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4.5.11 · D5Linear Algebra (Full)

Question bank — Pivot positions, free variables

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Before we begin, three words that appear everywhere below, each earned from the parent note:

  • Pivot column = a column that hosts a leading entry (the first nonzero number of a row) once the matrix is fully reduced. Picture a staircase: each step lands in one column, and that column is a pivot column.
  • Free variable = a variable whose column has no step landing in it — you get to hand-pick its value.
  • Rank = the number of pivots = the number of stairs in the staircase.
  • = the number of columns = the number of variables. The master count is .

True or false — justify

TF1. A matrix with more rows than columns () must have at least one free variable.
False — "extra rows" become zero rows after reduction, not free variables; if all columns pivot () there are free variables. Free count is , never tied to row count.
TF2. If a system has a free variable, it has infinitely many solutions.
Only if it is consistent. An inconsistent system (a row , ) has zero solutions even with free columns sitting right there.
TF3. Two different echelon forms of the same can put pivots in different columns.
False — intermediate echelon forms differ, but the RREF is unique, so the pivot positions (which columns) are determined by alone.
TF4. A square matrix always has exactly pivots.
False — only an invertible one does. A singular square matrix has , hence free variables and a whole flat of solutions to .
TF5. Adding a right-hand side can change which columns of are pivot columns.
False for the columns of — those pivots depend only on . What can do is make the augmented column a pivot column, which signals inconsistency, not a new free variable.
TF6. The number of free variables equals the number of zero rows in echelon form.
False — free variables count non-pivot columns (), which is a column fact; zero rows are a row fact. They coincide only by accident.
TF7. If has full column rank (), the equation has only the trivial solution.
True — free variables means nothing is left to roam, so is forced. This is exactly the column-independence condition.
TF8. A pivot in the augmented column means the system has one unique solution.
False — a pivot in the augmented column is the row ; it means no solution at all. Uniqueness is about 's columns all pivoting, a separate thing.
TF9. Every free variable adds one dimension to the solution set.
True when consistent — free variables give a -dimensional flat: point (), line (), plane (), matching the null-space dimension.
TF10. If the last column of is a non-pivot column, then is free.
True — "non-pivot column" is the definition of a free variable's column. The position (first, last, middle) is irrelevant; only pivot-vs-not matters.

Spot the error

SE1. "The matrix has 3 zero rows, so there are 3 free variables."
The error is confusing rows with columns. Free variables (non-pivot columns). A tall matrix can have many zero rows yet zero free variables if all columns pivot.
SE2. "There's a free variable, so infinitely many solutions — done."
The consistency check was skipped. If the augmented column pivots, the count describes a solution set that doesn't exist; you must confirm consistency first.
SE3. "I found an echelon form with a pivot in column 2, but the RREF has none there — one of them is wrong."
Neither is wrong. A pivot's column is preserved across all echelon forms of ; if they disagree, an arithmetic slip happened, not a genuine ambiguity — pivot columns are unique.
SE4. "rank and , and there's a free variable because one equation was redundant."
A redundant equation makes a zero row, not a free variable. With , : no free variables, redundancy or not.
SE5. ", so the system is consistent and has a unique solution."
Consistency is correct ( always works), but uniqueness is not guaranteed — if there are free variables and infinitely many solutions to the homogeneous system.
SE6. "This matrix has 4 columns and rank at most 2, so it has 4 free variables."
Free , and for any nonzero matrix, so at most free variables here — you subtracted the rank as if it were zero.
SE7. "Column 2 is all zeros in , so is a pivot variable."
An all-zero column can never host a leading entry, so it is a non-pivot column — is a free variable, the exact opposite.

Why questions

WHY1. Why does each pivot solve for exactly one variable and no more?
A leading entry sits in one column; the reduced row expresses that one variable in terms of the others, using up exactly one column and one equation — that is what makes it a rank-one contribution.
WHY2. Why is a subtraction and not something more complicated?
Every column is either a pivot column ( of them) or not; there is no third option. So the leftover count is just total minus pivots — pure bookkeeping.
WHY3. Why must consistency be checked separately from counting free variables?
The count describes the shape of the solution flat, but a flat can be empty. Consistency (no augmented-column pivot) decides whether that flat is populated at all.
WHY4. Why are pivot positions unique even though echelon forms are not?
All echelon forms of reduce further to the one and only RREF; the pivot positions survive that final reduction unchanged, so they are a fixed property of .
WHY5. Why does full column rank guarantee at most one solution?
forces free variables, so nothing is left to vary — any solution is pinned completely, giving uniqueness (existence still needs consistency).
WHY6. Why does the number of free variables equal the dimension of the null space?
Each free variable is an independent parameter you can dial; setting one to and the rest to builds one basis vector of the null space, so free count = null-space dimension = .
WHY7. Why can't the right-hand vector ever create a new pivot inside ?
Row reduction of operates on rows; the pivot columns of are set by 's own structure. only occupies the extra column, where it can at most become an inconsistency flag.

Edge cases

EC1. The zero matrix (size ): how many free variables for ?
Rank , so free — every variable is free, and the solution set is all of .
EC2. A matrix with : what happens?
If , is free (any value works, a line's worth in ); if the single row reads , inconsistent — no solution.
EC3. A wide matrix () that is consistent: can it ever have a unique solution?
No — at most pivots but columns, so free variables always remain; consistency then gives infinitely many, never one.
EC4. An identity matrix as the coefficient matrix: how many free variables?
Every column pivots, , so free variables — the solution is a single forced point.
EC5. A single equation : describe the solution set.
Both columns are non-pivot (rank ), so both variables are free — the solution set is the whole plane , a -dimensional flat.
EC6. A tall matrix () with but a nonzero augmented-column pivot: consistent or not?
Inconsistent — despite full column rank (which would give uniqueness if solvable), the augmented-column pivot forces a row, so there are zero solutions.
EC7. A homogeneous system with : can it be inconsistent?
Never — always solves it, so it is always consistent; with it has infinitely many solutions forming a flat through the origin.
Recall One-line self-test

The single sentence that dissolves most traps here ::: Free variables come from non-pivot columns, they describe the solution set's dimension only after consistency is confirmed, and pivot positions belong to alone.

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