Elimination, Pivots and Free Variables

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Preserve the solution set

Three elementary row operations preserve solutions of a linear system: swap rows, multiply a row by a nonzero scalar, and add a multiple of one row to another. Apply them to the entire augmented matrix $[A\mid\mathbf b]$ , including the right-hand side.

Echelon form places each leading nonzero entry, called a pivot, farther right than the pivot above it. Reduced row echelon form also makes each pivot $1$ and clears its column above and below. Nonpivot variable columns correspond to free variables. A row $[0\ \cdots\ 0\mid d]$ with $d\ne0$ proves inconsistency.

Visual guide

VISUAL GUIDEElimination preserves the intersection

x + y = 3
x − y = 1
After adding: x = 2

The lines x + y = 3 and x − y = 1 meet at (2, 1). Adding the equations produces 2x = 4, the vertical line x = 2. The transformed system retains the same solution point.

Worked example: a unique solution

For $x+2y=5$ , $2x+y=4$ , subtract twice the first row from the second. This gives $-3y=-6$ , so $y=2$ . Back-substitution gives $x=1$ . There is a pivot in both variable columns and no contradiction, so the solution is unique.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

What does an augmented row [0 0 | 1] tell you?

The system has a unique solution.

The system has no solution.

One variable is free.

Hint 1 · Find a starting point

Translate the row back into an equation.

Hint 2 · Take the next step

It says 0x+0y=1.

Show the reasoning

Answer: The system has no solution.

The impossible equation 0=1 makes the entire system inconsistent.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a family of solutions

For $x+2y-z=3$ and $2x+4y-2z=6$ , the second equation is redundant. Let $y=s$ and $z=t$ . Then

\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\0\\0\end{pmatrix}+s\begin{pmatrix}-2\\1\\0\end{pmatrix}+t\begin{pmatrix}1\\0\\1\end{pmatrix}.

This expresses every solution, not just one example. If the second right-hand side were $7$ , elimination would instead produce $0=1$ , so no solution would exist.

In floating-point computation, choosing a reasonably large pivot through row exchanges reduces avoidable numerical error. Exact symbolic elimination and finite-precision elimination have different practical concerns; a very small computed pivot may require a tolerance informed by the scale of the data.

Practice

Solve $x+y=3$ , $x-y=1$ .
Describe all solutions of $x+y+z=0$ .
Interpret the augmented row $[0\ 0\mid4]$ .

Show worked solutions

Adding equations gives $x=2$ , then $y=1$ .
Set $y=s,z=t$ : $(x,y,z)=(-s-t,s,t)$ .
It asserts $0=4$ , so the system is inconsistent.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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