Gaussian elimination — forward elimination, back substitution: Question bank

True or false — justify

Row operations change the solution set of a system.

False — the three elementary operations each preserve the solution set exactly; that is precisely why they are the only legal moves.

Multiplying a row by

0

is a valid elementary row operation.

False — scaling must be by a nonzero constant; scaling by

0

erases the equation and can destroy solutions (it is not reversible).

Swapping two rows can change which numbers are the solutions

x_{1}, \dots, x_{n}

False — a swap just reorders the equations; the same

x

still satisfies all of them, so the solution set is untouched.

Forward elimination always produces an upper-triangular matrix with every diagonal entry nonzero.

False — it produces upper-triangular shape, but a diagonal (pivot position) can end up zero, signalling a free variable or forcing a column skip.

A row of the form

[00 \dots 0 ∣ 0]

means the system has no solution.

False — an all-zero row (including RHS) is a redundant equation (

0 = 0

), which permits a free variable and hence infinitely many solutions.

A row of the form

[00 \dots 0 ∣ 5]

means the system has no solution.

True — it asserts

0 = 5

, a contradiction, so the system is inconsistent regardless of the other rows.

If a coefficient matrix is already upper-triangular, forward elimination has nothing to do.

True — every entry below the diagonal is already zero, so no multipliers are needed; you jump straight to back substitution.

Back substitution can be started from the top row.

False — only the bottom row isolates a single unknown; starting on top leaves that row still mixing several unknowns you don't yet know.

The multiplier

m_{ik} = a_{ik} / a_{k k}

can be negative.

True — its sign just reflects the sign of the entry being cleared relative to the pivot; a negative multiplier means you effectively add a scaled pivot row.

Gaussian elimination requires the matrix to be square.

False — it works on any

m \times n

system; squareness only guarantees a unique solution when a full set of pivots also exists.

Spot the error

"I triangularised

A

, then plugged the original

b

into back substitution."

Error —

b

must ride along through every row operation on the augmented matrix; using the untouched original RHS gives the wrong system.

"To clear

a_{ik}

I used

R_{i} \to R_{i} + \frac{a _{ik}}{a _{k k}} R_{k}

Error — the sign is wrong; it must be minus (

R_{i} \to R_{i} - m_{ik} R_{k}

) so the term

m_{ik} a_{k k}

cancels

a_{ik}

instead of doubling it.

"The pivot was

0

, so I divided anyway and got infinity, meaning infinitely many solutions."

Error — a zero pivot forbids division; you must swap in a lower row with a nonzero entry, or skip the column. Infinity here is a bug, not a conclusion.

"I picked the pivot from above the diagonal to make the numbers nicer."

Error — a pivot is chosen on or below the diagonal in its column; using a row above breaks the staircase already built and can reintroduce zeros you cleared.

"After getting

U

, I scaled row 3 by

- 1

but forgot to scale its RHS entry."

Error — scaling a row means scaling the whole row including the augmented column, otherwise that equation now says something false.

"Two rows became identical during elimination, so the system is inconsistent."

Error — identical rows produce an all-zero row after subtraction (

0 = 0

), which means redundancy and a free variable, not inconsistency.

"I did partial pivoting by swapping in the row with the smallest nonzero entry."

Error — partial pivoting swaps in the row with the largest magnitude entry; small pivots amplify rounding errors (see Pivoting and Numerical Stability).

Why questions

Why do we subtract a multiple of the pivot row rather than just deleting the offending entry?

Because we may only use the three legal operations, which keep the solution set intact; blanking an entry would silently change the equation.

Why is the multiplier defined as exactly

a_{ik} / a_{k k}

and nothing to memorise beyond that?

Setting the new entry

a_{ik} - m_{ik} a_{k k}

to zero and solving forces

m_{ik} = a_{ik} / a_{k k}

; the formula is literally the requirement "make this zero."

Why does back substitution go bottom-up instead of top-down?

Only the last equation of an upper-triangular system involves a single unknown; solving it frees the next-to-last, so each step upward reveals exactly one new variable.

Why do we bother with the augmented matrix instead of tracking

A

and

b

separately?

So a single row operation transforms coefficients and RHS together and consistently — packing them prevents the classic mistake of forgetting the RHS.

Why does a zero pivot with no nonzero entry below it in that column not stop the algorithm?

It means that variable is not pinned by any remaining equation — a free variable; you simply move to the next column and continue building the staircase.

Why is

R_{i} \to R_{i} - m R_{j}

reversible?

Because applying

R_{i} \to R_{i} + m R_{j}

undoes it exactly; reversibility is what guarantees no solutions were created or lost.

Why does the number of nonzero pivots matter beyond just solving?

It equals the rank of the matrix and, via the product of pivots, feeds the Determinant — elimination measures the system, not just solves it.

Why can partial pivoting improve accuracy even when the pivot is not exactly zero?

Dividing by a tiny pivot magnifies floating-point errors; swapping in a larger entry keeps the multipliers small and the arithmetic stable.

Edge cases

2 \times 2

system reduces to a single all-zero row plus one real equation — how many solutions?

Infinitely many: one genuine constraint on two unknowns leaves one free variable, a whole line of solutions.

Every diagonal pivot is nonzero after elimination on a square system — what does that guarantee?

A unique solution: full pivots mean full rank, nonzero Determinant, and back substitution determines each variable exactly.

The very first pivot

a_{11}

0

but a lower row has a nonzero first entry — what must you do before eliminating?

Swap that lower row up into the pivot position; only then is division by the pivot valid and column 1 can be cleared.

A system has more equations than unknowns (

m > n

) and elimination yields

[0 \dots 0 ∣ 7]

— verdict?

Inconsistent — an extra equation contradicts the rest; over-determined systems commonly have no exact solution.

A system has fewer equations than unknowns (

m < n

) and is consistent — can it have a unique solution?

No — with fewer pivots than unknowns at least one variable is free, forcing infinitely many solutions (never exactly one).

During elimination a pivot column below the pivot is already all zeros — what happens to the work?

Nothing to clear: the column is done, so you move to the next column; that pivot simply required no operations.

Connections

Row Echelon Form / Reduced Row Echelon Form — the target shapes these traps revolve around.
Solving Linear Systems — Consistency — the no/one/infinite verdicts formalised.
Pivoting and Numerical Stability — why swapping and pivot size matter.
LU Decomposition — where the multipliers get stored.
Rank of a Matrix / Determinant — what the pivots secretly measure.