What It Means to Preserve Solutions
A row operation changes the written equations in a linear system, but it should not change which values of the variables satisfy all of them. To make that claim precise, we will compare the solution sets before and after an operation. The key is not only to check that every old solution still works; we must also show that every solution of the changed system was already a solution of the original one. Reversing each elementary operation provides that second direction.
Work over \(\mathbb{R}\). A linear system with \(m\) equations and \(n\) unknowns can be written as \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\), \(\mathbf{x}\in\mathbb{R}^n\) is a column vector, and \(\mathbf{b}\in\mathbb{R}^m\). Its solution set is the set of all \(\mathbf{x}\in\mathbb{R}^n\) satisfying the equation. The augmented matrix \([\mathbf{A}\mid\mathbf{b}]\) records the coefficients and right-hand sides together: each row represents one equation.
Each operation has a direct interpretation for equations. Interchanging rows changes their order. Multiplying a row by a nonzero scalar multiplies both sides of that equation by the same nonzero number. Replacing a row adds a multiple of one equation to another. These descriptions suggest why the operations should work, but the proof must show equality of solution sets in both directions.
The Solution-Preservation Theorem
Proof. Write the \(i\)th equation as \(L_i(\mathbf{x})=b_i\), where \(L_i(\mathbf{x})=\sum_{k=1}^{n}a_{ik}x_k\). A vector \(\mathbf{x}\) is a solution precisely when all \(m\) equations hold. We check each allowed operation and show that it can be reversed.
If rows \(i\) and \(j\) are interchanged, the same two equations remain in the system, only in the opposite order. Thus any vector satisfying all the equations before the interchange satisfies all of them afterward, and any vector satisfying all of them afterward satisfies all of them before. The operation reverses itself by interchanging the same two rows again.
If row \(i\) is multiplied by \(c\ne0\), its equation becomes \(cL_i(\mathbf{x})=cb_i\). If \(L_i(\mathbf{x})=b_i\), multiplication by \(c\) gives the new equation. Conversely, if \(cL_i(\mathbf{x})=cb_i\), division by the nonzero scalar \(c\) gives \(L_i(\mathbf{x})=b_i\). All other equations are unchanged, so the solution sets are equal. The inverse operation multiplies row \(i\) by \(1/c\).
Finally, suppose row \(i\) is replaced by row \(i\) plus \(c\) times row \(j\), where \(i\ne j\). The new equation in position \(i\) is \(L_i(\mathbf{x})+cL_j(\mathbf{x})=b_i+cb_j\), while equation \(j\) is unchanged. If \(\mathbf{x}\) satisfies the original system, adding \(c\) times equation \(j\) to equation \(i\) shows that it satisfies the new equation. Conversely, if it satisfies the new system, it satisfies \(L_j(\mathbf{x})=b_j\); subtracting \(cL_j(\mathbf{x})=cb_j\) from the new equation gives \(L_i(\mathbf{x})=b_i\). The other equations were unchanged. This operation is reversed by \(R_i\to R_i-cR_j\), because row \(j\) has not changed. Each operation therefore preserves the solution set in both directions. \(\square\)
The reverse step for row replacement is essential. It works because the row being added to row \(i\)—row \(j\)—stays fixed. If row \(j\) were changed as part of the same step, the stated reverse operation would not necessarily undo the change. In an elementary row replacement, the two row indices must be different, and only the target row \(i\) is replaced.
Examples: Checking Each Type of Operation
Worked Example: Interchanging Two Equations
Consider the system \(x+2y=4\), \(3x-y=5\). Its augmented matrix is
The transformed system is \(3x-y=5\), \(x+2y=4\): it contains the same equations in reverse order. Solving the first original equation for \(x\) gives \(x=4-2y\). Substitution into the second gives \(3(4-2y)-y=5\), so \(12-7y=5\), \(y=1\), and \(x=2\). Thus the candidate solution is \((x,y)=(2,1)\).
Check: In the original system, \(2+2(1)=4\) and \(3(2)-1=5\). In the transformed system, \(3(2)-1=5\) and \(2+2(1)=4\). The row interchange reverses itself by swapping the rows again, as the theorem predicts.
Worked Example: Scaling an Equation by a Nonzero Scalar
Start with \(2x+4y=10\), \(x-y=2\), and multiply the first equation by \(\tfrac12\). Apply the operation to every entry in its augmented row:
The changed system is \(x+2y=5\), \(x-y=2\). Subtracting the second equation from the first gives \(3y=3\), so \(y=1\); then \(x-y=2\) gives \(x=3\).
Check: Substituting \((3,1)\) into the original equations gives \(2(3)+4(1)=10\) and \(3-1=2\). In the transformed equations, \(3+2(1)=5\) and \(3-1=2\). Reversing the scaling gives the original first row: \(\tfrac12[1\ \ 2\mid 5]=[\tfrac12\ \ 1\mid\tfrac52]\) is not the original row, so the inverse must be calculated carefully: the inverse operation multiplies the transformed row by \(2\), giving \([2\ \ 4\mid10]\). This confirms the operation is reversible.
Worked Example: Replacing a Row and Reversing It
Consider \(x+y=3\), \(2x-y=0\). Replace the second row by the second row minus twice the first. The entries change as follows: \((2,-1,0)-2(1,1,3)=(0,-3,-6)\).
The new equations are \(x+y=3\) and \(-3y=-6\). The second gives \(y=2\), and the first then gives \(x=1\).
Check: In the original system, \(1+2=3\) and \(2(1)-2=0\). In the transformed system, \(1+2=3\) and \(-3(2)=-6\). To reverse the row replacement, add twice row 1 to row 2. The changed second row becomes \((0,-3,-6)+2(1,1,3)=(2,-1,0)\), restoring the original augmented matrix.
Why a Sequence of Operations Also Works
In practice, a system is often changed through several elementary row operations. The one-step theorem extends to any finite sequence: each step preserves the solution set, so the set at the end equals the set at the beginning. This fact justifies using row operations one after another without tracking a new list of candidate solutions at each stage.
Proof. Let the systems in the sequence have solution sets \(S_0,S_1,\ldots,S_r\), where \(r\) is the number of operations. For each step from \(k-1\) to \(k\), the theorem gives \(S_{k-1}=S_k\). Applying these equalities in order gives \(S_0=S_1=\cdots=S_r\), including when \(r=0\), in which case the two systems are the same system. Therefore the initial and final solution sets are equal. \(\square\)
This statement concerns solutions, not the visual appearance of the matrices. The entries can change substantially while the solution set remains fixed. Also, the corollary does not say that every pair of matrices with the same solution set can be connected by elementary row operations; that is a different claim. Here, the guarantee applies when a finite sequence of the specified operations is actually performed.
Common Mistake: Scaling by Zero
A common error is to treat multiplication by zero as an allowed row scaling. For example, start with the one-equation system \(x=1\). Multiplying its row by zero gives \(0=0\), which is satisfied by every real \(x\), not just \(x=1\). The solution set has changed. The mistake is allowing a scaling factor with no multiplicative inverse: from \(0\cdot x=0\), division by zero cannot recover \(x=1\). Elementary row scaling therefore requires \(c\ne0\); its inverse is multiplication by \(1/c\).
Another useful check is to apply the proposed reverse operation to the entire augmented row, not only to the coefficients. In a row replacement, the right-hand side changes along with the coefficients. For instance, replacing the second row by \(R_2-2R_1\) changes \([2\ \ -1\mid0]\) into \([0\ \ -3\mid-6]\). Reversing only the coefficient entries would fail to restore the original equation \(2x-y=0\).
Why Reversibility Matters
Preserving solutions in both directions means no solution is lost and no new one is introduced. The row-swap inverse is another swap; the row-scaling inverse uses the reciprocal scalar; and the row-replacement inverse subtracts the same multiple of the unchanged source row. These explicit reverses are the reason elementary row operations are safe tools for transforming linear systems. When working with an augmented matrix, check that an operation is allowed, apply it to the full row, and identify its inverse. That gives a reliable way to justify each change.
Check Your Understanding
For each question, explain how the row operation or its inverse affects the solution set.
- Why does interchanging two equations preserve the solutions even when the equations are inconsistent?
- What operation reverses \(R_3\to -4R_3\), and why is that reverse allowed?
- For \(R_2\to R_2+5R_1\), write the reverse operation and state which row must remain unchanged.
- Starting from \(x+2y=5\), \(x-y=2\), scale the first equation by \(3\). Write the resulting augmented matrix and the inverse row operation.
- Explain why replacing a row by twice itself plus another row is not, as written, one of the elementary row replacements defined here.
