Question bank — Determinant of 2×2 matrix

False — scaling every side by $2$ scales the columns by $2$, so area scales by $2\times 2=4$; $\det (2A)=2^2\det (A)=4$ for a $2\times 2$ matrix.

False — $\det =1$ only says area and orientation are preserved; a rotation or a shear like $\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]$ also has $\det =1$ but is not the identity.

False — $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$ has no zeros yet $\det =4-4=0$; what matters is whether one column is a multiple of the other, not whether entries are zero.

False — a single row swap multiplies the determinant by $-1$ because it reverses the orientation of the two column vectors ($ad-bc\to cb-da$).

True — transposing swaps $b\leftrightarrow c$, and $ad-cb$ is the same number as $ad-bc$; area is unchanged.

False — the magnitude $|-3|=3$ gives the area factor (areas triple); the minus sign only reports an orientation flip, not shrinking.

False — the determinant is not additive. E.g. $A=B=I$ give $\det (A+B)=\det (2I)=4$ but $\det A+\det B=1+1=2$.

True — parallel columns span a line, not a parallelogram; the enclosed area is zero (see Linear independence).

True — since $\det (A)\det (A^{-1})=\det (I)=1$, we get $\det (A^{-1})=1/\det (A)$, which is why $A^{-1}$ can't exist when $\det (A)=0$.

True — both equal $\det (A)\det (B)$, and multiplication of numbers commutes, even though the matrix products $AB$ and $BA$ may differ.

The diagonals are subtracted in the wrong order; correct is main-diagonal-first, $ad-bc$. Their answer is the exact negative — a determinant computed with reversed orientation.

They computed the trace (sum of the diagonal), not the determinant. The determinant is $2\cdot 3-0\cdot 0=6$; trace and determinant are different quantities.

$\det =0$ means singular (columns linearly dependent), not empty. $\left[\begin{array}{cc}0& 1\\ 0& 2\end{array}\right]$ has $\det =0$ yet is far from the zero matrix.

The diagonal entries must be swapped ($a\leftrightarrow d$), not left in place; the correct adjugate is $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (see Inverse of a 2×2 matrix).

Multiplying $A$ by $5$ scales both columns by $5$, so area scales by $5^2$: $\det (5A)=25\det (A)$. The scalar comes out to the power equal to the matrix size.

$\det (AB)=\det (A)\det (B)=4\cdot 0=0$; any product involving a singular matrix is singular. Composing with a flattening map still flattens.

A zero determinant collapses the plane onto a line, so the map isn't reversible; different inputs land on the same output, leaving either no solution or infinitely many (this is the boundary case of Cramer's rule).

Rotation rigidly turns the plane without stretching or flipping, so area is preserved ($+$) with factor $1$ — no scaling and no orientation reversal.

The area magnitude comes from $|ad-bc|$; the sign records whether going column-1-then-column-2 sweeps counterclockwise ($+$) or clockwise ($-$), i.e. whether the plane was flipped like a pancake.

Determinant captures only the area-scaling-plus-flip summary, discarding shape and rotation; a shear and a rotation can both preserve area, giving equal determinants despite acting differently.

Dependent columns are parallel (one is a scalar multiple of the other), so the parallelogram they span has zero width and hence zero area — precisely $\det =0$ (see Linear independence).

Transposing reflects the parallelogram, preserving its area and its orientation relationship; a row swap re-labels which vector is "first," reversing the sweep direction and hence the sign.

$\det (A)=0$: a zero column contributes a zero-length side, so the parallelogram degenerates to a segment (or point) with no area.

Zero — identical columns are the most extreme parallel case; $3\cdot 5-3\cdot 5=0$, so the "parallelogram" is just a line.

It becomes $ad$ (the product of diagonal entries) because the axes stretch independently by $a$ and $d$, scaling area by $a\times d$ with no shearing.

No — invertibility is equivalent to $\det \ne 0$. If $B$ is singular then $\det (B)=0$, forcing $\det (A)=0$, which would make $A$ singular too.

It scales by $k$ once: $\det$ becomes $k\det (A)$, since only one side of the parallelogram is stretched by $k$, changing the area by that single factor.

Yes — invertibility only needs $\det \ne 0$. A negative determinant (say $-13$) means orientation flips, but the map is perfectly reversible.

A line segment or a single point — the collapse boundary between orientation-preserving ($\det >0$) and orientation-reversing ($\det <0$) transformations (see Transformations and scaling).