Contradictory Rows and Inconsistent Systems

When a Row Says Something Impossible

Row reduction can turn a system of equations into a form where its status is immediate. A particularly decisive outcome is a row whose variable coefficients are all zero but whose augmented entry is nonzero. That row represents an equation such as \(0=3\), which no choice of variables can satisfy. The key is to see why a row produced during reduction also rules out solutions of the original system: each row operation must preserve the system’s solution set.

For a system in variables \(x_1,\dots,x_n\), its augmented matrix records the coefficients and constants in the equations. For example, the row \([\,2\quad -1\mid 5\,]\) represents \(2x_1-x_2=5\). A vector \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\) is a solution if its entries satisfy every equation represented by the rows.

By contrast, a row with zero coefficients and a zero augmented entry represents \(0=0\). This equation holds for every choice of variables, so it is not a contradiction. The augmented entry matters: it is what distinguishes a row that imposes no restriction from one that makes the system impossible.

Why Row Operations Preserve Solutions

Row reduction operates on augmented matrices using three elementary row operations: interchange two rows, multiply a row by a nonzero scalar, or add a multiple of one row to another. In each operation, the augmented entry must be treated in the same way as the coefficient entries. The resulting matrix represents a system equivalent to the previous one.

Proof. Interchanging two rows only changes the order of two equations, so a vector satisfies all equations before the interchange exactly when it satisfies all equations afterward. Multiplying one row by a nonzero scalar replaces an equation \(L=b\) by \(cL=cb\), where \(c\ne0\). If \(L=b\), then \(cL=cb\); conversely, if \(cL=cb\), division by \(c\) gives \(L=b\). Finally, replacing one row by itself plus a multiple of another replaces an equation \(L_2=b_2\) by \(L_2+kL_1=b_2+kb_1\). Any vector satisfying the original two equations satisfies this new equation. Conversely, if it satisfies \(L_1=b_1\) and \(L_2+kL_1=b_2+kb_1\), subtracting \(kL_1=kb_1\) gives \(L_2=b_2\). Thus each operation preserves exactly the same solutions. Applying this reasoning at every step proves the claim for any finite sequence of row operations. \(\square\)

This equivalence is why a contradiction discovered after row reduction is decisive. The row does not need to have appeared in the original system: it is enough that it is obtained through valid operations, with the augmented entries updated correctly.

Proof. Suppose row reduction produces a row with all coefficient entries zero and augmented entry \(b\ne0\). Any solution of the reduced system would have to satisfy the equation \(0=b\). The left side equals zero for every assignment of the variables, while \(b\ne0\), so no vector satisfies that equation. The reduced system therefore has no solution. By the row-operation theorem, the reduced system and the original system have the same solution set. Hence the original system also has no solution. \(\square\)

Finding a Contradictory Row

In a larger system, the contradiction may appear only after combining several equations. The same principle applies: a row operation constructs an equivalent equation, and the final row can expose a disagreement among the original requirements.

A Zero Row Is Not Always a Contradiction

The phrase “zero row” can refer to two different rows in an augmented matrix. If every entry, including the augmented entry, is zero, the row says \(0=0\). If only the coefficient entries are zero while the augmented entry is nonzero, the row says \(0=b\) with \(b\ne0\), and the system is inconsistent. Inspect the entire row, not just the coefficients.

Common Pitfall: Changing Only the Coefficients

When performing a row operation, apply it to the augmented entry as well as to the coefficient entries. If the augmented entry is not updated according to the operation, the resulting row may fail to represent the equation obtained by that operation. A changed coefficient row alone is not enough to make a valid row operation on an augmented matrix.

For instance, begin with the consistent system in the previous example. A mistaken calculation might apply \(R_2\to R_2-2R_1\) to the coefficients but leave the augmented entry \(4\) unchanged:

This would falsely claim that the original system implies \(0=4\). The correction is to perform the same subtraction on the augmented entry: \(4-2(2)=0\). Therefore the valid result is \(\begin{bmatrix}1&1&\mid&2\\0&0&\mid&0\end{bmatrix}\), which records \(0=0\), as shown in the worked example.

Sometimes the augmented entry does not change during a row operation, and that can be correct. For example, if the entry being changed is already zero and the operation combines it with another zero entry, the result remains zero. The requirement is not that every entry must visibly change; it is that each entry, including the augmented one, receives the prescribed operation.

Using the Contradiction Carefully

A contradictory row provides a complete reason for inconsistency: every possible assignment of the variables makes its left side zero, but its right side is nonzero. The row-operation theorem transfers that conclusion from the reduced system back to the original one. This reasoning does not depend on the system having two, three, or any particular number of variables.

When checking work, verify both parts of the argument. First, check that each row operation was applied to the entire augmented row. Second, interpret the resulting row as an equation. A row \([\,0\ \cdots\ 0\mid b\,]\) with \(b\ne0\) rules out every solution; a row \([\,0\ \cdots\ 0\mid0\,]\) does not. The latter is simply an equation that is always true, so it cannot by itself establish inconsistency.