Determinants

🎯 Key Points

• Determinant of 2×2: |a b; c d| = ad − bc
• Determinant of 3×3 (expansion along row 1): a₁₁(M₁₁) − a₁₂(M₁₂) + a₁₃(M₁₃) where Mᵢⱼ = 2×2 minor
• Cofactor Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ (minor with sign); expansion = Σ aᵢⱼ Cᵢⱼ along any row or column
• Singular matrix: det(A) = 0 (no inverse); Non-singular: det(A) ≠ 0 (inverse exists)
• Adjoint: adj(A) = transpose of cofactor matrix; A⁻¹ = adj(A)/det(A)
• det(AB) = det(A)·det(B); det(Aⁿ) = (det A)ⁿ; det(kA) = kⁿ det(A) for n×n matrix
• Area of triangle with vertices (x₁,y₁),(x₂,y₂),(x₃,y₃) = ½|det| (absolute value)
• Cramer's rule: for AX=B, x₁=Δ₁/Δ, x₂=Δ₂/Δ (Δ=det(A), Δᵢ=det with i-th column replaced by B)

Properties of Determinants

• Interchange two rows/columns → determinant changes sign
• Two identical rows/columns → determinant = 0
• Multiply one row/column by k → determinant multiplied by k
• Add scalar multiple of one row to another → determinant unchanged
• det(A) = det(Aᵀ) — transpose doesn't change determinant

Worked Example: Evaluate 3×3 Determinant

Evaluate: |1 2 3; 4 5 6; 7 8 9|

Expand along R1: 1·|5 6; 8 9| − 2·|4 6; 7 9| + 3·|4 5; 7 8|

= 1·(45−48) − 2·(36−42) + 3·(32−35) = 1·(−3) − 2·(−6) + 3·(−3) = −3 + 12 − 9 = 0

Note: this is 0 because the rows are in arithmetic progression (linearly dependent).

Worked Example: Inverse using Adjoint

Find A⁻¹ for A = |2 1; 5 3|. det(A) = 6−5 = 1. adj(A) = |3 −1; −5 2|. A⁻¹ = adj(A)/1 = |3 −1; −5 2|.

Verify: A·A⁻¹ = |2 1; 5 3|·|3 −1; −5 2| = |6−5, −2+2; 15−15, −5+6| = |1 0; 0 1| = I ✓

Minors and Cofactors

• Minor Mᵢⱼ of element aᵢⱼ is the determinant left after deleting row i and column j.
• Cofactor Cᵢⱼ = (−1)^(i+j) Mᵢⱼ — the minor with a sign from the checkerboard pattern (+ − + / − + − / + − +).
• Expansion along any row or column gives the same value: det(A) = aᵢ₁Cᵢ₁ + aᵢ₂Cᵢ₂ + aᵢ₃Cᵢ₃.
• Key identity: the sum of products of one row's elements with the cofactors of a DIFFERENT row is 0 (e.g. a₁₁C₂₁ + a₁₂C₂₂ + a₁₃C₂₃ = 0).

Area of a Triangle

The area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) is:

Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

written as the determinant ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (absolute value, since area is non-negative). The three points are collinear exactly when this determinant equals 0.

Adjoint and Inverse of a Matrix

• adj(A) is the transpose of the cofactor matrix.
• Fundamental relation: A(adj A) = (adj A)A = det(A) I.
• If det(A) ≠ 0 (non-singular), the inverse exists: A⁻¹ = adj(A)/det(A).
• |adj A| = |A|ⁿ⁻¹ and adj(adj A) = |A|ⁿ⁻² A for an n×n matrix.

Solving Systems of Linear Equations and Consistency

Write a linear system as AX = B, where A is the coefficient matrix, X the variable column, and B the constant column.

• Matrix method: if det(A) ≠ 0, the unique solution is X = A⁻¹B.
• Cramer's rule: x = Δ₁/Δ, y = Δ₂/Δ, z = Δ₃/Δ, where Δ = det(A) and Δᵢ replaces the i-th column of A with B.
• Consistency: Δ ≠ 0 → unique solution (consistent). Δ = 0 with every Δᵢ = 0 → infinitely many solutions (consistent, dependent). Δ = 0 with some Δᵢ ≠ 0 → no solution (inconsistent).
• Homogeneous system AX = 0: Δ ≠ 0 gives only the trivial solution X = 0; Δ = 0 gives non-trivial (infinitely many) solutions.

Expansion of a Determinant (2×2 and 3×3)

• 2×2: |a b; c d| = ad − bc (product of the main diagonal minus product of the off-diagonal).
• 3×3 by cofactors: expand along any row or column, e.g. along R1: a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ — choose the row/column with the most zeros to minimise work.
• Sarrus (diagonal) rule for 3×3 only: add the three products going down-right and subtract the three going down-left: (a₁₁a₂₂a₃₃ + a₁₂a₂₃a₃₁ + a₁₃a₂₁a₃₂) − (a₁₃a₂₂a₃₁ + a₁₁a₂₃a₃₂ + a₁₂a₂₁a₃₃).
• The sign pattern for cofactors is the checkerboard + − + / − + − / + − +.

Evaluating Determinants Using Row and Column Operations

Instead of a full expansion, use elementary operations to create zeros, then expand along that row/column.

• Allowed without changing the value: adding a scalar multiple of one row (or column) to another, e.g. R₂ → R₂ − 2R₁.
• Interchanging two rows/columns multiplies the value by −1; taking a common factor k out of a row/column multiplies the value by k.
• Aim to make a row or column have two zeros so a single cofactor remains.
• Example: applying C₁ → C₁ + C₂ + C₃ often produces a common factor (like a+b+c) that can be pulled outside.

Equation of a Line Through Two Points

The line through A(x₁, y₁) and B(x₂, y₂) is the set of points P(x, y) collinear with A and B, so the area determinant vanishes:

|x y 1; x₁ y₁ 1; x₂ y₂ 1| = 0

Expanding this gives the standard equation of the line. The same vanishing determinant is the condition for three given points to be collinear.

Singular and Non-Singular Matrices

• Singular: det(A) = 0 — the inverse does NOT exist; rows/columns are linearly dependent.
• Non-singular: det(A) ≠ 0 — the inverse A^-1 = adj(A)/det(A) exists and is unique.
• Product rule: if either A or B is singular then AB is singular, since det(AB) = det(A)·det(B).
• A system AX = B has a unique solution exactly when the coefficient matrix is non-singular.