An Inconsistent-System Reduction Workshop

What a Contradictory Row Tells Us

When a system has no solution, the obstruction can be hidden among several equations. Row reduction combines equations in ways that reveal the obstruction directly: a row with zero coefficients on every unknown but a nonzero right-hand side asserts that zero equals a nonzero number. This workshop focuses on finding that row, interpreting exactly what it says, and checking that the contradiction really follows from the original system.

We consider linear systems over \(\mathbb{R}\), with unknowns such as \(x,y,z\). The augmented matrix records both the coefficients and the right-hand sides of the equations. For example, the row \(\begin{bmatrix}0&0&|&5\end{bmatrix}\) represents \(0x+0y=5\), or simply \(0=5\). The right-hand side is essential: a row of zero coefficients is contradictory only when its augmented entry is nonzero.

Here, unknowns are the values we seek. A parameter, when present, is a fixed scalar that may affect the equations but is not one of the unknowns being solved for. We will keep that distinction explicit: a claim that a system is inconsistent for every parameter value means that no assignment of the unknowns works, regardless of which permitted parameter value has been fixed.

Why Row Reduction Keeps the Same Solutions

Row reduction is useful for detecting inconsistency because each elementary row operation replaces equations by equivalent equations. The three allowed operations are swapping two rows, multiplying a row by a nonzero scalar, and adding a scalar multiple of one row to another. The corresponding equation changes are reversible, so they do not alter which assignments satisfy the entire system.

Proof. Consider one row operation at a time. Swapping two rows only changes the order of the equations, so an assignment satisfies the system before the swap exactly when it satisfies the system after the swap. Multiplying one equation by a nonzero scalar preserves its solutions, because the equation can be recovered by multiplying by the reciprocal scalar. For the third operation, replace equation \(E_i\) by \(E_i+cE_j\), leaving \(E_j\) unchanged. If an assignment satisfies both \(E_i\) and \(E_j\), it satisfies \(E_i+cE_j\). Conversely, if it satisfies \(E_i+cE_j\) and \(E_j\), subtracting \(cE_j\) recovers \(E_i\). Thus each operation preserves the solutions, and a sequence of such operations preserves them as well. \(\square\)

This gives a direct interpretation of a contradictory row. If row reduction produces \(0=c\) with \(c\ne0\), no assignment satisfies the reduced system. Since the row operations preserve the solution set, no assignment satisfies the original system either. The contradiction is not an artifact of changing the equations.

Proof. If such a row occurs, the reduced system contains the equation \(0=c\) for \(c\ne0\), which no assignment can satisfy. By the result just proved, the original system is inconsistent. For the converse, suppose the echelon form has no such row. Every row with all-zero coefficient entries must then have augmented entry zero, so it represents \(0=0\) and imposes no condition. Every other row has a leading nonzero coefficient in an unknown column. Set any unknowns in nonpivot columns equal to zero. Starting with the last nonzero row and proceeding upward, solve each pivot equation for its leading unknown: all unknowns to its right have already been assigned values, so this determines the pivot unknown. This produces values satisfying every nonzero row, while rows \(0=0\) are automatically satisfied. Hence the system has a solution. Therefore it is inconsistent exactly when a contradictory row occurs. \(\square\)

Finding the Contradiction in Practice

When a Parameter Is Present

A parameter may appear in the coefficients or right-hand sides while the unknowns remain the quantities being solved for. During row reduction, treat a fixed parameter as a scalar. If the contradiction does not depend on its value, the system is inconsistent for every permitted parameter choice. If the contradiction does depend on it, the resulting condition tells us which parameter values yield inconsistency.

Common Pitfall: Dropping the Augmented Entry

A zero row in the coefficient columns does not, by itself, mean that the row is harmless. The augmented entry determines whether that row says \(0=0\) or \(0=c\) for a nonzero \(c\). For example, a student might reduce the first worked example to \(\begin{bmatrix}1&2&|&3\\0&0&|&1\end{bmatrix}\), look only at the two zero coefficients in the second row, and incorrectly write “the second row is \(0=0\), so it can be ignored.” That calculation drops the entry \(1\) and changes the equation.

The correction is to read the full augmented row: its coefficient entries give \(0x+0y\), and its last entry gives \(1\). It represents \(0=1\), not \(0=0\). This distinction is the entire test for a contradictory row. By contrast, a row \(\begin{bmatrix}0&0&|&0\end{bmatrix}\) really does represent \(0=0\); it supplies no new restriction and is not a contradiction.

When finishing a reduction, check that the row operation was applied to the augmented entry as well as to every coefficient. Then translate any all-zero coefficient row into an equation before deciding what it means. This practice both identifies the exact contradiction and guards against mistaking an inconsistent system for one that merely has a redundant equation.