Question 941993
Use Cramer's rule,
{{{A=(matrix(3,3,2,-5,3,1,4,-2,1,-2,-4))}}}
{{{abs(A)=-68}}}
.
.
{{{A[x]=(matrix(3,3,1,-5,3,9,4,-2,-5,-2,-4))}}}
{{{abs(A[x])=-204}}}
.
.
{{{A[y]=(matrix(3,3,2,1,3,1,9,-2,1,-5,-4))}}}
{{{abs(A[y])=-136}}}
.
.
{{{A[z]=(matrix(3,3,2,-5,1,1,4,9,1,-2,-5))}}}
{{{abs(A[z])=-68}}}
So then,
{{{x=abs(A[x])/abs(A)=(-204)/(-68)=3}}}
{{{y=abs(A[y])/abs(A)=(-136)/(-68)=2}}}
{{{z=abs(A[z])/abs(A)=(-68)/(-68)=1}}}