Question 1073711
Using Cramer's rule,
{{{A=(matrix(2,2,
1,-2,
2,-1))}}}
{{{abs(A)=3}}}
.
.
.

{{{A[x]=(matrix(2,2,
5,-2,
6,-1))}}}
{{{abs(A[x])=7}}}
.
.
.
{{{A[y]=(matrix(2,2,
1,5,
2,6))}}}
{{{abs(A[y])=-4}}}
.
.
.
So then,
{{{x=abs(A[x])/abs(A)=7/3}}}
{{{y=abs(A[y])/abs(A)=-4/3}}}