Row Operations: Why the Solutions Stay the Same

Nablio · 8 minPublished September 11, 2026

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Shared linear algebra foundations for Concordia MATH 204 and McGill MATH 133. Course sequences and classroom timing may differ.

A solution of a system must satisfy every equation at the same time. Elimination replaces the equations with easier ones that have exactly the same shared solutions. It does not need to preserve the appearance of each equation.

That distinction explains why row operations work. Each permitted move can be reversed, so it neither loses an existing solution nor introduces a new one. Understanding this lets you check the reasoning as well as the arithmetic.

The lines x + 2y = 7 and 3x − y = 7 intersect at (3, 2). Subtracting three copies of the first equation from the second gives −7y = −14. — A solution must lie on both lines. Valid row operations preserve their shared solution, even when an individual line changes.

1. Use operations that can be reversed

You can swap two equations, multiply an equation by a nonzero number, or add a multiple of one equation to another. The reversibility matters: multiplying by 2 is undone by dividing by 2; adding three copies of one retained equation is undone by subtracting them. Multiplying an equation by zero is different. It destroys the equation’s information, and there is no inverse operation that recovers it. In an augmented matrix, the entries before the divider are coefficients and the final entry is the constant. A row operation must act on the entire equation, including that constant.

Worked example

Valid elementary row operations: • Swap two rows. • Multiply a row by a nonzero number. • Add a multiple of another row to a row. For example, dividing $\left[\begin{array}{cc|c}2&-4&10\end{array}\right]$ by 2 gives $\left[\begin{array}{cc|c}1&-2&5\end{array}\right]$ .

Common mistake. Replacing one row with a copy of another row can discard information. It is not the reversible operation of adding a multiple of another row to it.

Try it yourself

A student divides $\left[\begin{array}{cc|c}6&-9&15\end{array}\right]$ by 3 and writes $\left[\begin{array}{cc|c}2&-3&15\end{array}\right]$ . What is wrong? Would multiplying the original row by zero be valid instead?

Show answer and reasoning

The constant must also be divided by 3, giving $\left[\begin{array}{cc|c}2&-3&5\end{array}\right]$ . Multiplying by zero is not a valid elementary row operation: it would replace the constraint with $0=0$ , losing information that cannot be recovered.

2. Solve a system and check it

Solve $x+2y=7$ and $3x-y=7$ . Replace the second equation with itself minus three times the first. Any original solution satisfies this new equation. Conversely, because the first equation was retained, adding three copies of it back recovers the original second equation. Thus the new system has exactly the same solutions. Next, divide the second equation by $-7$ to obtain $y=2$ , then subtract twice that row from the first to obtain $x=3$ . Geometrically, elimination can replace a line while preserving the point common to both lines.

Worked example

Original augmented matrix: $\left[\begin{array}{cc|c}1&2&7\\3&-1&7\end{array}\right]$ $R_2\leftarrow R_2-3R_1$ : $\left[\begin{array}{cc|c}1&2&7\\0&-7&-14\end{array}\right]$ $R_2\leftarrow-\frac{1}{7}R_2$ : $\left[\begin{array}{cc|c}1&2&7\\0&1&2\end{array}\right]$ $R_1\leftarrow R_1-2R_2$ : $\left[\begin{array}{cc|c}1&0&3\\0&1&2\end{array}\right]$ Solution: $x=3$ , $y=2$ . Check: $3+2(2)=7$ and $3(3)-2=7$ .

Common mistake. Check the solution in the original equations. Substitution catches arithmetic errors and mistakes made when copying the system into a matrix.

Try it yourself

Solve $x-y=1$ and $2x+y=11$ using elimination, then check your answer.

Show answer and reasoning

Subtract twice the first equation from the second: $3y=9$ . Thus $y=3$ , and the first equation gives $x=4$ . Checking gives $4-3=1$ and $2(4)+3=11$ . Both original equations are satisfied.

3. Read each final row as an equation

Not every system ends with one value for each variable. A row $\left[\begin{array}{cc|c}0&0&3\end{array}\right]$ means $0=3$ : a contradiction, so the system has no solution. A row $\left[\begin{array}{cc|c}0&0&0\end{array}\right]$ means $0=0$ : that row adds no constraint. A zero row alone does not tell you how many solutions remain. Inspect the other rows for contradictions and determine whether any variables are free. A consistent system with a free variable has infinitely many solutions over the real numbers.

Worked example

$\left[\begin{array}{cc|c}0&0&3\end{array}\right]$ represents $0x+0y=3$ : impossible. $\left[\begin{array}{cc|c}0&0&0\end{array}\right]$ represents $0x+0y=0$ : always true.

Try it yourself

Both systems have first row $\left[\begin{array}{cc|c}1&2&5\end{array}\right]$ . System A has second row $\left[\begin{array}{cc|c}0&0&0\end{array}\right]$ ; system B has $\left[\begin{array}{cc|c}0&0&3\end{array}\right]$ . How many solutions does each have?

Show answer and reasoning

System A has infinitely many solutions. Choose any real number $t$ , set $y=t$ , and then $x=5-2t$ . System B has no solution because its second row requires the impossible equality $0=3$ .

Your next revision session

For your next system, write the row operation before computing it. Keep the constant in every calculation, translate the final rows back into equations, and check any proposed solution in the original system.
Rework a solved problem a day or two later, explaining aloud why each operation is reversible.

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Row Operations: Why the Solutions Stay the Same

1. Use operations that can be reversed

Worked example

Try it yourself

2. Solve a system and check it

Worked example

Try it yourself

3. Read each final row as an equation

Worked example

Try it yourself

Your next revision session

Official course outlines and descriptions

Connect these skills to your course

Work on the step you’re missing.

Free resources to work through

MATH 204: Matrix & Vector Fundamentals Before Your First Exam