(y))=((-5xx2+2xx9),(-3xx2+1xx9))#

#((x),(y))=((-10+18),(-6+9))#

#((x),(y))=((-10+18),(-6+9))#

#((x),(y))=((8),(3))#