Chapter 4 at a Glance

The determinant

Every square matrix AA carries one number det⁡A=∣A∣\det A = \vert A \vert. For order 11: ∣[a]∣=a\vert [a] \vert = a. For order 22: ∣a11a12a21a22∣=a11a22−a12a21\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}. For order 33: expand along any row or column — signed elements times 2×22 \times 2 minors — all six choices agree; pick the one with the most zeros.

Sign chessboard and first-row cofactor expansion of a three-by-three determinant

Minor MijM_{ij}: delete row ii, column jj. Cofactor Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}. Then:

  1. elements ×\times own cofactors =Δ= \Delta (any row or column),
  2. elements ×\times another row's cofactors =0= 0.

Area and collinearity

Δ=12∣x1y11x2y21x3y31∣\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

Take the absolute value; when the area is given, equate to ±\pm the area; collinear points ⇔\Leftrightarrow determinant =0= 0 (which also produces the equation of a line through two points).

Triangle area from a determinant and the collinearity zero-area test

Adjoint and inverse

adj⁡A\operatorname{adj} A = transpose of the cofactor matrix (2×22 \times 2 shortcut: swap the diagonal, negate the off-diagonal). The master identity and its offspring:

A(adj⁡A)=(adj⁡A)A=∣A∣I,A−1=1∣A∣adj⁡A(∣A∣≠0)A(\operatorname{adj} A) = (\operatorname{adj} A)A = \vert A \vert I, \qquad A^{-1} = \frac{1}{\vert A \vert}\operatorname{adj} A \quad (\vert A \vert \neq 0)

Singular: ∣A∣=0\vert A \vert = 0 (no inverse). Nonsingular: ∣A∣≠0\vert A \vert \neq 0 (invertible). Key formulas:

Formula Note
∣AB∣=∣A∣∣B∣\vert AB \vert = \vert A \vert \vert B \vert products multiply
∣kA∣=kn∣A∣\vert kA \vert = k^n \vert A \vert one kk per row
det⁡(A−1)=1/det⁡A\det(A^{-1}) = 1/\det A from AA−1=IAA^{-1} = I
∣adj⁡A∣=∣A∣n−1\vert \operatorname{adj} A \vert = \vert A \vert^{n-1} =∣A∣2= \vert A \vert^2 at order 33
(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} order reversed

Solving systems

Write the system as AX=BAX = B. If ∣A∣≠0\vert A \vert \neq 0: unique solution X=A−1BX = A^{-1}B. If ∣A∣=0\vert A \vert = 0: compute (adj⁡A)B(\operatorname{adj} A)B — nonzero means inconsistent; zero means infinitely many solutions or none.

Mistake checklist

  1. ∣A∣\vert A \vert is not a modulus — determinants can be negative, and only square matrices have them.
  2. ∣kA∣=k∣A∣\vert kA \vert = k\vert A \vert is wrong — the correct factor is knk^n, one per row.
  3. Cofactor ≠\neq minor — the (−1)i+j(-1)^{i+j} flip applies whenever i+ji + j is odd.
  4. Adjoint needs the transpose — cofactor matrix first, then flip across the diagonal.
  5. Given-area problems need ±\pm — solving with one sign loses half the answers.
  6. X=A−1BX = A^{-1}B, never BA−1BA^{-1} — premultiply, and state ∣A∣≠0\vert A \vert \neq 0 before inverting.
  7. ∣A+B∣≠∣A∣+∣B∣\vert A + B \vert \neq \vert A \vert + \vert B \vert — determinants respect products, not sums.
  8. 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.