After row-reducing a system of 4 equations in 5 unknowns to REF, you find exactly 3 pivot columns. How many free variables does this system have?
A3 — one per pivot
B4 — one per equation
C2 — one per non-pivot unknown (n − r = 5 − 3)
D1 — one per non-pivot row
Free variables correspond to non-pivot columns. With n = 5 unknowns and r = 3 pivots, there are n − r = 2 free variables. Each can take any real value independently, with the 3 basic variables determined by back-substitution. Option A confuses basic variables (pivot columns) with free variables.
Question 2 Multiple Choice
A student row-reduces a matrix and produces a different-looking REF than her classmate, who also correctly row-reduced the same matrix. What can they conclude?
AOne of them made an error — REF is unique
BBoth are correct — REF is not unique, but both have pivots in exactly the same columns
CREFs can differ only in their right-hand side values, not in pivot positions
DThe results are only comparable after both are further reduced to RREF
REF is not unique — different valid sequences of row operations can produce different-looking forms with different values in non-pivot positions. However, the pivot columns (and thus the rank) are invariants determined by the matrix itself, not by the reduction path. Both students will always agree on which columns are pivot columns, even if their REFs look different elsewhere.
Question 3 True / False
Two valid row reductions of the same matrix can produce different row echelon forms, yet both will identify the same columns as pivot columns.
TTrue
FFalse
Answer: True
REF is not unique — different operation sequences yield different-looking results. But the pivot columns are determined by the matrix itself: they correspond to the columns that contain the leading nonzero entries of each row, and this set is invariant regardless of reduction path. This invariance is why rank (the count of pivots) is a well-defined property of the matrix.
Question 4 True / False
A system with more unknowns than equations usually has infinitely many solutions.
TTrue
FFalse
Answer: False
Having more unknowns than equations guarantees at least one free variable IF the system is consistent — but the system might not be consistent. A zero row with a nonzero right-hand side (the equation 0 = c, c ≠ 0) is a contradiction that makes the system inconsistent with no solutions at all. More unknowns than equations is a necessary but not sufficient condition for infinitely many solutions.
Question 5 Short Answer
Why can a system of linear equations have exactly one solution, infinitely many solutions, or no solution — but never exactly two solutions?
Think about your answer, then reveal below.
Model answer: If a system is inconsistent (a zero row with nonzero RHS in REF), it has no solutions. If consistent, the solution set depends on free variables. Zero free variables means each unknown is uniquely determined by back-substitution: exactly one solution. One or more free variables means each can take any real value, generating a continuous family — infinitely many solutions. There is no configuration that produces a finite number greater than one, because any free variable ranges over all real numbers.
This follows directly from the structure of REF. The three cases — no solution, unique solution, infinitely many — are exhaustive and mutually exclusive. A 'exactly two solutions' scenario would require the solution set to be finite and greater than one, which cannot arise from linear equations: either all variables are pinned down (one solution) or at least one floats freely over ℝ (infinitely many).