Determinant -- Geometric & Algebraic

Press a square cookie cutter into a sheet of dough, then gently stretch the dough sideways. The square print becomes a slanted shape that is bigger, smaller, or squashed flat. The determinant is simply the number that tells you how much the area changed: an answer of $2$ means "twice as big," $0.5$ means "half as big," and $0$ means the shape collapsed completely onto a line.

A $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is a machine that moves every point of the plane. It turns the unit square (area $1$) into a parallelogram, and the determinant $\det A=ad-bc$ is exactly that parallelogram's area. Try a pure stretch $A=\begin{pmatrix}3&0\\0&2\end{pmatrix}$: it makes the $x$-direction $3\times$ wider and the $y$-direction $2\times$ taller, so the area grows by $3\times2=6$ -- and indeed $\det A=3\cdot2-0\cdot0=6$.

The columns of $A$ are exactly where the arrows $(1,0)$ and $(0,1)$ land; the sim draws them as $c_1$ and $c_2$. Their parallelogram has a signed area $ad-bc$: when it comes out negative the shape has been flipped like a mirror image (orientation reversed), which the sim signals by turning the patch from gold to red. When $ad=bc$ the two columns lie along the same line, the parallelogram has no width, and $\det A=0$ -- the matrix is singular and its action cannot be undone. The same idea climbs one dimension: a $3\times3$ determinant is the signed volume of the parallelepiped built from three column vectors, and scaling an entire $n\times n$ matrix by $k$ multiplies area/volume by $k^n$, so $\det(kA)=k^n\det A$.

Choose the Singular (det=0) preset and watch the gold parallelogram flatten onto a single line -- its area drops to zero. Pick Rotation (det=1), then press Animate: the square spins continuously yet $\det$ holds perfectly steady, because a rotation never changes area. Finally drag the a22 slider below zero and watch the filled region switch from gold to red the instant $\det A$ crosses zero -- that color flip is the orientation reversing.

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$: $\det A=ad-bc$. Geometric: vectors $(a,c)$ and $(b,d)$ form a parallelogram of area $|ad-bc|$. If $ad=bc$ the columns are proportional -- area is 0 -- singular.

Along row 1: $\det A=\sum_j(-1)^{1+j}a_{1j}M_{1j}$ where $M_{ij}$ is the $(i,j)$ minor. For 3x3: $\det A=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(\ldots)+a_{13}(\ldots)$. Can expand along any row or column (same result).

If $\det A\neq0$: $x_i=\det A_i/\det A$ where $A_i$ is $A$ with column $i$ replaced by $b$. Useful for understanding structure but inefficient ($O(n!)$ vs Gaussian elimination $O(n^3)$).

$\det A=\prod_i\lambda_i$ (product of all eigenvalues, with multiplicity). $\text{tr}(A)=\sum_i\lambda_i$. So $\det A=0\iff$ at least one eigenvalue is 0 $\iff A$ singular.

Row swap: $\det\to-\det$. Scale row by $k$: $\det\to k\cdot\det$. Add multiple of one row to another: $\det$ unchanged. These give an efficient $O(n^3)$ algorithm: reduce to triangular, $\det=$ product of diagonal entries times sign changes.

$\det\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}=1(45-48)-2(36-42)+3(32-35)=1(-3)-2(-6)+3(-3)=-3+12-9=0$. The columns are in arithmetic progression ($c_2-c_1=c_3-c_2=(1,1,1)$, so $c_3=2c_2-c_1$): linearly dependent, hence singular as expected.