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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.

The temporary vectorfor 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;