2×2
Second-Order Det.
Multiply diagonals, then subtract.
\(\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad - bc\)
3×3
Diagonal Rule
Sum of ↘ diagonals minus ↙ diagonals.
aei+bfg+cdh − ceg−fha−ibd
AREA
Triangle Area
Use vertices (a,b), (c,d), (e,f) in det.
\(A = \tfrac{1}{2}\left|\det\begin{pmatrix}a&b&1\\c&d&1\\e&f&1\end{pmatrix}\right|\)
KEY
Sign Matters!
Determinant can be negative → Area is always |A| (absolute value).
Area = |det| · ½ ≥ 0