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#
