How do you solve the system #4x+8y=7, 3x-3y=0# using matrix equation?

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Answer 1 · 2017-04-11T12:49:13

#x=7/12# and #y=7/12#

In matrix notation, the system of linear equations

#4x+8y=7# and #3x-3y=0# can be written as

#((4,8),(3,-3))((x),(y))=((7),(0))#

i.e. #A((x),(y))=((7),(0))#, then

#((x),(y))=A^(-1)((7),(0))#,

where #A^(-1)# is inverse of #A=((a,b),(c,d))# and for a 2X2 matrix is defined as #A^(-1)=1/(ad-bc)((d,-b),(-c,a))#

Hence inverse of #((4,8),(3,-3))# is #1/(-12-24)((-3,-8),(-3,4))#

i.e. #1/(-36)((-3,-8),(-3,4))=((1/12,2/9),(1/12,-1/9))#

and #((x),(y))=((1/12,2/9),(1/12,-1/9))((7),(0))=(7/12,7/12)#

i.e. #x=7/12# and #y=7/12#