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))
