Question 1039067
Use Cramer's rule,
{{{A=(matrix(4,4,
1,1,-3,2,
2,-1,2,-3,
3,-2,1,-4,
-4,1,-3,1))}}}
{{{abs(A)=-35}}}
,
,
,
{{{A[w]=(matrix(4,4,
0,1,-3,2,
0,-1,2,-3,
0,-2,1,-4,
0,1,-3,1))}}}
{{{abs(A[w])=0}}}
,
,
,
{{{A[x]=(matrix(4,4,
1,0,-3,2,
2,0,2,-3,
3,0,1,-4,
-4,0,-3,1))}}}
{{{abs(A[x])=0}}}
,
,
,
{{{A[y]=(matrix(4,4,
1,1,0,2,
2,-1,0,-3,
3,-2,0,-4,
-4,1,0,1))}}}
{{{abs(A[y])=0}}}
,
,
,
{{{A[z]=(matrix(4,4,
1,1,-3,0,
2,-1,2,0,
3,-2,1,0,
-4,1,-3,0))}}}
{{{abs(A[z])=0}}}
,
,
,
So then,
{{{w=abs(A[w])/abs(A)=0/35=0}}}
{{{x=abs(A[x])/abs(A)=0/35=0}}}
{{{y=abs(A[y])/abs(A)=0/35=0}}}
{{{z=abs(A[z])/abs(A)=0/35=0}}}
.
.
.
The only other solution is where the system is dependent, where one row is a multiple of other rows.