NapkinCalc

Linear Algebra A — Matrices & Determinants

The determinant — the area factor

continues from lesson 1 — values defined earlier in the course stay live here

ELI5: the determinant crushes a 2×2 machine down to one telling number — the factor by which it scales area. det = 5 → areas come out 5× bigger; det = 0 → the machine flattens space onto a line, and flattening can never be undone. For [[a, b], [c, d]] the formula is ad − bc.

dM=det(M)d_{M} = \mathrm{det}\left(M\right) = 55 det = 2·3 − 1·1 = 5: areas grow 5×
Mflat:=[2,4;1,2]M_{flat} := [2, 4; 1, 2] = [[2,4],[1,2]][[2, 4], [1, 2]] second row is just half the first…
dflat=det(Mflat)d_{flat} = \mathrm{det}\left(M_{flat}\right) = 00 …so it collapses space: det = 0
✓ pass dflat==0anddM==5d_{flat} == 0 and d_{M} == 5 one machine stretches, one flattens

Real-world hook: a zero determinant is the math behind a singular (un-invertible) system — sensors that secretly measure the same thing, equations that don't actually pin down a unique answer.

Try it yourself: compute the determinant of [[5, 2], [3, 4]] (use ad − bc).

detyou=14det_{you} = 14 ✏️ Your turn: find det[[5, 2], [3, 4]] with ad − bc. The check confirms it equals 5·4 − 2·3.
✓ pass abs(detyou−5⋅4−2⋅3)<10−9\mathrm{abs}\left(det_{you} - 5 \cdot 4 - 2 \cdot 3\right) < 10^{-9} green when your determinant equals ad − bc

Where next: Linear Algebra B uses the determinant to undo a matrix and to solve systems of equations.