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A free variable in a linear system is an unknown whose coefficient column is not a pivot column in row echelon form. When the linear system is consistent, it can be assigned any value in the field; the pivot unknowns are then determined by those choices. This use of “free” differs from a free variable in mathematical logic.

Subtracting the first equation from the third gives $4x-4z=6$, so $z=x-3/2$. The second equation then becomes $(\alpha -2)x=\beta -3$. The third fixes $y=7(1-x)/5$. Therefore the three possibilities areFor the first case the coefficient determinant is $20(\alpha -2)\ne 0$, confirming uniqueness. At $\alpha =2$ its matrix rank is two. The augmented matrix has the same matrix rank only for $\beta =3$, giving one free variable in a linear system; otherwise the linear system is inconsistent.

The required matrix form of the linear system isSubtract the first row from the third and expand the determinant along the resulting row:Thus the linear system has a unique solution, for every real $\mu$, precisely when $\boxed{\lambda \ne 1,2}$.

For $\lambda =0$, the second equation gives $x=\mu$. Subtracting the first equation from the third gives $-y=1$, and the first equation then determines $z$:For $\lambda =1$, the first and third left-hand sides are identical but their right-hand sides are $1$ and $2$. HenceFor $\lambda =2$, subtracting the first equation from the third gives $y=1$. The first equation now gives $x+2z=0$, while the second requires $x+2z=\mu$. ConsequentlyThe free variable in a linear system in the consistent singular case describes a line of solutions. These direct reductions distinguish an inconsistent linear system from one with infinitely many solutions, which a zero determinant alone cannot do.

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

Work over the rational numbers, allowing unknowns in the real numbers when interpreting the geometry. Consider four equations in three unknowns:The third row of $A$ is the sum of its first two rows, and the fourth is their difference, second minus first. The same relations must hold on the right: consistency requires $\tau =1+4=5$ and $3=4-1$. The latter condition already holds. Gaussian elimination makes this compatibility test explicit while exposing the remaining freedom.

Apply the elementary row operations in this order, using each row's current value:then $R_3\leftarrow R_3-R_2$ and $R_4\leftarrow R_4-R_2$. The resulting augmented matrix is in row echelon form for every $\tau$:When $\tau \ne 5$, the third row has its pivot element in the augmented column. It says $0=\tau -5$, so the linear system has no solution. When $\tau =5$, that entire row vanishes and only two pivot elements remain. More equations than unknowns therefore do not force a unique solution: some equations may repeat information.

The final operation $R_1\leftarrow R_1-2R_2$ reduces the coefficient block toThe product of all six elementary matrices isDirect matrix multiplication verifies both displayed transformations. Its determinant is one, so this transformation is reversible. In particular, the third row of $E$ extracts the discrepancy between the third equation and the sum of the first two.

For $\tau =5$, $[R\mid Eb]$ is the reduced row echelon form of the full augmented matrix. For $\tau \ne 5$, only its coefficient block is reduced: scale the third row by $1/(\tau -5)$ and clear the entries above its augmented pivot element. The full reduced row echelon form is thenThe last column is part of the matrix when testing the echelon conditions. Calling $[R\mid Eb]$ fully reduced before clearing that column would miss this distinction.