Consistency as a Question About Reachability
For a matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\), the coefficient matrix determines which right-hand sides can occur. Each product \(\mathbf{A}\mathbf{x}\) is a linear combination of the columns of \(\mathbf{A}\). Thus, asking whether the system has a solution is the same as asking whether \(\mathbf{b}\) can be formed from those columns. This viewpoint is useful when comparing several systems that share the same coefficient matrix: each right-hand side can be tested against the same column space.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\), and let \(\mathbf{b}\in\mathbb{R}^m\). The column space of \(\mathbf{A}\) is \(\operatorname{Col}\mathbf{A}=\operatorname{span}\{\mathbf{a}_1,\dots,\mathbf{a}_n\}\), the set of all linear combinations of its columns. The system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent if there is at least one \(\mathbf{x}\in\mathbb{R}^n\) satisfying the equation; otherwise it is inconsistent. The matrix-vector product is \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\).
Proof. Suppose first that the system is consistent. Then some \(\mathbf{x}=(x_1,\dots,x_n)^\top\in\mathbb{R}^n\) satisfies \(\mathbf{A}\mathbf{x}=\mathbf{b}\). By the definition of matrix-vector multiplication, \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), which is a linear combination of the columns of \(\mathbf{A}\). Hence \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\). Conversely, suppose \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\). By the definition of column space, there are scalars \(x_1,\dots,x_n\) such that \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). Set \(\mathbf{x}=(x_1,\dots,x_n)^\top\). The formula for matrix-vector multiplication gives \(\mathbf{A}\mathbf{x}=\mathbf{b}\), so the system is consistent. Both implications hold, proving the equivalence. \(\square\)
The theorem is an exact criterion, not merely a geometric description. To test whether \(\mathbf{b}\) belongs to the column space, one can seek coefficients \(x_1,\dots,x_n\) such that \(x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n=\mathbf{b}\). This is precisely the system \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Row-reducing its augmented matrix gives a practical way to decide whether such coefficients exist.
How Augmented Row Reduction Detects Inconsistency
Proof. Each elementary row operation replaces the system by an equivalent system: swapping two equations preserves the solutions, multiplying an equation by a nonzero scalar preserves its solutions, and adding a multiple of one equation to another preserves its solutions. Therefore row reduction preserves consistency. A row \([0\ \cdots\ 0\mid c]\), where \(c\ne0\), represents the equation \(0=c\), which no \(\mathbf{x}\) can satisfy. Thus, if such a row appears, the system is inconsistent. In the other direction, suppose row reduction ends without such a row. The resulting system has no equation with zero coefficients on the variables and nonzero right-hand side. Assign zero to every free variable, if there are any. Each remaining equation with a pivot variable then determines that variable from variables already assigned or determined; solving from the last pivot row upward gives values satisfying every row. If there are no pivot variables, the absence of a contradictory row means that every row is \(0=0\), and any choice of variables, including all zeros, is a solution. Hence the reduced system, and therefore the original system, is consistent. \(\square\)
Worked Example: A Right-Hand Side Inside or Outside the Column Space
Let \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\). Decide whether the systems with right-hand sides \(\mathbf{b}_1=(2,-1,1)^\top\) and \(\mathbf{b}_2=(2,-1,4)^\top\) are consistent.
For \(\mathbf{b}_1\), augment and row-reduce:
The last row represents \(0=0\), so there is no contradiction. The first two rows give \(x_1=2\) and \(x_2=-1\). Thus the system is consistent, and the theorem says \(\mathbf{b}_1\in\operatorname{Col}\mathbf{A}\).
Check: \(\mathbf{A}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}2\\-1\\2+(-1)\end{bmatrix}=\begin{bmatrix}2\\-1\\1\end{bmatrix}=\mathbf{b}_1\).
For \(\mathbf{b}_2\), the augmented matrix reduces as follows. First subtract \(R_1\) from \(R_3\); the new third row is \((0,1\mid 2)\). Then subtract the new \(R_2\) from \(R_3\):
The last row says \(0=3\), so the system is inconsistent and \(\mathbf{b}_2\notin\operatorname{Col}\mathbf{A}\). The contradiction also reflects the column structure: any linear combination \(x_1(1,0,1)^\top+x_2(0,1,1)^\top\) has third coordinate equal to the sum of its first two coordinates. For \(\mathbf{b}_2\), those coordinates would require \(4=2+(-1)\), which is false.
Classifying Several Right-Hand Sides
Suppose the coefficient matrix is fixed, but the right-hand side varies. Each system is a separate membership question: for each \(\mathbf{b}_j\), determine whether \(\mathbf{b}_j\in\operatorname{Col}\mathbf{A}\). If the right-hand sides are placed as columns of a matrix \(\mathbf{B}\), the systems can be written together as \(\mathbf{A}\mathbf{X}=\mathbf{B}\), where each column of \(\mathbf{X}\) contains the unknown coefficients for the corresponding column of \(\mathbf{B}\).
Proof. Write the columns of \(\mathbf{X}\) as \(\mathbf{x}_1,\dots,\mathbf{x}_q\). By the definition of matrix multiplication by columns, \(\mathbf{A}\mathbf{X}=[\mathbf{A}\mathbf{x}_1\ \cdots\ \mathbf{A}\mathbf{x}_q]\). Therefore \(\mathbf{A}\mathbf{X}=\mathbf{B}\) holds exactly when \(\mathbf{A}\mathbf{x}_j=\mathbf{b}_j\) for every \(j\). If such an \(\mathbf{X}\) exists, each of these equations has a solution, so each system is consistent and each \(\mathbf{b}_j\) lies in \(\operatorname{Col}\mathbf{A}\) by the consistency theorem. Conversely, if every \(\mathbf{b}_j\in\operatorname{Col}\mathbf{A}\), that theorem gives, for each \(j\), a vector \(\mathbf{x}_j\) with \(\mathbf{A}\mathbf{x}_j=\mathbf{b}_j\). Put these vectors into the columns of \(\mathbf{X}\). Then \(\mathbf{A}\mathbf{X}=\mathbf{B}\). \(\square\)
Worked Example: Classifying Three Right-Hand Sides at Once
Use \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\) and consider the columns \(\mathbf{b}_1=(1,2,3)^\top\), \(\mathbf{b}_2=(1,1,3)^\top\), and \(\mathbf{b}_3=(0,0,0)^\top\). Classify their systems together by reducing \([\mathbf{A}\mid\mathbf{B}]\), where \(\mathbf{B}=[\mathbf{b}_1\ \mathbf{b}_2\ \mathbf{b}_3]\).
The augmented matrix has one right-hand-side column for each system. Subtract \(R_1\) from \(R_3\), then subtract the new \(R_2\) from \(R_3\):
The final row gives \(0=0\) for the first and third right-hand sides, but gives \(0=1\) for the second. Thus \(\mathbf{A}\mathbf{x}_1=\mathbf{b}_1\) and \(\mathbf{A}\mathbf{x}_3=\mathbf{b}_3\) are consistent, while \(\mathbf{A}\mathbf{x}_2=\mathbf{b}_2\) is inconsistent. In particular, there is no single \(\mathbf{X}\) satisfying \(\mathbf{A}\mathbf{X}=\mathbf{B}\), because its second column would have to solve the inconsistent second system.
The first two rows show that the solutions for the consistent systems are \(\mathbf{x}_1=(1,2)^\top\) and \(\mathbf{x}_3=(0,0)^\top\). Their products verify the classification:
For the second column, the first two equations would force \(x_1=1\) and \(x_2=1\); then the third coordinate of \(\mathbf{A}\mathbf{x}\) would be \(x_1+x_2=2\), not the required \(3\). This directly confirms its inconsistency.
Right-Hand Sides Depending on a Parameter
A symbolic right-hand side can reveal exactly which choices are reachable. Row-reducing with a parameter often produces a final condition such as \(0=c(t)\). The system is consistent precisely for parameter values making that condition true. The condition comes from the equations; it should not be guessed from a numerical example.
Worked Example: Find the Consistent Parameter Values
For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\) and \(\mathbf{b}(t)=(t,2,5)^\top\), determine all real \(t\) for which \(\mathbf{A}\mathbf{x}=\mathbf{b}(t)\) is consistent.
Augment and eliminate the third-row entries in the coefficient columns:
The last row represents \(0=3-t\). It is compatible exactly when \(3-t=0\), or \(t=3\). For \(t=3\), the first two rows give \(\mathbf{x}=(3,2)^\top\). For any \(t\ne3\), the last row is a contradiction, so the system is inconsistent.
Check: At \(t=3\), \(\mathbf{A}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}3\\2\\3+2\end{bmatrix}=\begin{bmatrix}3\\2\\5\end{bmatrix}=\mathbf{b}(3)\). The parameter condition is therefore attained by an actual solution.
Common Pitfalls
Do not change the right-hand side when row-reducing. If the goal is to test \(\mathbf{A}\mathbf{x}=\mathbf{b}\), apply every row operation to the entire augmented matrix \([\mathbf{A}\mid\mathbf{b}]\). Reducing only \(\mathbf{A}\) and leaving \(\mathbf{b}\) unchanged does not produce an equivalent system, because the equations on the left and right would no longer have undergone the same operations.
Row operations on \(\mathbf{A}\) alone do not preserve its column space. For instance, the column space of \(\begin{bmatrix}1\\1\end{bmatrix}\) is the line spanned by \((1,1)^\top\). Applying \(R_2\to R_2-R_1\) changes that column to \((1,0)^\top\), whose span is a different line. Row reduction is valid for solving a system because the same operation is applied to \(\mathbf{b}\), producing an equivalent equation; it does not mean that the row-reduced matrix has the original column space.
One consistent right-hand side does not make all right-hand sides consistent. In the multiple-right-hand-side example, the first and third systems have solutions, but the second does not. The criterion is applied to each column of \(\mathbf{B}\). A simultaneous equation \(\mathbf{A}\mathbf{X}=\mathbf{B}\) is consistent only if every one of those individual systems is consistent.
The column-space criterion separates the coefficient matrix from the requested output. The matrix determines the set of reachable right-hand sides, and consistency asks whether a particular target belongs to that set. For several targets, augmenting by several columns lets one classify them in a single row reduction, while preserving the distinction between each system’s own solution.
Check Your Understanding
For each question, use the column-space criterion and verify any claimed solution by substitution.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), determine whether \(\mathbf{b}=(3,-2,1)^\top\) is in \(\operatorname{Col}\mathbf{A}\).
- Find all real \(t\) for which \(\mathbf{A}\mathbf{x}=(t,1,4)^\top\) is consistent for that same matrix \(\mathbf{A}\).
- State the condition on the columns of \(\mathbf{B}\) that is equivalent to the existence of \(\mathbf{X}\) satisfying \(\mathbf{A}\mathbf{X}=\mathbf{B}\).
- Explain why a row \([0\ \cdots\ 0\mid -2]\) in a reduced augmented matrix proves inconsistency.
- Give a one-sentence reason why row operations must be applied to the right-hand side as well as the coefficient matrix.
