In this example, the extended coefficient matrix contains an upper triangular matrix. $$\left(\begin{array}{ccccccc|c}\alpha_{11}& \alpha_{12}&\ldots&\alpha_{1r}&\alpha_{1,r+1}&\ldots&\alpha_{1n}&\beta_1\\ 0& \alpha_{22}&\ldots&\alpha_{2r}&\alpha_{2,r+1}&\ldots&\alpha_{2n}&\beta_2\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0& 0 &\ldots&\alpha_{rr}&\alpha_{r,r+1}&\ldots&\alpha_{rn}&\beta_r\\ \vdots& \vdots &\ldots&0&0&\ldots&0&\beta_{r+1}\\ \vdots& \vdots &\ldots&\vdots&\vdots&\ldots&\vdots&\vdots\\ 0& \ldots &\ldots&0&0&\ldots&0&\beta_m\\ \end{array}\right)$$
In this SLE, there exists an $r\in \{1,\ldots,\min(m,n)\},$ for which $\alpha_{jj}\neq 0$ for $j=1,\ldots,r,$ and $\alpha_{jj}=0$ for $j=r+1,\ldots,m.$
We have seen already in the previous example that if $\beta_j\neq 0$ for at least one of the "zero-lines" $j=r+1,\ldots m,$ then the whole SLE has no (simultaneous) solution. So let us assume $\beta_j\neq 0$ for all $j=r+1,\ldots m.$ In this case, the solution of the SLE can be found as follows:
(Please note that if $r < n$, the simultaneous solution has $n-r$ degrees of freedom. Otherwise (if $r=n$), the solution is unique, as the following steps show:)
$$x_r:=\frac 1{\alpha_{rr}}\left(\beta_r-\sum_{j=r+1}^n \alpha_{rj}x_j\right)$$ For $r=n$ the equation has the form $$x_n:=\frac {\beta_n}{\alpha_{nn}}$$ 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 $r-2, r-3,\ldots, 1.$
This solving approach is called backward substitution and can be written as the following formula:
$$x_i:=\frac 1{\alpha_{ii}}\left(\beta_i-\sum_{j=i+1}^n \alpha_{ij}x_j\right),\quad\quad\text{for }i=r,r-1,\ldots,1.$$
Definitions: 1
Examples: 2
Sections: 3
