Every square matrix A carries one number detA=∣A∣. For order 1: ∣[a]∣=a. For order 2: a11a21a12a22=a11a22−a12a21. For order 3: expand along any row or column — signed elements times 2×2 minors — all six choices agree; pick the one with the most zeros.
MinorMij: delete row i, column j. CofactorAij=(−1)i+jMij. Then:
elements × own cofactors =Δ (any row or column),
elements × another row's cofactors =0.
Area and collinearity
Δ=21x1x2x3y1y2y3111
Take the absolute value; when the area is given, equate to ± the area; collinear points ⇔ determinant =0 (which also produces the equation of a line through two points).
Adjoint and inverse
adjA = transpose of the cofactor matrix (2×2 shortcut: swap the diagonal, negate the off-diagonal). The master identity and its offspring:
A(adjA)=(adjA)A=∣A∣I,A−1=∣A∣1adjA(∣A∣=0)
Singular: ∣A∣=0 (no inverse). Nonsingular: ∣A∣=0 (invertible). Key formulas:
Formula
Note
∣AB∣=∣A∣∣B∣
products multiply
∣kA∣=kn∣A∣
one k per row
det(A−1)=1/detA
from AA−1=I
∣adjA∣=∣A∣n−1
=∣A∣2 at order 3
(AB)−1=B−1A−1
order reversed
Solving systems
Write the system as AX=B. If ∣A∣=0: unique solution X=A−1B. If ∣A∣=0: compute (adjA)B — nonzero means inconsistent; zero means infinitely many solutions or none.
Mistake checklist
∣A∣ is not a modulus — determinants can be negative, and only square matrices have them.
∣kA∣=k∣A∣ is wrong — the correct factor is kn, one per row.
Cofactor = minor — the (−1)i+j flip applies whenever i+j is odd.
Adjoint needs the transpose — cofactor matrix first, then flip across the diagonal.
Given-area problems need ± — solving with one sign loses half the answers.
X=A−1B, never BA−1 — premultiply, and state ∣A∣=0 before inverting.
∣A+B∣=∣A∣+∣B∣ — determinants respect products, not sums.
Skew-symmetric zero rule needs odd order — at even order the determinant can be nonzero.
The 15 questions below are a fast pass over the whole chapter at recall level. Each explanation names the section to revisit if it feels shaky.
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