Question 929890
Use Cramer's rule.
{{{A=(matrix(3,3,
1,1,4,
1,1,2,
1,-4,-8))}}}
{{{abs(A)=-10}}} 
Since the determinant is not zero, there is a unique solution.
{{{A[x]=(matrix(3,3,
10,1,4,
4,1,2,
36,-4,-8))}}}
{{{abs(A[x])=40}}} 
{{{A[y]=(matrix(3,3,
1,10,4,
1,4,2,
1,36,-8))}}}
{{{abs(A[y])=-20}}} 
{{{A[z]=(matrix(3,3,
1,1,10,
1,1,4,
1,-4,36))}}}
{{{abs(A[z])=-30}}}
.
.
.
{{{x=abs(A[x])/abs(A)=(40)/(-10)=-4}}}
.
.
{{{y=(-20)/(-10)=2}}}
.
.
{{{z=abs(A[z])/abs(A)=(-30)/(-10)=3}}}