Recognizing Inconsistency as Failed Span Membership

When a Target Cannot Be Made from the Columns

A system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) asks whether some combination of the columns of \(\mathbf{A}\) equals the target vector \(\mathbf{b}\). Sometimes the equations expose an obstruction: one equation says that a certain quantity is zero, while the same system forces it to be nonzero. In row reduction, that conflict appears as a row of zeros in the coefficient part and a nonzero entry in the right-hand side.

Throughout, matrices and vectors have real entries. If \(\mathbf{A}\in\mathbb{R}^{m\times n}\), write its columns as \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The span of these columns is the set of all their linear combinations:

By the definition of matrix-vector multiplication, \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). Thus solving \(\mathbf{A}\mathbf{x}=\mathbf{b}\) means finding a combination of the columns equal to \(\mathbf{b}\). The system is consistent if it has at least one solution, and inconsistent if it has no solution.

A row such as \([0\quad 0\quad 0\mid 1]\) is not a solution with free variables. It asserts \(0=1\), which no choice of variables can satisfy. The key is to inspect the augmented matrix, including its final column: a zero row in the coefficient part alone does not tell us whether the system is consistent.

Reading the Contradiction in a Reduced Matrix

Row reduction is useful because elementary row operations preserve the solution set of an augmented system. Swapping two rows merely changes their order; multiplying a row by a nonzero number gives an equivalent equation; and adding a multiple of one row to another replaces an equation by an equivalent one when the other equation is retained. Each operation is reversible, so it does not add or remove solutions. The standard result that row operations preserve augmented-system solutions is proved in the later material on that subject.

Consequently, a contradiction found after row reduction was already present in the original system, even if its meaning was less visible there. Conversely, if the reduced system has no contradictory row, it is consistent: assign zero to each free variable and determine the pivot variables by working upward through the equations. A row with all coefficient entries zero must then have right-hand side zero; otherwise it would be a contradictory row.

Proof. First consider the reduced augmented matrix. If it contains a row \([0\ \cdots\ 0\mid c]\) with \(c\ne0\), the corresponding equation is \(0=c\), so the reduced system has no solution. Since row operations preserve the solution set, the original system has no solution either.

For the converse, suppose the reduced augmented matrix has no contradictory row. In each nonzero row, the first nonzero entry is a pivot in a coefficient column; a pivot in the augmented column would give a row whose coefficient entries are all zero and whose final entry is nonzero, contrary to the assumption. Assign zero to every variable in a nonpivot column. Starting from the last pivot row and working upward, choose each pivot variable to satisfy its row. Rows with all coefficient entries zero have right-hand side zero, since there is no contradictory row, so they impose no additional restriction. This constructs a solution. Therefore the original system is consistent. Finally, \(\mathbf{A}\mathbf{x}\) is a linear combination of the columns with coefficients given by \(\mathbf{x}\). Hence a solution exists exactly when \(\mathbf{b}\) is in their span, proving the equivalent statement. \(\square\)

The theorem is a test and an interpretation at once. The row certifies that no coefficients can produce the target, while the column-span statement describes the same failure without referring to a particular row reduction.

Worked Examples of Inconsistency

A Common Mistake: Dropping the Right-Hand Side

A frequent error is to reduce only \(\mathbf{A}\), then treat a zero row in the reduced coefficient matrix as proof that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent. The conclusion does not follow: the same row operations must be applied to the right-hand side. For example, take \(\mathbf{A}=\begin{bmatrix}1&1\\2&2\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}1\\3\end{bmatrix}\). Reducing the coefficient matrix gives

It would be wrong to conclude that the system is consistent just because the second row is zero: that calculation omitted the transformed right-hand side. Reducing the augmented matrix gives the correct calculation:

The resulting equation \(0=1\) proves inconsistency. In the original system, the first equation gives \(x_1+x_2=1\), so twice its left side equals \(2\), whereas the second equation requires that doubled left side to equal \(3\). The augmented column records exactly this conflict.

Using the Test Carefully

When a reduced augmented matrix contains a contradictory row, it gives a conclusive negative answer: the system has no solution, and the target is not in the column span. The row itself need not appear at the bottom of an intermediate matrix; row operations can move it, and further reduction can put it in the final position. What matters is that its coefficient entries are all zero while its right-hand entry is nonzero.

Keep the two equivalent viewpoints distinct. The row-reduction viewpoint finds a contradiction in equations; the column-space viewpoint says no linear combination of the columns equals \(\mathbf{b}\). Neither says that the columns themselves are zero, nor does inconsistency mean there are “too few variables” in every case. The examples show that even a system with several variables can fail because its equations impose incompatible requirements.