Exercises — Geometric interpretation — signed volume

WHAT: we judge sign from the turn and size from the box. WHY: turning from "right" ($\mathbf{a}$) to "up" ($\mathbf{b}$) is a CCW quarter-turn → positive. WHAT IT LOOKS LIKE: an upright $2\times 3$ rectangle — the shaded (mint) box in Figure 1, the one bounded by the right-pointing and up-pointing arrows — with area $6$. So the answer is $+6$. Check: $2\cdot 3-0\cdot 0=6$. ✓

(a) The box is flattened onto a line (2D) or plane (3D) — zero volume. (b) The columns are linearly dependent (one is a combination of the others). (c) The map is non-invertible — it squashes information, so it cannot be undone. See Invertibility and the inverse matrix and Linear independence and rank.

False. The magnitude $|-5|=5$ is the physical volume; the minus sign only records that orientation was reversed (a mirror flip). Negative determinants are ordinary and meaningful.

WHAT: apply $a_1{b}_2-a_2{b}_1$ with $a_1=4,a_2=2,b_1=1,b_2=3$. $4\cdot 3-2\cdot 1=12-2=10.$ Positive → orientation preserved; the box has area $10$.

$1\cdot 6-2\cdot 3=6-6=0$. The second column $(3,6)=3\cdot (1,2)$ is parallel to the first, so the parallelogram collapses to a line — zero area, dependent columns.

WHY the triple product: $\mathbf{b}\times \mathbf{c}$ gives a vector whose length is the base-parallelogram's area and whose direction is the base's normal; dotting with $\mathbf{a}$ multiplies by the height along that normal → base × height = volume (see Cross product and scalar triple product). First $\mathbf{b}\times \mathbf{c}$ with $\mathbf{b}=(0,3,0),\mathbf{c}=(1,1,1)$: $\mathbf{b}\times \mathbf{c}=(3\cdot 1-0\cdot 1,0\cdot 1-0\cdot 1,0\cdot 1-3\cdot 1)=(3,0,-3).$ Then dot with $\mathbf{a}=(1,0,2)$: $1\cdot 3+0\cdot 0+2\cdot (-3)=3-6=-3.$ Signed volume $-3$: magnitude $3$, orientation reversed (a left-handed frame). See Orientation and handedness of bases.

WHY $c^n$: multiplying the whole matrix by $2$ scales all three columns, and each scaling multiplies the volume by $2$: $2\times 2\times 2=2^3$. $\det (2A)=2^3\cdot 4=8\cdot 4=32.$

Notation: write $\det (\mathbf{a},\mathbf{b})$ for the signed area of the box with edge columns $\mathbf{a},\mathbf{b}$ — this is exactly $\det \left(\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right)$, just viewed as a function of its two columns. The multilinearity axiom (recap): the parent note established that signed area is multilinear — it is additive in each column separately, $\det (\mathbf{u}+\mathbf{w},\mathbf{b})=\det (\mathbf{u},\mathbf{b})+\det (\mathbf{w},\mathbf{b})$ (and likewise in the second slot). This is what lets us break a sum inside one slot into two separate determinants — it is the concrete engine of the expansion below. The alternating axiom (recap): the parent note also established that signed area is alternating — a box with two equal edges is flat, so $\det (\mathbf{v},\mathbf{v})=0$ for any $\mathbf{v}$. Algebra: apply the alternating axiom to the vector $\mathbf{a}+\mathbf{b}$ in both slots, so $\det (\mathbf{a}+\mathbf{b},\mathbf{a}+\mathbf{b})=0$. Now expand by multilinearity — split the first slot's sum, then split each resulting second slot's sum: $\det (\mathbf{a},\mathbf{a})+\det (\mathbf{a},\mathbf{b})+\det (\mathbf{b},\mathbf{a})+\det (\mathbf{b},\mathbf{b})=0.$ The two equal-argument terms $\det (\mathbf{a},\mathbf{a})$ and $\det (\mathbf{b},\mathbf{b})$ vanish by the alternating axiom, leaving $\det (\mathbf{a},\mathbf{b})=-\det (\mathbf{b},\mathbf{a})$. Geometry: swapping the edges reverses the sense of the turn (CCW ↔ CW), so the same box gets the opposite sign — a mirror flip. Numeric check: $\det \left(\begin{array}{cc}4& 1\\ 2& 3\end{array}\right)=10$ while $\det \left(\begin{array}{cc}1& 4\\ 3& 2\end{array}\right)=1\cdot 2-3\cdot 4=-10$. ✓

WHY this formula for $\det$: $a_1{b}_2-a_2{b}_1$ with $a_1=\cos \theta ,a_2=\sin \theta ,b_1=-\sin \theta ,b_2=\cos \theta$: $\cos \theta \cdot \cos \theta -\sin \theta \cdot (-\sin \theta )={\cos }^2\theta +{\sin }^2\theta =1.$ Interpretation: a rotation never stretches or flips anything — a spun square keeps its area ($|\det |=1$) and its handedness (sign $+$). The Pythagorean identity is exactly the algebra of "rigid turn."

$\det S=1\cdot 1-0\cdot k=1$ for every $k$. Geometry: the sheared unit square becomes a parallelogram with the same base $1$ and same height $1$ — only the top slides horizontally. Area = base × height is untouched by the slide, so the determinant stays $1$ no matter how extreme the shear.

WHY multiplicativity: doing map $B$ then map $A$ scales volume by $\det B$ then by $\det A$, so the composite scales by the product. Since $A{A}^{-1}=I$ and $\det I=1$: $\det A\cdot \det (A^{-1})=1\Rightarrow \det (A^{-1})=\frac{1}{\det A}.$ This fails exactly when $\det A=0$ — a flattened map cannot be undone, so no inverse exists (Invertibility and the inverse matrix). Example: if $\det A=5$, then $\det (A^{-1})=\frac{1}{5}$.

WHY the Jacobian determinant: locally every smooth map looks linear, and its linear part $J$ scales tiny areas by $|\det J|$ — that is precisely the factor that appears in the change-of-variables integral (Change of variables and the Jacobian). $\det J=2\cdot 3-0\cdot 1=6,\text{new area}=4\cdot 6=24.$ Orientation is preserved (sign $+$).

WHY the product: the determinant equals the product of eigenvalues, because along each eigendirection the map scales by that eigenvalue, and volume is the product of the per-direction stretches (Eigenvalues — product equals determinant). $\det A={\lambda }_1{\lambda }_2=3\cdot (-2)=-6.$ Magnitude $6$: areas are stretched sixfold. Sign $-$: orientation is reversed — the single negative eigenvalue reflects the plane along its eigendirection.

WHAT: shift $P$ to the origin by using edge vectors $\mathbf{u}=Q-P$ and $\mathbf{v}=R-P$. WHY: translation does not change area, and the two edges $\mathbf{u},\mathbf{v}$ span a parallelogram whose signed area is $\det [\mathbf{u}\mathbf{v}]$. The triangle is exactly half that parallelogram (the diagonal cuts it in two congruent triangles). The absolute value discards orientation because a physical triangle's area is positive. Compute: $\mathbf{u}=Q-P=(3,0)$, $\mathbf{v}=R-P=(1,4)$. $\det \left(\begin{array}{cc}3& 1\\ 0& 4\end{array}\right)=3\cdot 4-0\cdot 1=12,\text{Area}=\frac{1}{2}|12|=6.$

Logic: three points are collinear exactly when the triangle they form has area $0$, i.e. the edge vectors $\mathbf{u}=Q-P$ and $\mathbf{v}=R-P$ are parallel → their determinant is $0$. $\mathbf{u}=(2,4)$, $\mathbf{v}=(4,8)$: $\det \left(\begin{array}{cc}2& 4\\ 4& 8\end{array}\right)=2\cdot 8-4\cdot 4=16-16=0.$ The determinant is $0$, so the points are collinear — the "triangle" is flat.

We need $\det =1$ with both columns longer than the unit length. Try $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right),\det A=2\cdot 1-1\cdot 1=1.$ Column lengths are $\sqrt{2^2+1^2}=\sqrt{5}\approx 2.24$ and $\sqrt{1^2+1^2}=\sqrt{2}\approx 1.41$ — both exceed $1$. Determinant $1$ means area- and orientation-preserving, yet it is not a rotation (rotations keep every length; here the columns are stretched to $\sqrt{5}$ and $\sqrt{2}$, so lengths change — a rotation is impossible). What map is it, then? It is a general area-preserving shearing map. Concretely, factor it into elementary shears: $\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right),$ a horizontal shear followed by a vertical shear (each has determinant $1$, and $1\cdot 1=1$). So $A$ slides space sideways, then slides it upward, distorting the unit square into a slanted parallelogram of the same area $1$ and the same handedness — a pure shear-composite, never a rigid rotation. Any integer matrix with $\det =1$ (an element of $S{L}_2(\mathbb{Z})$) gives such a map.

Signed area of edges $\mathbf{a},\mathbf{b}$ in 2D ::: $a_1{b}_2-a_2{b}_1$ $\det (cA)$ for $n\times n$ ::: $c^n\det A$ (sign flips if $c<0$ and $n$ is odd) $\det (A^{-1})$ ::: $1/\det A$ (needs $\det A\ne 0$) $\det (AB)$ ::: $\det A\cdot \det B$ Triangle area from vertices ::: half the absolute value of the determinant of its two edge vectors Collinearity test for three points ::: edge-vector determinant equals $0$