What does a pivot actually tell us?

After elimination, a system can become easier to understand even before you finish solving it. Consider these rows:

$$\left[\begin{array}{ccc|c} 1&1&1&6\\ 0&1&-1&2\\ 0&0&0&0 \end{array}\right].$$

The first three columns multiply $x,y,z$. The last column contains the right-hand sides. Read the rows aloud: $x+y+z=6$, $y-z=2$, and $0=0$.

Does the last row mean every variable is zero, or that it imposes no further condition?

Keep a record of your investigation

Use the displayed matrix with third row $[0,0,0\mid0]$ as the reference. Before each comparison, write whether solutions exist and whether a choice remains free. Keep the first two rows fixed and replace only the third constraint. Compare the cases at step 0, since choosing a case resets the row operations. Record the last row and its meaning before reducing each case or revealing its interpretation.

Compare three endings

Start with Repeated constraint. Predict whether choosing $z$ would determine the other two variables. Use Next operation to subtract the second row from the first.

Return to step 0 with Undo step. Predict the effect of replacing only the last row by $[0,0,0\mid1]$, then select Conflicting constraint. Record your interpretation before opening Check the interpretation.

Return to Repeated constraint at step 0. Predict the effect of replacing its last row with $z=1$, then select One more constraint. Compare the initial matrices before stepping through the remaining operations.

Read the rows before solvingReset

Repeated constraintConflicting constraintOne more constraint

Step 0 of 1: Starting echelon form

Augmented matrix. Outlined entries lead their rows.

x	y	z	equals
1	1	1	6
0	1	-1	2
0	0	0	0

Undo stepNext operation

Check the interpretation

Pivots in x and y. No pivot in z: choose z, then recover y and x. The zero row says 0 = 0 and adds no constraint.

Compare existence and freedom

The row $[0,0,0\mid0]$ says $0=0$, which is always true. The first two equations leave $z$ free.

The row $[0,0,0\mid1]$ says $0=1$, which is impossible. No choice of variables can solve the full system.

The row $[0,0,1\mid1]$ says $z=1$. Then $y=3$ and $x=2$: the extra constraint removes the free choice without causing a contradiction.

The staircase and its leading entries

A matrix is in row echelon form when its nonzero rows sit above its zero rows, each leading entry is to the right of the leading entry above it, and entries below each leading entry are zero.

The leading nonzero entry of an echelon row is its pivot. In a variable column, it identifies a leading variable that can be recovered using that equation. A variable column without a pivot corresponds to a free variable, provided the system is consistent.

In our starting matrix, the first row leads in $x$ and the second leads in $y$. Neither row leads in $z$. So we can choose $z$, use the second row to recover $y$, and then use the first to recover $x$.

A zero coefficient is not a zero variable value. It says that variable does not contribute to that equation.

Reduced form makes the reading easier

A reduced row echelon form also makes each pivot 1 and removes all other entries in its pivot column, including those above it. For the repeated-constraint case, subtracting row 2 from row 1 gives

$$\left[\begin{array}{ccc|c} 1&0&2&4\\ 0&1&-1&2\\ 0&0&0&0 \end{array}\right].$$

Now the relationships are immediate: $x=4-2z$ and $y=2+z$. The solution set did not change; its description became simpler.

The number of pivots in the coefficient matrix is its rank. Here it is 2. There are three variables, so a consistent system has one free variable. A leading entry in the right-hand-side column is not a fourth variable pivot: it signals a contradiction.

Read first, calculate second

Consider a different echelon matrix:

$$\left[\begin{array}{ccc|c} 1&2&0&5\\ 0&0&1&-1 \end{array}\right].$$

Which variables lead? Which can you choose? What changes if we append $[0,0,0\mid4]$?

Check the interpretation

The pivots are in $x$ and $z$. The variable $y$ is free, $z=-1$, and $x=5-2y$. Appending the contradictory row makes the whole system inconsistent; we must not report a solution family just because a variable column lacks a pivot.

The next lesson turns a free choice into a complete rule for every solution.

Design your own comparison

On paper, keep the first two reference rows and choose your own replacement for the third row. Aim for exactly one solution. Predict which variable gains a pivot, then reduce the matrix and check your solution in all three equations.

Next change only the third row to produce a contradiction. Explain why an empty variable column cannot rescue that system.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

A row $[0,0,c\mid d]$ with $c\ne0$ fixes $z=d/c$, after which $y=2+z$ and $x=4-2z$. A row $[0,0,0\mid d]$ with $d\ne0$ is impossible. Your own rows may differ; check consistency before counting free variables.

Question 1 of 30 checked · 0 correct

An echelon row reads [0, 0, 0 | 0]. What does it say?

Every variable equals zero.The row imposes no additional restriction.There are no solutions.

Check answer

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