# Proof

• By hypothesis, $u_1,\ldots,u_n$ be a solution of the inhomogeneous system of linear equations, i.e. $$\sum_{\nu=1}^n \alpha_{\mu\nu}u_\nu=\beta_\mu\quad\quad\text{ for all }\,\mu=1,\ldots,m.\quad\quad ( \dagger )$$
• The same holds for other solutions (if any) $w_1,\ldots,w_n$: $$\sum_{\nu=1}^n \alpha_{\mu\nu}w_\nu=\beta_\mu\quad\quad\text{ for all }\,\mu=1,\ldots,m.\quad\quad ( \dagger \dagger )$$
• Therefore, subtracting the equations of both systems $( \dagger )$ and $( \dagger \dagger )$ line by line from each other results in $$\sum_{\nu=1}^n \alpha_{\mu\nu}(u_\nu -w_\nu)=0\quad\quad\text{ for all }\,\mu=1,\ldots,m,$$ which means that $h_\nu:=u_\nu-w_\nu$ for all $\nu=1,\ldots,n$ is a solution of the corresponding homogeneous system.
• Conversely, if $h_1,\ldots,h_n$ is any solution of the corresponding homogeneous system, then we have $$\sum_{\nu=1}^n \alpha_{\mu\nu}h_\nu=0\quad\quad\text{ for all }\,\mu=1,\ldots,m.\quad\quad (\dagger \dagger \dagger )$$
• Therefore, adding the equations of both systems $( \dagger \dagger\dagger )$ and $( \dagger )$ line by line results in $$\sum_{\nu=1}^n \alpha_{\mu\nu}(u_\nu + h_\nu)=\beta_\mu\quad\quad\text{ for all }\,\mu=1,\ldots,m,$$ which means that $w_\nu:=u_\nu+h_\nu$ for all $\nu=1,\ldots,n$ is a solution of the corresponding inhomogeneous system.
• Alltogether, it follows that all solutions of the inhomogeneous system are given by $w_\nu:=u_\nu\pm h_\nu$ for all $\nu=1,\ldots,n$, where $u_1,\ldots,u_n$ is one of its solutions and $h_1,\ldots,h_n$ is a solution the corresponding homogeneous system.

Github: ### References

#### Bibliography

1. Knabner, P; Barth, W.: "Lineare Algebra - Grundlagen und Anwendungen", Springer Spektrum, 2013