Determinants, Matrices and Solving Linear Equations

The 2×2 determinant

• For A = [[a, b], [c, d]], det A = |A| = ad − bc.
• Example: det [[3, 5], [2, 4]] = (3)(4) − (5)(2) = 12 − 10 = 2.
• The determinant is a number, never a matrix.

The 3×3 determinant

• Expand along any row or column using cofactors with alternating signs (+ − +).
• Expanding along the first row: |A| = a(ei − fh) − b(di − fg) + c(dh − eg).
• Example: A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]] → 1(0 − 24) − 2(0 − 20) + 3(0 − 5) = −24 + 40 − 15 = 1.
• Practical tip: expand along the row or column containing the most zeros — it shortens the work without changing the answer.

Properties of determinants

Addition and scalar multiplication

• Matrices of the same order are added entry by entry: (A + B)_ij = a_ij + b_ij.
• Scalar multiplication multiplies every entry: (kA)_ij = k·a_ij.
• Different orders cannot be added: a 2×3 plus a 3×2 is undefined.

Multiplication

• A (m×n) can multiply B (n×p) only when the inner dimensions agree; the product is m×p.
• Entry (i, j) of AB = row i of A · column j of B.
• Example: [[1, 2], [3, 4]] × [[2, 0], [1, 5]] = [[1·2 + 2·1, 1·0 + 2·5], [3·2 + 4·1, 3·0 + 4·5]] = [[4, 10], [10, 20]].
• Matrix multiplication is not commutative: AB ≠ BA in general.

Transpose and identity

• Aᵀ is formed by turning rows into columns; (AB)ᵀ = BᵀAᵀ.
• The identity matrix I has 1s on the diagonal and 0s elsewhere, and AI = IA = A.

When an inverse exists

• A square matrix A has an inverse A⁻¹ only if det A ≠ 0; such a matrix is called non-singular.
• If det A = 0 the matrix is singular and has no inverse.
• Definition: AA⁻¹ = A⁻¹A = I.

The 2×2 formula

• For A = [[a, b], [c, d]] with det A ≠ 0: A⁻¹ = (1/det A) × [[d, −b], [−c, a]].
• The matrix [[d, −b], [−c, a]] is the adjugate (transpose of the cofactor matrix).
• Example: A = [[3, 5], [2, 4]], det A = 2 → A⁻¹ = ½ [[4, −5], [−2, 3]] = [[2, −2.5], [−1, 1.5]].
• Check: A A⁻¹ = [[3·2 + 5·(−1), 3·(−2.5) + 5·1.5], [2·2 + 4·(−1), 2·(−2.5) + 4·1.5]] = [[1, 0], [0, 1]] = I.

Useful identities

• (A⁻¹)⁻¹ = A  ·  (AB)⁻¹ = B⁻¹A⁻¹  ·  (Aᵀ)⁻¹ = (A⁻¹)ᵀ  ·  det(A⁻¹) = 1/det A.

The inverse-matrix method

• Write the system as AX = B, then X = A⁻¹B, provided det A ≠ 0.
• Example: 2x + 3y = 7 and x − 2y = −7.
  A = [[2, 3], [1, −2]], det A = −4 − 3 = −7, A⁻¹ = (−1/7)[[−2, −3], [−1, 2]].
• X = A⁻¹B = (−1/7)[[−2·7 + (−3)(−7)], [−1·7 + 2(−7)]] = (−1/7)[[7], [−21]] = [[−1], [3]], so x = −1 and y = 3.

Cramer's rule

• For AX = B with det A ≠ 0: x = D_x/D, y = D_y/D, z = D_z/D, where D = det A and D_x is D with the x-column replaced by B.
• Same system: D = −7, D_x = |[[7, 3], [−7, −2]]| = −14 + 21 = 7 → x = 7/(−7) = −1.
• D_y = |[[2, 7], [1, −7]]| = −14 − 7 = −21 → y = (−21)/(−7) = 3.
• Both methods give the same answer; Cramer's rule is faster for a single unknown, the inverse method is better when several right-hand sides share the same A.

Interpreting the determinant

1. det [[5, 2], [3, 1]] → 5 − 6 = −1
2. det [[1, 0, 2], [3, 1, 0], [0, 4, 1]] → 1(1 − 0) − 0 + 2(12 − 0) = 25
3. For which k is [[k, 4], [2, k]] singular? → k² − 8 = 0 → k = ±2√2
4. A = [[1, 2], [3, 4]]; find A⁻¹ → det = −2 → [[−2, 1], [1.5, −0.5]]
5. If |A| = 3 and |B| = 5, find |AB| and |2A| for 2×2 matrices → 15 and 12
6. Solve 3x + y = 9, x − y = −1 by Cramer → D = −4, D_x = −8, D_y = −12 → x = 2, y = 3
7. Solve x + 2y + z = 4, 2x − y + 3z = 9, 3x + y − z = 2 → x = 1, y = 1, z = 1
8. If AB = I, what is B in terms of A? → B = A⁻¹