The x and y would be the scores of the spectrum (0,3) , and we have to find them, knowing that the
first loading is (2,-1), and the second (-1,2), so x and y will be the solutions of the equation:
2x - y = 0
-x + 2y = 3
In this case there is a solution, and the residual is cero, but in the case of real spectra there are more variables than two and we try to fit as best as possible the solution to the unknown with the linear combinations of the loadings multiplied by the scores, and the residual is the residual vector e.