How do you solve #x+4y-z =3#, #2x+3y-2z =1# and #-x+2y+3z =7# using matrices?

1 Answer
Narad T.
Feb 6, 2017

The answer is #((x),(y),(z))=((1),(1),(2))#

Explanation:

We rewrite the equations with a matrix

#((1,4,-1),(2,3,-2),(-1,2,3))((x),(y),(z))=((3),(1),(7))#

Let matrix #A=((1,4,-1),(2,3,-2),(-1,2,3))#

We must calculate the inverse of #A#, #A^-1#

We calculate the determinant of #A#

#detA=|(1,4,-1),(2,3,-2),(-1,2,3)|#

#=1*|(3,-2),(2,3)|-4*|(2,-2),(-1,3)|-1*|(2,3),(-1,2)|#

#=1*(13)-4(4)-1*(7)#

#=13-16-7=-10#

#detA!=0#, so the matrix is invertible

The matrix of cofactor is

#C=((|(3,-2),(2,3)|,-|(2,-2),(-1,3)|,|(2,3),(-1,2)|),(-|(4,-1),(2,3)|,|(1,-1),(-1,3)|,-|(1,4),(-1,2)|),(|(4,-1),(3,-2)|,-|(1,-1),(2,-2)|,|(1,4),(2,3)|))#

#=((13,-4,7),(-14,2,-6),(-5,0,-5))#

We calculate the transpose of #C#

#C^T=((13,-14,-5),(-4,2,0),(7,-6,-5))#

#A^-1=C^T/detA=-1/10((13,-14,-5),(-4,2,0),(7,-6,-5))#

and

#((x),(y),(z))=-1/10((13,-14,-5),(-4,2,0),(7,-6,-5))*((3),(1),(7))#

#=-1/10((-10), (-10),(-20))=((1),(1),(2))#

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