In order to solve (10.28), we first determine the vector Z such that

(10.31)

and then y such that

(10.32)

Since the vector y thus determined satisfies (10.28).

In order to carry out this method, we note that (10.30) is equivalent to the relations

(10.33)

(10.34)

(10.35)

(10.36)

Relation (10.36) immediately yields relations (10.34) and (10.35) may be rearranged to yield and recursively, as follows:

(10.37)

(10.38)

(10.39)

The algorithm (10.38) and (10.39) breaks down when . If this is the case, then, denoting by the determinant of the matrix A, we have, and A is singular.