Reducing Systems With Several Right-Hand Sides

One Reduction for Several Systems

Suppose several linear systems have the same coefficients but different right-hand sides. Reducing each system separately repeats the same work on the coefficient matrix. Instead, place all the right-hand sides beside that matrix and perform one shared row reduction. Each row operation must be applied across the entire matrix, including every right-hand-side column. The resulting columns then show which systems are consistent and, when they are, what their solutions are.

Throughout, scalars are in \(\mathbb{R}\). Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) be a coefficient matrix and let \(\mathbf{b}_1,\dots,\mathbf{b}_q\in\mathbb{R}^m\) be right-hand-side columns. The \(j\)th system is \(\mathbf{A}\mathbf{x}=\mathbf{b}_j\). We collect the right-hand sides into \(\mathbf{B}=\begin{bmatrix}\mathbf{b}_1&\cdots&\mathbf{b}_q\end{bmatrix}\), and write the combined augmented matrix as \(\begin{bmatrix}\mathbf{A}&\vert&\mathbf{B}\end{bmatrix}\). A system is consistent if it has at least one solution; otherwise it is inconsistent.

The shared reduction does not combine the unknown vectors of the different systems. Each right-hand-side column is still read separately: the entries in that column determine the corresponding system’s equations after reduction. What is shared is the sequence of row operations, not the solution or the consistency outcome.

Why the Shared Reduction Works

A row operation is reversible: a row swap undoes itself, scaling a row by a nonzero scalar is undone by scaling by its reciprocal, and adding a multiple of one row to another is undone by subtracting that multiple. Therefore, applying one row operation to an augmented matrix preserves the solution set of each represented system, provided it acts on the coefficient entries and on that system’s right-hand-side entry in the same row. Repeating this reasoning shows that a complete row-reduction sequence preserves every system’s solution set.

Proof. Fix one right-hand-side column and consider the corresponding system. Since each row operation is reversible and is applied to the coefficient entries and this right-hand-side column together, the original system and its reduced system have exactly the same solutions. In the reduced system, a row with all zero coefficients and a nonzero right-hand entry asserts \(0=c\) for some \(c\ne0\), which no vector can satisfy. Thus the system is inconsistent if such a row occurs. Conversely, suppose no such row occurs. In echelon form, any row with no nonzero coefficient has right-hand entry zero, so it imposes only \(0=0\). Assign zero to each free variable. Starting at the last pivot row and proceeding upward, choose the pivot variable to satisfy that row; each choice is possible because its pivot coefficient is nonzero and the variables to its right have already been assigned. This constructs a solution. Therefore the system is consistent. The argument applies separately to each right-hand-side column, proving the test for all columns. \(\square\)

This criterion makes the final consistency check column-specific. A row may contain zero coefficients and a nonzero entry in one right-hand-side column, but zero in another. The first system is inconsistent, while that row causes no contradiction for the second. A single reduction can therefore reveal different outcomes among systems that share the same coefficients.

Reading Several Outcomes from One Reduction

The example illustrates the main efficiency: the coefficient part is reduced only once, but the right-hand-side columns retain their distinct information. In particular, do not decide that every system is inconsistent just because one column produces a contradiction. Check each right-hand-side entry in every zero-coefficient row.

Consistent Systems Can Have Free Variables

The consistency test tells whether solutions exist; it does not by itself say whether a solution is unique. If a reduced system has free variables, a consistent right-hand side can give infinitely many solutions. The same coefficient pattern is used for every right-hand side, while the constants in the pivot equations change from column to column.

A Second Shared Reduction

Common Pitfall: Updating Only Some Columns

A row operation on a combined augmented matrix must be performed on every right-hand-side column. For the matrix in the preceding example, after \(R_2\to R_2-2R_1\), the second right-hand-side entry must become \(5-2(2)=1\). If it is mistakenly left as \(5\), the incorrect intermediate matrix has second right-hand-side column \(\begin{bmatrix}2\\5\end{bmatrix}\) beside the reduced coefficient matrix \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Reading it gives \(x_2=5\), \(x_1=-3\). But substitution into the original system gives \(\mathbf{A}\begin{bmatrix}-3\\5\end{bmatrix}=\begin{bmatrix}2\\9\end{bmatrix}\), not \(\begin{bmatrix}2\\5\end{bmatrix}\).

The correction is to apply the operation to the entire row: \(5-2(2)=1\). The next operation \(R_1\to R_1-R_2\) then gives the correct solution \(\begin{bmatrix}1\\1\end{bmatrix}\) for the second system. A quick check by multiplying each proposed solution by the original coefficient matrix is a reliable way to catch a missed update.

A Reliable Procedure

This procedure separates two questions that are easy to conflate: whether a system has a solution, and what its solution set is. The zero-coefficient rows answer the first question for each right-hand side. The pivot and free variables in the reduced equations answer the second. Because the reduction is shared, the coefficient structure is exposed once, while every column still receives its own consistency decision and solution check.