Chapter 4 Quiz — Gaussian Elimination and Row Reduction
Twelve quick checks on the algorithm and what it reveals. Try each before opening the answer. These are conceptual — reach for a pencil only on the short ones.
Q1. Name the three elementary row operations.
Answer
**Swap** two rows; **scale** a row by a nonzero constant; **add a multiple of one row to another**. All three are reversible, which is why none changes the solution set.Q2. Why is "scale a row by $0$" not an allowed elementary operation?
Answer
Because it is **not reversible** — multiplying an equation by $0$ turns it into the vacuous $0 = 0$, destroying information, and you cannot undo it (there is no number you can multiply $0=0$ by to recover the original equation). Reversibility is exactly what guarantees the solution set is preserved, so the zero scaling is banned.Q3. What is the goal of forward elimination, and what is the goal of back-substitution?
Answer
Forward elimination creates zeros **below** the diagonal, driving the system to a triangular (row echelon) form. Back-substitution then solves that triangular system from the **bottom row up**, one new unknown at a time. Forward elimination is the hard part; solving the triangle is easy.Q4. During elimination you find a $0$ sitting in the current pivot position, but there is a nonzero entry below it in the same column. What do you do — and is this an error?
Answer
**Swap** in a lower row that has a nonzero entry in that column, then continue. It is **not** an error — pivoting (swapping) is a normal, expected step. It would only be a problem if you tried to divide by the zero pivot. (If *no* row had a nonzero entry there, the column simply yields no pivot — a free variable, still not an error.)Q5. A matrix is in REF with pivots $2, -3, 5$. Is it in RREF? What two things must change to make it RREF?
Answer
**No.** For RREF, (1) every pivot must be **$1$** (so scale each row by the reciprocal of its pivot), and (2) each pivot must be the **only nonzero entry in its column** (so clear the entries *above* each pivot, not just below). REF only guarantees the zeros below.Q6. True or false: a given matrix can have several different row echelon forms but only one reduced row echelon form.
Answer
**True.** Different legal sequences of row operations can produce different REFs (same staircase pattern, different numbers), but the **RREF is unique** — determined entirely by the original matrix. This uniqueness is why RREF serves as a canonical fingerprint and underlies the rank and the four-subspaces theory later.Q7. After reducing $[A\mid\mathbf{b}]$, the last row is $0\ 0\ 0 \mid 4$. How many solutions does the system have?
Answer
**None** — the system is **inconsistent**. That row reads $0 = 4$, which no values of the variables can satisfy. Equivalently, $\operatorname{rank}([A\mid\mathbf{b}]) > \operatorname{rank}(A)$.Q8. A different reduction ends with the last row $0\ 0\ 0 \mid 0$, and the matrix has $3$ variables but only $2$ pivots. How many solutions, and what is the dimension of the solution set?
Answer
**Infinitely many.** The all-zero row is a harmless redundant equation, and $3$ variables $-$ $2$ pivots $= 1$ **free variable**, so the solution set is a **line** (dimension $1$) — a particular solution plus all multiples of one direction vector.Q9. Fill in the existence/uniqueness rule. With $n$ variables: a unique solution requires $\operatorname{rank}(A) = \operatorname{rank}([A\mid\mathbf{b}]) = \rule{1cm}{0.4pt}$; infinitely many requires the two ranks equal but $\rule{1cm}{0.4pt}$ than $n$; no solution requires the two ranks to be $\rule{1cm}{0.4pt}$.
Answer
Unique: both ranks equal **$n$** (no free variables). Infinitely many: both ranks equal but **less** than $n$ (at least one free variable). No solution: the two ranks **unequal** (a pivot in the augmented column).Q10. Why does the chapter insist you should not solve $A\mathbf{x}=\mathbf{b}$ by first computing $A^{-1}$ and then forming $A^{-1}\mathbf{b}$?
Answer
Because computing $A^{-1}$ is **more work** than a single elimination (it is itself an elimination, plus an extra matrix–vector multiply) and is **numerically less stable**. The professional rule is "never invert a matrix to solve a system; eliminate." numpy's `solve` follows this — it runs a pivoted LU elimination, not an inversion.Q11. Is row reduction "the point" of linear algebra? Explain in one or two sentences.
Answer
**No.** Row reduction is a **tool** — for solving $A\mathbf{x}=\mathbf{b}$ and for *revealing a matrix's structure* (its rank, pivot columns, free-variable directions). The point of linear algebra is **linear transformations** — what matrices *do* to space — and elimination is one of our sharpest instruments for studying them, not the subject itself.Q12. When elimination produces a row of all zeros (including the augmented entry), the chapter calls it the algebraic fingerprint of what?