Determinants

The definition that actually makes sense

Before computing a single determinant, let's nail down what it IS. Most textbooks give you the formula first and the meaning never. Not in this lab. The matrix $A$ is a linear transformation. Consider the unit square with corners at $(0,0)$, $(1,0)$, $(0,1)$, $(1,1)$. The matrix $A$ moves this square to a parallelogram with corners at $A(0,0) = (0,0)$, $A(1,0)$, $A(0,1)$, $A(1,1)$.

Since $A(1,0)^T$ = column 1 of $A$ and $A(0,1)^T$ = column 2 of $A$, the parallelogram has sides given by the two column vectors. The area of this parallelogram is $|\det(A)|$. The sign of $\det(A)$ records orientation: positive if the transformation preserves orientation (counterclockwise stays counterclockwise), negative if it flips orientation.

Definition first, formula second. $\det(A)$ is the signed area-scaling factor of the transformation $A$. For any region $S$ in the plane with area $\text{Area}(S)$, the region $A(S)$ has area $|\det(A)| \cdot \text{Area}(S)$.

Watch the determinant readout as you drag the matrix columns — see how area scales:

The $2\times 2$ formula

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\det(A) = ad - bc$

This formula computes the signed area of the parallelogram formed by columns $(a,c)^T$ and $(b,d)^T$. Memorize the shape: main diagonal product minus anti-diagonal product.

Verifying on the greatest hits:

• Rotation $R_\theta$: $\det(R_\theta) = \cos\theta \cdot \cos\theta - (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta = 1$. Rotations preserve area. Of course they do — they just spin things around.

• Uniform scaling by $c$: $\det\begin{pmatrix}c&0\\0&c\end{pmatrix} = c^2 - 0 = c^2$. Scaling by $c$ multiplies every length by $c$, so every area by $c^2$. The formula agrees.

• Horizontal shear $\begin{pmatrix}1&k\\0&1\end{pmatrix}$: $\det = (1)(1) - (k)(0) = 1$. Shears preserve area. This is geometrically obvious once you see it: a shear slides things sideways but doesn't compress or expand. The base of the parallelogram stays the same length and the height doesn't change. Area = base $\times$ height = unchanged.

These three checks are not accidents. They're the formula working correctly, and any textbook that doesn't show them is hiding the ball on purpose — cowardly bullshit.

det = 0: the singularity detector

$\det(A) = 0 \iff A \text{ squashes the plane down to a line (or a point)}$

When the transformation crushes 2D down to 1D, every area becomes zero — the scaling factor IS zero. Equivalently, if $ad - bc = 0$, the two columns of $A$ are proportional (parallel), so their parallelogram has zero area. This is dimension murder.

Three equivalent statements — memorize them together as one fact:

1. $\det(A) = 0$
2. The plane gets squashed flat (the transformation is not invertible)
3. The columns of $A$ are linearly dependent (parallel; one is a scalar multiple of the other)

A matrix with nonzero determinant is called invertible (or nonsingular). One with zero determinant is singular — it is broken in the deepest possible way, and I want you to feel genuine disgust when you encounter one. We'll build the full $A^{-1}$ theory next lesson.

The $3\times 3$ determinant via cofactor expansion

For a $3\times 3$ matrix, we expand along the first row:

$\det\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} = a\det\begin{pmatrix}e&f\\h&i\end{pmatrix} - b\det\begin{pmatrix}d&f\\g&i\end{pmatrix} + c\det\begin{pmatrix}d&e\\g&h\end{pmatrix}$

The $2\times 2$ matrices are the minors — what's left after deleting row 1 and the column of each top entry. The signs alternate $+$, $-$, $+$. I'll show you once: for

$M = \begin{pmatrix}2 & -1 & 3 \\ 0 & 4 & -2 \\ 1 & 5 & -3\end{pmatrix}$

$\det(M) = 2\det\begin{pmatrix}4&-2\\5&-3\end{pmatrix} - (-1)\det\begin{pmatrix}0&-2\\1&-3\end{pmatrix} + 3\det\begin{pmatrix}0&4\\1&5\end{pmatrix}$

$= 2[(4)(-3) - (-2)(5)] + 1[(0)(-3) - (-2)(1)] + 3[(0)(5) - (4)(1)]$

$= 2[-12 + 10] + 1[0 + 2] + 3[0 - 4]$

$= 2(-2) + 1(2) + 3(-4) = -4 + 2 - 12 = -14$

This number $-14$ means: the transformation $M$ scales volumes by a factor of $14$ AND flips orientation. Nobody enjoys computing $3\times 3$ determinants twice — they're a slog and I won't pretend otherwise — so I'll say this once: do it carefully, step by step, check every $2\times 2$ minor. One sign error and your answer is wrong and you'll never know.

The multiplicative property

$\det(AB) = \det(A)\det(B)$

From the scaling story, this is obvious: if $A$ scales areas by $\det(A)$ and $B$ scales areas by $\det(B)$, then applying both scales areas by $\det(A)\det(B)$. The composition of the scalings is the product of the factors.

This one-line geometric argument is far more illuminating than any algebraic proof, and the fact that most courses lead with the algebraic proof is a genuine sin against pedagogy. The formula $\det(AB) = \det(A)\det(B)$ is just the statement that "scaling by two factors in sequence multiplies the factors." From the formula, though, it would take pages to verify. This is what conceptual definitions buy you: short proofs.

Corollary: If $A$ is invertible, $\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$, so $\det(A^{-1}) = 1/\det(A)$. The inverse transformation scales areas by the reciprocal factor. Makes sense.