What a Check Must Establish
A row reduction is a chain of arithmetic steps, and one incorrect entry can change the system being solved. Checking a result therefore has two parts: first, verify that each displayed matrix follows from the preceding one by the stated row operation; second, test the claimed solution description in the original equations. The original system matters: a solution of an incorrectly reduced system need not solve the problem that was asked.
Throughout, the systems are over \(\mathbb{R}\). An augmented matrix records the coefficients of a system followed by its right-hand sides. For example, the augmented matrix \(\begin{bmatrix}1&2&|&5\\3&7&|&17\end{bmatrix}\) represents \(x+2y=5\) and \(3x+7y=17\). Row operations act on the entire row, including the right-hand-side entry. The three elementary row operations are swapping two rows, multiplying a row by a nonzero scalar, and adding a scalar multiple of one row to another.
The final clause is essential. Substitution can show that every vector in a proposed list or family is a solution, but that alone does not show that the list or family contains all solutions. A complete check must address both inclusion directions.
Why Valid Row Operations Preserve Solutions
Proof. A solution assigns values to all variables so that every equation holds. Swapping two rows only changes the order of two equations, so the values satisfying all equations before the swap satisfy all equations after it, and conversely. Multiplying one equation by a nonzero scalar preserves exactly the values that satisfy it: if \(q=0\), then \(cq=0\); and if \(cq=0\) with \(c\ne0\), then \(q=0\). Finally, suppose an equation is replaced by itself plus \(c\) times another equation. Any values satisfying both original equations satisfy the replaced equation. Conversely, the original equation is recovered by subtracting \(c\) times the unchanged other equation from the replaced equation. Thus this operation is reversible as well. Each elementary operation preserves the solution set, so the result follows for each of the three types. \(\square\)
For a sequence of row operations, apply this result one step at a time: every intermediate system has the same solutions as the preceding one. This gives a practical audit rule. If even one transition is not the result of its stated operation, the preservation argument no longer applies to the claimed chain.
Worked Example: Locate an Arithmetic Error
Consider the system \(x+2y=5\), \(3x+7y=17\). A proposed first step is to apply \(R_2\to R_2-3R_1\), but the displayed result is claimed to be \(\begin{bmatrix}1&2&|&5\\0&1&|&3\end{bmatrix}\). Check that transition and correct it.
The original augmented matrix is
The first row stays unchanged. For the second row, subtract three times the first row entry by entry: \(3-3(1)=0\), \(7-3(2)=1\), and \(17-3(5)=2\). Therefore the correct transition is
The claimed right-hand-side entry \(3\) is an arithmetic error: \(17-15=2\), not \(3\). Continue from the corrected matrix by applying \(R_1\to R_1-2R_2\). Its first row becomes \((1,\,2,\,|\,5)-2(0,\,1,\,|\,2)=(1,\,0,\,|\,1)\), so
This gives \(x=1\), \(y=2\). Check in the original equations: \(1+2(2)=5\) and \(3(1)+7(2)=3+14=17\). Both hold. If the incorrect entry \(3\) had been used instead, the resulting equations would have produced a different answer, so the final substitution is a useful additional check.
Checking a Parameterized Solution Set
A parameterized description uses one or more free parameters to list candidate solutions. To verify it fully, check that every parameter choice gives a solution and then establish that an arbitrary solution must arise from some parameter choice. When the reduction is valid, the second part can often be read from the reduced equations; it can also be derived directly from the original equations.
Proof. Suppose first that the proposed description equals \(S\). Then every \(\mathbf{x}(t)\) belongs to \(S\), giving (1). Also, every \(\mathbf{x}\in S\) belongs to the proposed description, so it equals \(\mathbf{x}(t)\) for some \(t\in T\), giving (2). Conversely, suppose (1) and (2) hold. Condition (1) says that every vector in the proposed description is in \(S\). Condition (2) says that every vector in \(S\) is in the proposed description. The two sets therefore have exactly the same elements. \(\square\)
Worked Example: Verify an Infinite Family and Its Completeness
Consider the system \(x+2y-z=4\), \(2x+5y+z=9\). Check the reduction and determine whether the proposed family \((x,y,z)=(2+7t,\,1-3t,\,t)\), \(t\in\mathbb{R}\), gives exactly all solutions.
Begin with the augmented matrix. Apply \(R_2\to R_2-2R_1\). The new second row is \((2-2,\,5-4,\,1-2(-1),\,|\,9-8)=(0,1,3,|,1)\). Thus
Next apply \(R_1\to R_1-2R_2\). The first row becomes \((1,\,2-2,\, -1-6,\,|\,4-2)=(1,0,-7,|,2)\), giving
The reduced equations are \(x-7z=2\) and \(y+3z=1\). Set the free variable \(z=t\). Then \(x=2+7t\) and \(y=1-3t\), which gives the proposed family. To check the first direction, substitute this family into the original equations:
So every parameter value gives a solution. To check completeness, take any solution of the original system. Subtract twice its first equation from its second; this gives \(y+3z=1\), so \(y=1-3z\). Substituting that into the first equation gives \(x+2(1-3z)-z=4\), hence \(x=2+7z\). Therefore that solution has the proposed form with \(t=z\). The family is exactly the solution set. The row operations and the direct substitution agree.
Checking an Inconsistent Result
A proposed result may say that a system has no solutions. In that case, there is no candidate vector to substitute. Instead, check that the contradiction follows from the original equations. A valid row reduction can expose a contradiction, but the original equations provide a direct way to confirm it.
Worked Example: Confirm That There Are No Solutions
Consider \(x+y=1\), \(2x+2y=3\). A proposed reduction gives a row \(0x+0y=1\). Check the row operation and confirm the conclusion from the original system.
The augmented matrix is \(\begin{bmatrix}1&1&|&1\\2&2&|&3\end{bmatrix}\). Applying \(R_2\to R_2-2R_1\) gives the new second row \((2-2,\,2-2,\,|\,3-2)=(0,0,|,1)\):
The last row says \(0=1\), which no values of \(x\) and \(y\) can satisfy. Check the same contradiction in the original equations: twice the first equation gives \(2x+2y=2\), whereas the second equation requires \(2x+2y=3\). These requirements cannot both hold, so the original system has no solutions. The computed reduction and the direct check agree.
A Reliable Audit Routine
Use a fixed sequence of checks rather than relying on the appearance of a final matrix. The following routine separates arithmetic accuracy from the interpretation of the result.
Confirm that each equation is represented by the correct coefficients and right-hand side, in the same variable order throughout.
Read the stated row operation, leave untouched rows unchanged, and recompute every entry in the changed row. Include the augmented entry.
Translate the final rows into equations carefully. Verify signs, coefficients, and which variables are free.
Substitute each proposed solution form into the original equations. For a parameterized family, also show that every original solution can be expressed using the parameters.
A compact audit table can help identify which kind of check has been completed:
| Claim being checked | What to calculate | What the check establishes |
|---|---|---|
| A row operation is correct | Recompute each changed entry from the stated operation | The displayed transition is arithmetically valid |
| A vector is a solution | Substitute its coordinates into every original equation | That vector satisfies the original system |
| A parameterized family is a solution set | Verify every parameter choice and derive the form for an arbitrary solution | The family contains exactly all solutions |
| A system is inconsistent | Derive a contradiction from the original equations | No vector can satisfy the original system |
Common Pitfall: Verifying Only One Direction
Suppose a student claims that \(x=t\), \(y=t\), for \(t\in\mathbb{R}\), describes the solutions of \(2x+y=3\), \(x-y=0\). Substituting the proposed family into the first equation gives \(2t+t=3\), or \(3t=3\). This holds only when \(t=1\), not for every real \(t\). For example, \(t=0\) gives \((x,y)=(0,0)\), and \(2(0)+0=0\ne3\). So the proposed family fails even the first inclusion check.
The actual solution can be found directly: \(x-y=0\) implies \(x=y\), and then \(2x+y=3\) becomes \(3x=3\), so \(x=1\) and \(y=1\). Substitution confirms \(2(1)+1=3\) and \(1-1=0\). This example also shows why testing a single parameter value is inadequate: a family claimed for all \(t\) must satisfy the system for every permitted \(t\). More generally, testing a few members cannot replace a proof covering every parameter and every possible solution.
Check Your Understanding
For each question, show the arithmetic or substitution that supports your conclusion.
- For \(x+3y=7\), \(2x+7y=16\), verify the result of applying \(R_2\to R_2-2R_1\) to the augmented matrix.
- A proposed step changes \(\begin{bmatrix}1&-2&|&4\\3&1&|&8\end{bmatrix}\) to \(\begin{bmatrix}1&-2&|&4\\0&7&|&0\end{bmatrix}\) using \(R_2\to R_2-3R_1\). Is the step correct? Check all three changed entries.
- For \(x+y=4\), \(x-y=0\), test the proposed solution \((x,y)=(2,2)\) by substitution into both original equations.
- For \(x+2y=5\), is \((x,y)=(5-2t,t)\), \(t\in\mathbb{R}\), the complete solution set? Check both that the family solves the equation and that every solution has this form.
- Explain why checking that every vector in a proposed family solves a system does not, by itself, prove that the family lists all solutions.
