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Given an inconsistent system of linear equations, it seems always possible to make it consistent by introducing new variables and adding them in convenient places.

For a baby example, given the system of equations \begin{align*} x + y &= 0 \\ x + y &= 1 \end{align*} one can make it consistent by adding a new variable $w$ to get \begin{align*} x + y &= 0 \\ w + x + y &= 1 \end{align*} (I shouldn't have added it to both equations, however.)

Is there a systematic method to achieve this? How many variables do I need to add to a system of $n$ equations in $n$ unknowns? Is there a "degree of inconsistency" (e.g. the number of equations I need to take out to get a consistent system of equations)?

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    $\begingroup$ One standard way is the method of least squares. It will add one variable $w_i$ to each equation, then give you a solution where the vector $(w_1, w_2,\ldots,w_n)$ is as short as possible. $\endgroup$ – Arthur Feb 27 '18 at 14:18
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Write your system like this:

$$a_{11}x_{1} + a_{12}x_{1} + \cdots + a_{1n}x_{n} = b_1$$ $$a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} = b_2$$ $$\vdots$$ $$a_{m1}x_{1} + a_{m2}x_{2} +\cdots + a_{mn}x_{n} = b_m$$

Or equivalently as $Ax = b$:

$$\underbrace{\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}}_{x} = \underbrace{\begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}}_{b}$$

This system being inconsistent is equivalent to the fact that $b$ is not in the linear span of the columns of the matrix $\{A_1, A_2, \ldots, A_n\} \subseteq \mathbb{R}^m$ where $A_j = \begin{pmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj}\end{pmatrix}$.

In particular, the set $\{A_1, A_2, \ldots, A_n\}$ does not span the space $\mathbb{R}^m$, meaning that $$\dim \operatorname{span}\{A_1, A_2, \ldots, A_n\} < m$$

Adding additional unknowns to the system in the way you propose amounts to adding more vectors to this set, until the set is large enough so that $b$ is contained in its linear span.

Namely, adding an extra unknown to the $i$-th equation augments the matrix like this:

$$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & 0\\ a_{21} & a_{22} & \cdots & a_{2n} & 0\\ \vdots & \vdots & \cdots & \vdots & \vdots\\ a_{i1} & a_{i2} & \cdots & a_{in} & 1\\ \vdots & \vdots & \cdots & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} & 0\\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ w\end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}$$

The set of columns is now $\{A_1, \ldots, A_n, e_j\}$. Since the unit vectors $\{e_1, \ldots, e_m\}$ span $\mathbb{R}^m$, it is clear that if you add enough of them into $\{A_1, \ldots, A_n\}$, you will achieve that $b$ is in the span of the matrix columns.

If adding unknowns multiplied by a scalar is allowed, then you can simply add $b$ into the set $\{A_1, \ldots, A_n\}$ so that you have

$$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & b_1\\ a_{21} & a_{22} & \cdots & a_{2n} & b_2\\ \vdots & \vdots & \cdots & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_m\\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ w\end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m\end{pmatrix}$$

which is trivially solvable with $x_1 = \cdots = x_m = 0$ and $w = 1$.

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If you allow coefficients for the new variables, you can always achieve this with one extra variable. Indeed, given the system $$a_{11}x_{1} + a_{12}x_{2} +\cdots + a_{1n}x_{n} = b_1 \\ a_{21}x_{1} + a_{22}x_{2} +\cdots + a_{2n}x_{n} = b_2\\ ...\\ a_{m1}x_{1} + a_{m2}x_{2} +\cdots + a_{mn}x_{n} = b_m\\$$ the new system $$a_{11}x_{1} + a_{12}x_{2} +\cdots + a_{1n}x_{n}+b_1w = b_1 \\ a_{21}x_{1} + a_{22}x_{2} +\cdots + a_{2n}x_{n}+b_2w = b_2\\ ...\\ a_{m1}x_{1} + a_{m2}x_{2} +\cdots + a_{mn}x_{n}+b_mw = b_m\\$$ is always consistent.

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