This step is called forward substitution. Proceeding with these substitutions, we obtain the general expressions for the new coefficients $$ \begin{equation} a_{jk}^{(m+1)} = a_{jk}^{(m)}-\frac{a_{jm}^{(m)}a_{mk}^{(m)}}{a_{mm}^{(m)}} \quad j,k=m+1,\dots,n, \tag{17} \end{equation} $$ with \( m=1,\dots,n-1 \) and a right-hand side given by $$ \begin{equation} w_j^{(m+1)} =w_j^{(m)}-\frac{a_{jm}^{(m)}w_m^{(m)}}{a_{mm}^{(m)}}\quad j=m+1,\dots,n. \tag{18} \end{equation} $$ This set of \( n-1 \) elimations leads us to an equations which is solved by back substitution. If the arithmetics is exact and the matrix \( \mathbf{A} \) is not singular, then the computed answer will be exact.
Even though the matrix elements along the diagonal are not zero, numerically small numbers may appear and subsequent divisions may lead to large numbers, which, if added to a small number may yield losses of precision. Suppose for example that our first division in \( (a_{22}-a_{21}a_{12}/a_{11}) \) results in \( -10^{-7} \) and that \( a_{22} \) is one. one. We are then adding \( 10^7+1 \). With single precision this results in \( 10^7 \).