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))#