4.5.10 · D5Linear Algebra (Full)
Question bank — Row echelon form and reduced row echelon form
This bank leans on ideas from Gaussian Elimination, Gauss-Jordan Elimination, Rank of a Matrix, Free and Basic Variables, Solving Systems of Linear Equations, Linear Independence, Matrix Inverse via RREF and Elementary Matrices — but every answer is reasoned from the definitions, not memorised.
True or false — justify
A matrix can be in REF but not RREF.
True — REF only demands the staircase and zeros below pivots; RREF adds leading 1's and zeros above, so a REF with a pivot equal to or a nonzero entry above a pivot is not yet RREF.
The RREF of a given matrix is unique.
True — no matter which valid sequence of row operations you choose, you always land on exactly one RREF for that matrix; the pivot positions and the cleaned entries are forced.
The REF of a given matrix is unique.
False — the staircase shape (which columns hold pivots) is forced, but the actual numbers are not; different scalings and replacements give different REFs, e.g. pivots vs .
If two matrices have the same RREF, they have the same solution set.
True — RREF is reached by solution-preserving row operations, so matrices sharing an RREF are row-equivalent and describe the same system's solutions.
Every matrix has an RREF.
True — Gauss–Jordan always terminates: each column either gets a pivot or is skipped, and the process is finite, so an RREF always exists (even the all-zero matrix, whose RREF is itself).
The zero matrix is already in RREF.
True — it has no non-zero rows, so conditions 1–5 hold vacuously (no pivots to check); nothing needs cleaning.
A row of all zeros counts as a pivot row.
False — a pivot is the first non-zero entry of a row, and a zero row has none, so it contributes no pivot and no rank.
Adding a multiple of a row to itself (i.e. ) is a legal elementary row operation.
False — the legal "replace" op is with ; using the same row multiplies it by , which is really a scale and is only reversible if — mislabelling it invites accidentally zeroing a row.
Scaling a row by is an allowed operation.
False — scaling requires ; multiplying by turns an equation into and is not reversible, so it can destroy solution information.
If a matrix has more rows than columns, it cannot be in RREF.
False — extra rows simply become zero rows at the bottom; RREF handles any shape, tall or wide.
A pivot in the last column of an augmented matrix means the system has no solution.
True — that pivot represents a row like , an impossible equation, so the system is inconsistent.
A pivot in the last column of a coefficient (unaugmented) matrix means no solution.
False — in a plain coefficient matrix the last column is just another variable's column; a pivot there is perfectly fine and simply makes that variable basic.
Row-equivalent matrices always have equal rank.
True — rank equals the number of pivots, and row operations never change the pivot count, so equivalent matrices share their rank.
Spot the error
"I reached a staircase with zeros below every pivot, so this is the RREF."
Error — that is only REF; RREF also needs every pivot equal to and zeros above each pivot, which you have not yet done.
"My REF has a as a pivot, so I made a mistake somewhere."
No mistake — REF permits any non-zero pivot value; only RREF forces pivots to be exactly .
"There's a pivot in column 3 and a nonzero number two rows below and to the left in column 1, so the staircase is fine."
Error — the staircase requires each pivot strictly right of the one above; a nonzero entry left-and-below a pivot means an earlier pivot was skipped and the form is not REF.
"The matrix has 3 variables and rank 3, so it must have infinitely many solutions."
Error — full rank equal to the number of variables (with a consistent system) gives a unique solution, not infinitely many; free variables appear only when rank is less than the variable count.
"I got a zero row, so the system is inconsistent."
Error — a zero row () is a redundant, always-true equation, not a contradiction; inconsistency needs a pivot in the augmented column ().
"Both matrices reduced to the same staircase shape, so they're the same matrix."
Error — same pivot positions means same rank and same pivot columns, but the actual entries and hence the two matrices can differ; only identical RREFs force identical row spaces.
"To count the rank I counted every row that isn't the very bottom row."
Error — rank counts non-zero rows in REF (equivalently, the number of pivots), regardless of position; bottom rows can be non-zero and top rows could in principle be zero before sorting.
Why questions
Why must all-zero rows sit at the bottom in REF?
Because the pivots must march down-and-right; a zero row in the middle would break the strictly-descending staircase of pivot columns.
Why do we clear entries above pivots for RREF but not for plain REF?
Clearing below is enough to isolate variables for back-substitution (REF's job); clearing above too means each variable appears in exactly one equation, so the answer can be read off with no substitution.
Why do the three row operations never gain or lose a solution?
Each one is reversible — you can undo a swap, un-scale by , or subtract back the added multiple — and a reversible transformation maps the solution set onto itself exactly.
Why does having fewer pivots than variables produce infinitely many solutions (when consistent)?
Each pivot-free column is a free variable you can set to any value, and every such choice yields a valid solution, so the solution set is a whole line/plane rather than a point.
Why are the pivot columns of the original matrix linearly independent?
Row operations preserve linear dependence relations among columns, and in RREF the pivot columns are distinct standard basis vectors (only entry a ), which are visibly independent — see Linear Independence.
Why can Gauss–Jordan be used to compute a matrix inverse?
Augmenting and reducing to applies the exact inverse operations to , so the right block becomes ; this only works when reduces fully to (i.e. is invertible).
Why do we swap a non-zero entry into the pivot position instead of just dividing?
If the current pivot slot is you cannot divide by it, so you swap up a row that has a non-zero entry there — the swap avoids division by zero and keeps the staircase forming.
Edge cases
What is the RREF of the zero matrix?
It is the zero matrix itself — no pivots exist, so all conditions hold vacuously and nothing changes.
What does REF look like for a single-row matrix like ?
It is already in REF (one pivot in column 3, no rows below to worry about); scaling to get a leading would additionally make it RREF.
If a coefficient matrix reduces to the identity, how many solutions does the system have?
Exactly one — three pivots for three variables leaves no free variables, so the solution is the unique vector read off the augmented column.
A consistent system has rank with variables; how big is the solution set?
There are free variables, so the solutions form a 2-dimensional family (a plane's worth), infinitely many.
Can an inconsistent system still have a well-defined REF/RREF?
Yes — the reduction always completes; the inconsistency simply shows up as a pivot in the augmented column (a row), and the RREF is still unique.
What is the rank of an matrix whose RREF has non-zero rows?
The rank is , since rank equals the number of pivots equals the number of non-zero rows in any REF/RREF — bounded by .
Is a matrix with one nonzero row, , in RREF?
No — the pivot is , not ; scaling by to get makes it RREF (the entry above/beside is fine since it is not in a pivot column).
Recall One-line summary of the traps
REF is a staircase (not unique, any pivot value); RREF cleans above too (unique, leading 1's); zero rows and pivots-in-the-augmented-column decide redundant vs impossible; and rank = pivots = non-zero rows always.