Practice: Solving and Characterising Linear Systems

Reduce. Subtract $2\times$ row 1 from row 2, and row 1 from row 3:

Rows 2 and 3 are identical, so subtracting one from the other leaves $0=0$, redundancy, not contradiction.

Pivots. There is a pivot in the $x$ column and a pivot in the $z$ column. The $y$ column has none, so $y$ is free.

The case. A variable column without a pivot, and no row of the form $0=c$ with $c\ne 0$: infinitely many solutions.

The solution set. From $z=1$ and $x=6-y-2z=4-y$:

A particular solution plus a direction to slide along, not a single point.

Check. Take $y=3$, giving $(1,3,1)$. First equation: $1+3+2=6$. Third: $1+3+3=7$. Both hold.

Note. Three equations in three unknowns, and not a unique solution, because the third equation repeated information already in the first two. The count of equations predicted nothing; the pivots decided it.

Step 1. Scale row 1 by $\frac{1}{2}$ to get a leading $1$:

Reason: multiplying by a nonzero constant is reversible, multiply by $2$ to undo it, so every solution of the old row solves the new one and conversely.

Step 2. Row 2 $-4\times$ row 1, and row 3 $+2\times$ row 1:

Reason: adding a multiple of one row to another is reversible by subtracting the same multiple. A point satisfying both original rows satisfies the combination, and nothing new is admitted.

Step 3. Row 3 $-$ row 2:

This is echelon form: pivots in $x$, $y$ and $z$.

Check. Back-substitute: $z=2$, then $y=2$, then $x=1-4+2=-1$. In the original third equation: $-2(-1)-3(2)+7(2)=2-6+14=10$. It holds.

What the reduction did not do. It did not decide anything about the solution set beyond producing the pivots. Reading the case off this form is the next question, not part of the arithmetic.

What the student did. Replacing BOTH rows with their sum is not an elementary row operation. The legal move adds a multiple of one row to another and keeps the row it was added to; this one destroys two equations and puts one in their place. Information is discarded, and an operation that discards information is not reversible. There is no way to recover the original pair from $3x+6y=15$.

Why the answer came out right anyway. The two equations were genuinely the same constraint, so there was no information to lose. The method was wrong and the system was forgiving.

A system where it fails.

This has the unique solution $(1,1)$. Replace both rows with their sum: $2x=2$, so $x=1$, and $y$ is now unconstrained. The method reports infinitely many solutions $(1,y)$. The second equation, which pinned $y$, was thrown away.

The correct move. Subtract $2\times$ row 1 from row 2, KEEPING row 1:

Row 1 survives, and the $0=0$ row records that the second equation added nothing. Same verdict, reached by a step that could be undone.

The error. The count of equations against unknowns does not decide the case. It says how many statements were written down, not how many independent restrictions they impose.

Why more equations need not mean no solution. Equations can repeat one another. Five equations in three unknowns may reduce to three independent ones with a unique solution, or to two with a free variable and infinitely many solutions. The extra rows collapsing to $0=0$. Nothing about the count prevents that.

A concrete case. $x=1$, $y=2$, $z=3$, $x+y=3$, $y+z=5$ has five equations and three unknowns, and the solution $(1,2,3)$ satisfies all of them. Overdetermined by the count; perfectly consistent in fact.

What would decide it. Reduce the augmented matrix and read the pivot structure. A row of the form $000\mid c$ with $c\ne 0$ means no solution. A pivot in every variable column means exactly one. A variable column without a pivot means infinitely many. That is decisive and it takes less time than arguing from shape.

The same error in reverse. Fewer equations than unknowns does not guarantee infinitely many solutions either. The system can still be inconsistent, as $x+y=1$ together with $2x+2y=5$ shows with two equations and two unknowns.

Version A — exactly one solution. Every variable column carries a pivot, so the constraints pin the plan down completely. There is nothing to choose.

Version B — no solution. Reduction produces a row asserting $0=c$ for some nonzero $c$: the constraints contradict one another. No adjustment of the quantities will help, because the contradiction is among the requirements themselves rather than in the search.

Version C — infinitely many solutions. At least one variable column has no pivot, so there is a free variable and the solution set is a flat: a particular plan plus a direction along which the plan can slide. That two plans share a cost is the visible symptom; the cause is that the objective does not change along that direction.

What to report in case C. Not a single plan. The correct report describes the set. A particular solution together with the direction or directions generated by the free variables, so the reader can see the whole family of equally good plans and choose among them on grounds the model does not capture, such as robustness or convenience. Reporting one member of an infinite family silently makes that choice for them, and hides that any choice was available.

Why this matters. The freedom in case C is usually the most useful output the model produces: it says where the plan can be adjusted at no cost, which is exactly the room a planner wants when circumstances change.

What did not decide these outcomes. All three versions have equality constraints and the same shape; the counts of equations and unknowns are identical across them. Only the pivot structure after elimination separates one plan from no plan from a family of plans, which is why the shape of the model predicts nothing about which of the three a planner will get.