We can express this by splitting the equations up into a coefficient matrix with a vector for the variables and a vector for the solutions.
Avec(u) = vec(v)
where A is the coefficient matrix. Assuming A is invertible we can left multiply both sides of the equation by the inverse of A, denoted A^(-1) to obtain
Ivec(u) = A^(-1)vec(v)
where I is the n x n identity matrix.
For a 2xx2 matrix
A = ((a,b),(c,d))
The inverse of A is given by:
A^(-1) = (1)/(ad-bc)((d,-b),(-c,a))
In the context of this problem we have
((1,-3),(-1,1))((x),(y)) = ((-5),(1))
so A = ((1,-3),(-1,1))
A^(-1) = -1/2((1,3),(1,1))
Just to check that this is correct:
A^(-1)A = -1/2((1,3),(1,1))((1,-3),(-1,1))
=-1/2((-2,0),(0,-2)) = ((1,0),(0,1)) as required
so we have:
((x),(y)) = -1/2((1,3),(1,1)) ((-5),(1))
((x),(y)) = -1/2((-2),(-4)) = ((1),(2))
therefore x = 1 and y = 2