A similar example, in which the extended coefficient matrix contains a lower triangular matrix. $$\left(\begin{array}{lllllll|c}0& 0&\ldots&0&\ldots&\ldots&0&\beta_1\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0& 0&\ldots&0&\ldots&\ldots&0&\beta_{r-1}\\ \alpha_{r1}& \alpha_{r2} &\ldots&\alpha_{rr}&0&\ldots&0&\beta_r\\ \alpha_{r+1,1}&\alpha_{r+1,2}&\ldots&\alpha_{r+1,r}&\alpha_{r+1,r+1}&\ldots&0&\beta_{r+1}\\ \vdots& \vdots &\ldots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \alpha_{m1}& \ldots &\ldots&\alpha_{mr}&\alpha_{m,r+1}&\ldots&\alpha_{mn}&\beta_m\\ \end{array}\right)$$
In this SLE, there exists an $r\in \{1,\ldots,\min(m,n)\},$ for which $\alpha_{jj}= 0$ for $j=1,\ldots,r-1$ and $\alpha_{jj}\neq 0$ for $j=r,\ldots,m.$
Similarly as the previous example, if $\beta_j\neq 0$ for at least one of the "zero-lines" $j=1,\ldots r-1,$ then the whole SLE has no (simultaneous) solution. Thus, assume $\beta_j\neq 0$ for all $j=1,\ldots, r-1.$ In this case, the solution of the SLE can be found as follows:
(Please note that if $r > 1$, the simultaneous solution has $r-1$ degrees of freedom. Otherwise (if $r=1$), the solution is unique, as the following steps show:)
$$x_r:=\frac 1{\alpha_{rr}}\left(\beta_r-\sum_{j=1}^{r-1} \alpha_{rj}x_j\right)$$ For $r=1,$ the equation has the form $$x_1:=\frac {\beta_1}{\alpha_{11}}$$ 1. Since $x_r$ is now known, we can subtitute it in the $r+1$-th equation and find a value for the unknown $x_{r+1}.$ 1. We can repeat this process for all remaining unknowns, i.e. those with the indices $r+2, r+3,\ldots, \min(m,n).$
This solving approach is called forward substitution and can be written as the following formula:
$$x_i:=\frac 1{\alpha_{ii}}\left(\beta_i-\sum_{j=1}^{\min(m,n)} \alpha_{ij}x_j\right),\quad\quad\text{for }i=1,2,\ldots,\min(m,n).$$