Coefficient Matrices and Augmented Matrices

Two Matrices, One Linear System

A linear system has coefficients and right-hand sides, and those play different roles. The coefficient matrix records how the variables occur; the right-hand side records the values those expressions must equal. Keeping the two parts distinct is useful, but separating them during row reduction can lose the very information needed to decide whether the system has solutions.

Consider a system of \(m\) equations in \(n\) variables over \(\mathbb{R}\). Write its variable vector as the column \(\mathbf{x}\in\mathbb{R}^n\), its right-hand-side vector as \(\mathbf{b}\in\mathbb{R}^m\), and its coefficient matrix as \(\mathbf{A}\in\mathbb{R}^{m\times n}\). The system is written \(\mathbf{A}\mathbf{x}=\mathbf{b}\).

For example, the system \(x_1+2x_2=3,\;2x_1+4x_2=6\) has coefficient matrix \(\begin{bmatrix}1&2\\2&4\end{bmatrix}\) and augmented matrix \(\begin{bmatrix}1&2&\mid&3\\2&4&\mid&6\end{bmatrix}\). The coefficient matrix describes the left sides of the equations, but by itself it does not say whether those left sides must equal \(3,6\), or some other values.

Row Operations Must Include the Right-Hand Side

Each row of the augmented matrix represents one equation. Consequently, an elementary row operation on a system must act on the whole row, including its right-hand side. Swapping equations, multiplying an equation by a nonzero number, or adding a multiple of one equation to another produces an equivalent system: the new equations have exactly the same solutions as the old ones.

Proof. It is enough to verify that each allowed row operation preserves the solutions, because a sequence of operations is a succession of such steps. Swapping two rows only changes the order of two equations, so a vector satisfies the system before the swap exactly when it satisfies the system after it. Multiplying one row by a nonzero scalar \(c\) replaces an equation \(L(\mathbf{x})=d\) by \(cL(\mathbf{x})=cd\). The new equation follows from the old one; conversely, division by \(c\ne0\) recovers the old equation. Finally, replacing one row by itself plus \(c\) times another row replaces \(L_1(\mathbf{x})=d_1\) by \(L_1(\mathbf{x})+cL_2(\mathbf{x})=d_1+cd_2\). Any solution of the old equations satisfies this new equation. Conversely, subtracting \(c\) times the unchanged second equation recovers \(L_1(\mathbf{x})=d_1\). Thus each operation preserves the solution set, and so does their sequence. \(\square\)

This result explains why the vertical bar in \([\mathbf{A}\mid\mathbf{b}]\) is a visual separator, not a boundary beyond which row operations stop. If the coefficients in a row change, the right-hand side in that row must change according to the same operation. Reducing \(\mathbf{A}\) alone can be useful for studying its pivot positions, but it does not, on its own, carry out the equivalent-system operations for a particular \(\mathbf{b}\).

What the Right-Hand Side Tells You

A row with zeros in all coefficient columns reads \(0=c\). If \(c\ne0\), it is a contradiction; if \(c=0\), it is an equation that imposes no condition on the variables. The distinction is invisible in the coefficient part alone, because both rows have zeros there.

Proof. If such a row occurs, its equation is \(0=c\) with \(c\ne0\), which no vector \(\mathbf{x}\) can satisfy. The reduced system, and therefore the original system by the preceding theorem, is inconsistent. For the converse, suppose there is no such row. Every row with zero coefficients then has right-hand side zero, so it gives \(0=0\). Each remaining row has at least one nonzero coefficient. In echelon form, back-substitution determines the pivot variables in order from the last nonzero row upwards. Assign every free variable the value \(0\); each pivot variable can then be assigned the value required by its row. This produces a solution to all nonzero rows, while the zero rows impose no further conditions. The reduced system is therefore consistent, and the original system is consistent as well. \(\square\)

Reducing Coefficients While Tracking the Right-Hand Side

Sometimes it is convenient to describe row reduction as operations on the coefficient matrix. That is valid for solving a system only if the effect of the same operations on \(\mathbf{b}\) is retained. One way to do this is to write the augmented matrix and reduce it as a whole. Another is to keep a separate column beside the coefficient matrix and perform every row operation on both.

What Can Be Learned from the Coefficient Matrix Alone?

Reducing \(\mathbf{A}\) alone can still answer questions about the coefficient columns, such as which columns have pivots. But the reduction only describes transformed coefficients. For a particular system \(\mathbf{A}\mathbf{x}=\mathbf{b}\), the rows of those transformed coefficients must be paired with the right-hand sides produced by the same row operations. Appending the original \(\mathbf{b}\) after reducing \(\mathbf{A}\) is not generally valid: the reduced coefficient rows may no longer be paired with the original right-hand-side entries.

Common Mistake: Leaving the Last Column Alone

A frequent error is to perform the row operation on the coefficient entries but not on the right-hand side. For the inconsistent system \(x_1+2x_2=3,\;2x_1+4x_2=7\), the correct operation is \(R_2\to R_2-2R_1\) on the augmented matrix:

A wrong calculation would change the second row’s coefficient entries to zeros but leave its right-hand side as \(7\), giving \([0\ \ 0\mid 7]\). That is not the result of the row operation: the new right-hand side must be \(7-2(3)=1\). Although both nonzero values would signal a contradiction in this example, the incorrect row is still not equivalent to the original equation. The correction is to apply each operation to the entire row, including every entry after the vertical bar.

Keep three checks in mind when solving a system:

• Keep the final column visibly separate from the coefficient columns, but include it in every row operation.
• Do not treat a pivot in the final column as a pivot variable; it signals a contradictory row when the coefficient entries in that row are all zero.
• Substitute a proposed solution into the original equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\), not merely into a partially reduced system.