Question bank — Properties of determinants

False. Linearity (P5) works one row at a time with the other rows frozen; adding whole matrices changes every row at once, so it can't factor. Counterexample $A=B=I_2$: $\det (2I_2)=4\ne 1+1$.

False. Scaling $A$ by $k$ scales all $n$ rows, and each contributes one factor of $k$ (P4), so $\det (kA)=k^n\det A$.

False. A zero row forces $\det =0$ (P7), but the converse fails: $\det =0$ only means the rows are linearly dependent, e.g. $\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=0$ with no zero row.

True. By P1 every row rule is a column rule, and swapping rows flips the sign (P2), so swapping columns does too.

Only square matrices have determinants, so the premise is fine only for square $A$; for those P1 gives $\det (A^{\top })=\det A$ because reindexing the permutation sum by ${\sigma }^{-1}$ keeps every signed term.

True. By the product rule $\det (AB)=\det A\det B=\det B\det A=\det (BA)$, since the two numbers commute even when the matrices do not.

False. The permutation sum uses off-diagonal entries too. $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|=-1\ne 0$.

False. "Slide Saves" (P6) requires adding a multiple of a different row. $R_i\to R_i+k{R}_i=(1+k)R_i$ scales the row, so by P4 the determinant is multiplied by $(1+k)$.

False in general. $-A=(-1)A$ so $\det (-A)=(-1{)}^n\det A$: unchanged for even $n$, negated for odd $n$.

True. From $A{A}^{-1}=I$ and the product rule, $\det A\cdot \det (A^{-1})=\det I=1$, so $\det (A^{-1})=1/\det A$ (which also shows $\det A\ne 0$ is required for an inverse to exist). See Inverse of a Matrix via Adjoint.

The swap was ignored. The slide (P6) is safe, but the swap (P2) multiplies the running value by $-1$; forgetting that flips the sign of the final answer.

If the $3$ came from one row, dividing by $3$ is right (P4). But if you factored $3$ from every row you pulled out $3^3=27$, so you must divide by $27$, not $3$.

Proportional still kills it: factor the $5$ out of $R_2$ (P4) to get two equal rows, which force $\det =0$ (P3). Dependence, not literal equality, is the trigger.

Powers are products, not scalings. By P9, $\det (A^2)=\det A\cdot \det A=(\det A{)}^2$, not $2\det A$.

Each swap flips the sign (P2). If you swap during a computation you must track a factor of $-1$ per swap; otherwise the "easier" route gives the wrong sign.

Inverting has nothing to do with sign flips. The correct relation is $\det (A^{-1})=1/\det A$ from the product rule.

Every term in the permutation sum picks exactly one entry from that row; with a zero factor available in each term, all terms vanish (this is exactly P7, i.e. P4 with $k=0$).

Any permutation that dips off the diagonal must grab a zero entry from the empty triangle, killing that term; only the identity permutation survives, giving $a_{11}{a}_{22}\cdots a_{nn}$ (P8).

By additivity (P5) it splits into the original plus $k$ times a determinant with two equal rows; the second is $0$ (P3), so the value is "saved" — that is precisely how P6 is proved. See the sliding-card picture below.

Because $\det (A^{\top })=\det A$ (P1): transposing turns rows into columns without changing the value, so any statement provable for rows transfers verbatim to columns.

The swap changes nothing (rows were identical) yet must negate the value (P2), so $\det =-\det$, and the only number equal to its own negative is $0$.

The magnitude tells how area/volume stretches, and the sign records whether orientation is preserved $(+)$ or mirror-flipped $(-)$. The picture below shows a positive-area square and its flipped, negative-area mirror image — see Linear Transformations & Scaling of Area.

By convention it is $1$ — the permutation sum has exactly one term, the empty product, which equals $1$. This keeps rules like $\det (AB)=\det A\det B$ and cofactor expansion working when a submatrix runs out of rows.

Just $a$ itself — the permutation sum has a single identity term $a_{11}$, so there is no cross-multiplication.

$1$, because it is triangular with all diagonal entries $1$ (P8); this is the "no stretching, no flip" baseline.

The two rows are dependent, so the unit square is squashed onto a line: zero area, no inverse. This is exactly the Area of a Triangle using Determinants going flat.

It becomes a zero row, and $\det =0$ (P4 with $k=0$, which is P7); the map collapses one direction entirely.

Yes — positivity of entries says nothing about dependence. $\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=0$ has equal rows despite every entry being positive (P3).

Only $n=1$, where $\det (kA)=ka=k\det A$; for all $n\ge 2$ the correct factor is $k^n$, so the naive rule is a genuine trap.