The system can be solved in the usual manner by introducing a temporary vector :

We have a choice between several equivalent ways of solving the system:

The first and fourth formulae are not suitable since they require both multiplication and division with ; the difference between the second and third is only one of ease of coding. In this section we use the third formula; in the next section we will use the second for the transpose system solution.

Both halves of the solution have largely the same structure as the matrix vector multiplication.

for i = 1, n sum = 0 for j = row_ptr(i), diag_ptr(i)-1 sum = sum + val(j) * z(col_ind(j)) end; z(i) = pivots(i) * (x(i)-sum) end; for i = n, 1, (step -1) sum = 0 for j = diag(i)+1, row_ptr(i+1)-1 sum = sum + val(j) * y(col_ind(j)) y(i) = z(i) - pivots(i) * sum end; end;The temporary vector

Mon Nov 20 08:52:54 EST 1995