Question bank — Solving 2×2 systems using Cramer's rule

False. It only applies when $D\ne 0$; if $D=0$ the two lines are parallel or identical, so there is no unique point to hand back and the division $D_x/D$ is undefined.

True — but only when $D\ne 0$. Then $x=D_x/D=0/D=0$. If $D=0$ as well, $x$ is not zero, it is undefined by this rule, so the statement can trap you.

False. $D=0$ only means "no unique solution." It splits into two cases: no solution (parallel distinct lines) or infinitely many (same line), decided by whether $D_x,D_y$ are also zero.

False. Swapping the rows flips the sign of $D$, $D_x$, and $D_y$ all at once, so every ratio $D_x/D$ and $D_y/D$ is unchanged — same lines, same crossing point.

False. It replaces the first column, because $x$'s coefficients live down column 1. "Variable = column," never row.

True. With $D\ne 0$ both ratios are $0/D=0$, so $(0,0)$ is the unique solution — this is exactly a homogeneous system whose only answer is the origin, see Consistency of linear systems.

False. Scaling equation 1 by $5$ multiplies the top row of every determinant by $5$, so $D,D_x,D_y$ all scale by $5$; the ratios and the "$D=0$ or not" verdict are untouched.

False. $D\ne 0$ guarantees the two lines cross at exactly one point, so a solution always exists and is unique. Failure only happens when $D=0$.

The error is dividing by zero — $D_x/0$ is undefined, not "big." Once $D=0$ you must stop and inspect $D_x,D_y$ to classify the system instead.

Wrong column. $y$ owns column 2, so $D_y$ swaps column 2 for the constant column; swapping column 1 there gives you $D_x$ by mistake.

A sign flip — the rule is main-diagonal product minus anti-diagonal product, in that order. This student negated $D$; if they don't also negate $D_x,D_y$ the answers come out with wrong signs.

You cannot cancel the zero denominator into the numerator. $-8/0$ is undefined; $D=0$ with $D_x\ne 0$ is the signature of no solution, not a value.

Identical lines overlap everywhere, giving infinitely many solutions, not one. Here $D=D_x=D_y=0$, and you describe the solution as a whole line, e.g. $x=c_1-b_{1}y$ with $y$ free.

$D$ is built purely from the coefficient matrix $A$ (the $a$'s and $b$'s). Pulling the constants in turns $D$ into $D_x$ or $D_y$ and wrecks the whole ratio.

A row collapsing to $0=0$ is a true redundant statement — it means the two equations are the same line, giving infinitely many solutions ($D=D_x=D_y=0$), the opposite of inconsistent.

Because if $D=0$ Cramer's rule cannot give a value; computing $D$ first lets you stop early and switch to classifying the degenerate case rather than dividing by zero.

Elimination of $y$ leaves $(a_1{b}_2-a_2{b}_1)x=c_1{b}_2-c_2{b}_1$; the right side $c_1{b}_2-c_2{b}_1$ is the determinant of $A$ with column 1 swapped for the constants — the algebra manufactures the column swap for you.

$D$ is the signed area of the parallelogram spanned by the two coefficient rows; zero area means the rows point along the same direction, i.e. the two lines have equal slope and are parallel or identical.

$D=0$ is common to both degenerate cases. Only $D_x$ and $D_y$ break the tie: all-zero means the same line (infinite), any nonzero means parallel distinct lines (none).

Column swapping is an antisymmetry of the determinant — it reflects the parallelogram, reversing its orientation and hence the sign of the signed area. This is why column order in $D_x,D_y$ matters.

The inverse Inverse of a 2×2 matrix carries a factor $1/\det A=1/D$, and multiplying it against $c$ reproduces the $D_x/D$ and $D_y/D$ ratios term by term — same $1/D$, same numerators.

Scaling row 1 by $k$ multiplies $D,D_x,D_y$ all by the same $k$, so the ratios $D_x/D$ and $D_y/D$ divide the $k$ out — geometrically you rewrote the same line, not moved it.

Then $D_x=D_y=0$ automatically. If $D\ne 0$ the only solution is $x=y=0$ (the origin); if $D=0$ there are infinitely many solutions through the origin.

$D=D_x=D_y=0$, so infinitely many solutions; you keep one equation and let one variable run free, e.g. $y$ free and $x$ expressed from it.

The top row of $A$ is all zeros, forcing $D=0$; the "equation" is either impossible ($c_1\ne 0$, no solution) or vacuous ($c_1=0$, redundant) — it never contributes a usable line for a unique intersection.

No. $D\ne 0$ because the lines are not parallel; the intersection merely happens to sit at the origin, which shows up as $D_x=D_y=0$ while $D$ stays nonzero.

Yes, $D$ is a signed area, so it can be negative; the sign only records the orientation of the coefficient rows. Nothing is wrong — the ratios $D_x/D,D_y/D$ still give the correct unique solution.

Yes. A tiny but nonzero $D$ still means the lines cross once, so the system has a genuine unique solution; treating it as $0$ falsely declares the system degenerate.