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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.

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You are solving two systems of linear equations at the same time. Since the two systems have the same coefficient matrix, this is possible by putting two columns next to the coefficient matrix, just like you did.

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.

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Since the matrix of coefficients of your systems are the same we can conclude that if one system has a unique solution the other one also has a unique solution.

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 $A^{-1}$

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