From a System to an Augmented Matrix
A linear system can have one solution, no solutions, or more than one solution. This workshop focuses on the first case: reducing an augmented matrix until its unique solution can be read off, then substituting that answer into the original equations. The method applies to square systems and rectangular systems, where the number of equations and unknowns differ.
Write a system with \(m\) equations and \(n\) unknowns as \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\), \(\mathbf{x}\in\mathbb{R}^n\), and \(\mathbf{b}\in\mathbb{R}^m\). Its augmented matrix is \([\mathbf{A}\mid\mathbf{b}]\): the coefficient matrix with the right-hand-side column appended. Row operations on this augmented matrix apply the same equation operations to the coefficients and constants. As established in the earlier material on row operations, each elementary row operation is reversible and preserves the system’s solution set.
A pivot in a variable column is a leading entry in one of the first \(n\) columns of an echelon form. A variable column without a pivot corresponds to a free variable. A row with all zero coefficients and a nonzero constant represents an impossible equation, such as \(0=3\), so the system is inconsistent.
Proof. Row operations preserve the solution set, so it is enough to examine the system in RREF. If the system is inconsistent, it has no solutions and therefore does not have exactly one. If it is consistent, no row gives an impossible equation. When every variable column has a pivot, there are no free variables. Each pivot equation determines its variable from the right-hand side and any variables in columns to its right; in RREF, those other variable entries in a pivot row are zero. Thus every variable is fixed, and consistency ensures that the resulting values satisfy all rows. There is exactly one solution. Conversely, suppose the system is consistent but some variable column has no pivot. Its variable is free. Assigning it one value or a different value, while keeping any other free variables fixed, gives two different choices for the free variables. The pivot variables are determined by those choices, and each choice satisfies the pivot equations; the zero rows are also satisfied by consistency. The resulting solutions differ in the free variable, so the solution is not unique. Therefore consistency together with a pivot in every variable column is necessary and sufficient. \(\square\)
This criterion is a guide to the calculation: reduce the whole augmented matrix, check for consistency, and look for a pivot in every variable column. A rectangular system may have extra zero rows and still have one solution, provided every variable has a pivot.
Worked Reduction: A Square System
Worked Example: Three Equations and Three Unknowns
Solve the system and verify the answer:
Write its augmented matrix. First use \(R_2\to R_2-2R_1\) and \(R_3\to R_3-R_1\):
Swap rows 2 and 3, then apply \(R_3\to R_3+3R_2\):
Scale row 3 by \(-\tfrac17\). Then apply \(R_2\to R_2+2R_3\), \(R_1\to R_1-R_3\), and \(R_1\to R_1-R_2\), in that order:
There is a pivot in every variable column and no contradictory row. The reduced equations give \(x=1\), \(y=2\), and \(z=-1\). Check the answer in the original equations: \(1+2-1=2\), \(2(1)-2+(-1)=-1\), and \(1+2(2)-(-1)=6\). All three equations hold.
Worked Reduction: More Equations Than Unknowns
Worked Example: A Rectangular System with a Redundant Equation
Solve the four-equation system in two unknowns:
Form the augmented matrix. Apply \(R_2\to R_2-2R_1\), \(R_3\to R_3-3R_1\), and \(R_4\to R_4-4R_1\):
Swap rows 2 and 3, then apply \(R_4\to R_4-R_2\). Scale row 2 by \(-\tfrac17\), and finally apply \(R_1\to R_1-2R_2\):
Both variable columns have pivots, and the remaining rows are zero rows, not contradictory rows. The unique solution is \(x=2\), \(y=1\). Check all four original equations: \(2+2(1)=4\), \(2(2)+4(1)=8\), \(3(2)-1=5\), and \(4(2)+1=9\). Each equation is satisfied, including the one that became a zero row during reduction.
Worked Reduction: Pivoting After a Row Swap
Worked Example: A Rectangular System That Needs a Swap
Solve the system
The first variable column begins with a zero, so swap rows 1 and 2. Then apply \(R_3\to R_3-2R_1\):
Apply \(R_3\to R_3+R_2\), scale row 2 by \(\tfrac12\), and apply \(R_1\to R_1-R_2\):
Both variable columns have pivots, so the solution is \(x=2\), \(y=3\). Check the original equations: \(2(3)=6\), \(2+3=5\), and \(2(2)=4\). The row swap changed the order of the equations, not their solution set.
What the Pivot Pattern Says About the Shape
The examples illustrate two useful consequences of the criterion. In a square system, having a pivot in every variable column means the reduced coefficient block is the identity. In a rectangular system with more equations than unknowns, extra rows can reduce to zeros without affecting uniqueness, as long as all variable columns have pivots and the system is consistent.
Proof. By the unique-solution criterion, uniqueness is equivalent to a pivot in each of the \(n\) variable columns. In an \(n\)-row matrix, these \(n\) pivots occupy all \(n\) rows, one per pivot row. In RREF each pivot is \(1\), and every other entry in its column is \(0\). Since there are \(n\) pivot columns and \(n\) rows, the coefficient block is \(\mathbf{I}_n\); the augmented column is some vector \(\mathbf{c}\). Conversely, in \([\mathbf{I}_n\mid\mathbf{c}]\) every variable column has a pivot and no row is contradictory, so the unique-solution criterion gives exactly one solution. \(\square\)
At the other extreme, a consistent system with more unknowns than equations cannot have a unique solution. There can be at most one pivot per equation row, so if \(n>m\), at least one of the \(n\) variable columns lacks a pivot. The unique-solution criterion then rules out uniqueness. This observation does not say that every rectangular system has a unique solution: the pivot pattern and consistency must both be checked.
Common Pitfall: Reducing Only the Coefficients
Row operations must be applied to the entire augmented matrix, including the right-hand-side column. For example, consider \(x+y=3\) and \(2x+y=4\). The correct augmented reduction begins
The second row gives \(y=2\), and the first then gives \(x=1\). A wrong calculation would apply the row operation to the coefficient entries but leave the right-hand side unchanged, producing \(\begin{bmatrix}1&1&3\\0&-1&4\end{bmatrix}\) and the false conclusion \(y=-4\). The correction is to subtract twice the entire first row, including its constant: \(4-2(3)=-2\), not \(4\). Check the correct answer: \(1+2=3\) and \(2(1)+2=4\).
A second check is structural: a zero row \([0\ \cdots\ 0\mid 0]\) is harmless, whereas \([0\ \cdots\ 0\mid c]\) with \(c\ne0\) is contradictory. Do not treat every zero coefficient row as evidence that a variable is free; free variables correspond to variable columns without pivots.
A Reliable Reduction Routine
Build \([\mathbf{A}\mid\mathbf{b}]\), keeping the right-hand side as the last column.
Use row swaps, nonzero row scalings, and row replacements, carrying out each operation across every column.
Look for a contradictory row, then check whether every variable column has a pivot. Both conditions are needed for a unique solution.
Read the variable values from the reduced equations and substitute them into each original equation.
Verification is not a substitute for reduction: a candidate that satisfies the equations is a solution, but the pivot pattern establishes whether any other solution exists. A complete solution therefore reports both the values and the reason they are unique.
Check Your Understanding
For each system, reduce the augmented matrix before deciding what the solution set is.
- For \(x+y=4\), \(2x-y=2\), find the solution and check both equations.
- Can a consistent system with \(3\) equations and \(4\) unknowns have exactly one solution? Explain using the pivot criterion.
- A \(4\times 2\) augmented matrix reduces to two pivot rows followed by two zero rows. What additional condition must hold for its system to have a unique solution?
- Explain why \([0\ 0\mid 5]\) and \([0\ 0\mid 0]\) have different meanings in an augmented matrix.
- In the reduction of a system, a row operation replaces \(R_2\) with \(R_2-3R_1\). Which entries must be updated, and why?
