Rewrite the equation in matrix form
#((1,-1,2),(-1,0,3),(2,1,0))((x),(y),(z))=((-5),(0),(1))#
Let #A=((1,-1,2),(-1,0,3),(2,1,0))#
We must find the inverse of matrix #A#
Let's calculate the
#Det A# #= | (1,-1,2), (-1,0,3), (2,1,0) | #
#=1* | (0,3), (1,0) |+1* | (-1,3), (2,0) | +2* | (-1,0), (2,1) | #
#=-3-6-2=-11#
As #Det A!=0#, matrix A is invertible
Now, we calculte the matrix of co-factors
#C=(( | (0,3), (1,0) | , -| (-1,3), (2,0) | , | (-1,0), (2,1) | ),( -| (-1,2), (1,0) | , | (1,2), (2,0) | , -| (1,-1), (2,1) | ),( | (-1,2), (0,3) | , -| (1,2), (-1,3) | , | (1,-1), (-1,0) | ))#
#=((-3,6,-1),(2,-4,-3),(-3,-5,-1))#
We calculate the transpose of matrix #C#
#C^T=((-3,2,-3),(6,-4,-5),(-1,-3,-1))#
The inverse is
#A^-1=C^T/detA=-1/11*((-3,2,-3),(6,-4,-5),(-1,-3,-1))#
#=((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))#
Verification, by doing #A*A^-1#
#A*A^-1=((1,-1,2),(-1,0,3),(2,1,0))*((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))#
#=((1,0,0),(0,1,0),(0,0,1))=I#
Now, we can solve our equation
#((x),(y),(z))=((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))*((-5),(0),(1))#
#=((-12/11),(35/11),(-4/11))#