# How do I interpret the results of Gaussian elimination of two linear systems that have the same coefficient matrix?

The systems of linear equations $$x_1 - x_2 + 2x_3 - x_4= 6$$ $$x_1 - x_3 + x_4 = 4$$ $$2x_1 + x_2 + 3x_3 - 4x_4 = -2$$ $$-x_2 + x_3 - x_4 = 5$$ and $$x_1 - x_2 + 2x_3 - x_4= 1$$ $$x_1 - x_3 + x_4 = 1$$ $$2x_1 + x_2 + 3x_3 - 4x_4 = 2$$ $$-x_2 + x_3 - x_4 = -1$$ give $$\begin{bmatrix}1&-1&2&-1&|&6&1\\1&0&-1&1&|&4&1\\2&1&3&-4&|&-2&2\\0&-1&1&-1&|&5&-1\end{bmatrix}$$ $$R_1 + (-1)R_2\rightarrow R_2$$ $$R_1 + (-12)R_3\rightarrow R_3$$ $$\begin{bmatrix}1&-1&2&-1&|&6&1\\0&-1&3&-2&|&2&0\\0&-3/2&1/2&1&|&7&0\\0&-1&1&-1&|&5&-1\end{bmatrix}$$ $$R_2 + (-2/3)R_3\rightarrow R_3$$ $$R_2 + (-1)R_4\rightarrow R_4$$ $$\begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&-1&3&-2&|&2&0\\ 0&0&8/3&-8/3&|&-8/3&0\\ 0&0&2&-1&|&3&1\end{bmatrix}$$ $$(-1)R_2 \rightarrow R_2$$ $$(3/8)R_3 \rightarrow R_3$$ $$R_3 + (-8/6)R_4 \rightarrow R_4$$ $$\begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&1&-3&2&|&-2&0\\ 0&0&1&-1&|&-1&0\\ 0&0&0 &-4/3&|&4/3&-4/3\end{bmatrix}$$ $$(-3/4)R_4 \rightarrow R_4$$ $$\begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&1&-3&2&|&-2&0\\ 0&0&1&-1&|&-1&0\\ 0&0&0&1&|&-1&1\end{bmatrix}$$

The problem is that I don't understand how $$x_4$$ can be both $$-1$$ and $$1$$, or how that applies to the rest of the equations.

In the first system of equations, $$x_4=-1$$; and in the second system of equations, $$x_4=1$$. You should not mix the solutions of the two systems.
Solutions are not the same unless the vectors on the right hand side of equations are also the same . Your example is providing two different values for $$x_4$$ because you are solving two different systems simultaneously.
There is a linearity relation for solutions of systems with the same coefficient matrix and that comes from $$x=A^{-1}b$$ and the linearity properties of the matrix $$A^{-1}$$