#math/linear-algebra A system $Ax=b$ is `inconsistent` when it does not contain a solution. This exactly when $b$ is not in the column span of $A$. A system is inconsistent if and only if there is a pivot in the last column of an echelon form of the augmented matrix $[A\mid b]$. Equivalently, there is a row $ \begin{pmatrix}0 & 0 & \cdots & 0 & \mid & c\end{pmatrix},\qquad c\neq 0, $ which represents the contradiction $0=c$. Hence we can conclude that $Ax=b$ is false (no solutions exist). A system is `consistent` if it is not inconsistent. > [!Note] On Free Variables > A free variable is simply one where if we set it to any scalar value, the system still holds—not that its value has no effect on the system. For example $x_{1}+x_{2}=3$, setting $x_{2}=t$ means $x_{1}=3-t$ and all solutions are $(3-t,t)$. Having one free variable means a one-dimensional kernel. There is a notion of 'standard free-variable' due to row reduction, as we opt to pick the non-pivot. ## Remarks **I.** If a solution exists, then it is unique if and only if there are no free variables. In other words, the echelon form of the coefficient matrix has a pivot in every column. This is injective, but not necessarily surjective. **II.** Equation $Ax=b$ is consistent for all right sides $b$ if and only if echelon form of the coefficient matrix has a pivot in each row. This is surjective, but not necessarily injective. **III.** Equation $Ax=b$ has a unique solution for any right side $b$ if and only if echelon form of the coefficient matrix $A$ has a pivot in every column and every row. This is bijective.