--- title: "Determinant identity: 2(ab + bc + ca)³" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:determinant-identity-2-ab-bc-ca" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-2-ab-bc-ca site: BCT Engineering --- # Determinant identity: 2(ab + bc + ca)³ Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2076 Asoj, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-2-ab-bc-ca - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Prove that [(b+c)² c² b²; c² (c+a)² a²; b² a² (a+b)²] (determinant) = 2(ab + bc + ca)³. ## Short answer Multiply the matrix by T = [[−1,1,1],[1,−1,1],[1,1,−1]] on left and right, then use the matrix determinant lemma. The determinant reduces to 2(ab+bc+ca)³. ## Answer Let s = ab + bc + ca, σ = a + b + c, and let M be the matrix of the determinant ```text Δ = | (b+c)² c² b² | | c² (c+a)² a² | | b² a² (a+b)² | ``` #### Step 1: Combine rows and columns with the matrix T Let T = [−1 1 1; 1 −1 1; 1 1 −1]. Multiplying M by T on the left performs the row operations R₁ → R₂ + R₃ − R₁, R₂ → R₃ + R₁ − R₂, R₃ → R₁ + R₂ − R₃ (all applied together). Since |T| = 4, this multiplies the determinant by 4. M is symmetric, so doing the same combinations on the columns (multiplying by T on the right) multiplies it by another 4. Hence N = TMT satisfies |N| = 16Δ. Using (b+c)² − c² − b² = 2bc and similar differences, the entries of N are ```text N = 2 [ 2a²+s 2ab−s 2ac−s ] [ 2ab−s 2b²+s 2bc−s ] [ 2ac−s 2bc−s 2c²+s ] ``` (for example N₁₁ = 2(2a² + ab + ac + bc), N₁₂ = 2(ab − ac − bc) = 2(2ab − s)). #### Step 2: Evaluate the 3 × 3 determinant inside N Let v = (a, b, c)ᵀ, e = (1, 1, 1)ᵀ, J = eeᵀ. The bracketed matrix equals ```text Q = s(2I − J) + 2vvᵀ ``` Put K = 2I − J. Its eigenvalues are 2 − 3 = −1, 2, 2, so |K| = −4, and K⁻¹ = ½(I − J) (check: (2I − J)(I − J) = 2I − 3J + J² = 2I, since J² = 3J). By the matrix determinant lemma, |B + 2vvᵀ| = |B|(1 + 2vᵀB⁻¹v) with B = sK (taking s ≠ 0; the identity is polynomial, so it then holds for all a, b, c): ```text |B| = s³ · (−4) vᵀB⁻¹v = (1/s)·½(vᵀv − (eᵀv)²) = (a²+b²+c² − σ²)/(2s) = −2s/(2s) = −1 |Q| = −4s³ (1 + 2(−1)) = 4s³ ``` #### Step 3: Combine |N| = 2³ |Q| = 8 · 4s³ = 32 s³, so Δ = |N|/16 = 2s³. **Δ = 2(ab + bc + ca)³.** ∎ (Spot check: a = b = c = 1 gives rows (4,1,1), (1,4,1), (1,1,4) with determinant 54 = 2 · 3³.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the determinant of T = [-1 1 1; 1 -1 1; 1 1 -1]? A. 1 B. 4 C. -4 D. 0 Answer: B. |T| = 4, so the row combinations multiply the determinant by 4. (Key point: Matrix T) 2. Why does the factor become 16? A. The matrix is symmetric, so the same combination applied to columns gives another factor 4 B. T is applied twice to rows C. The determinant is cubed D. Each entry is doubled Answer: A. Rows give 4 and columns give another 4, so |N| = 16Δ. (Key point: Both rows and columns) 3. K = 2I - J has eigenvalues -1, 2, 2. What is |K|? A. 4 B. 3 C. -4 D. -1 Answer: C. The product of the eigenvalues is -1 x 2 x 2 = -4. (Key point: Matrix determinant lemma) 4. With |Q| = 4s^3 and |N| = 8|Q|, what is Δ? A. s^3 B. 4s^3 C. 32s^3 D. 2s^3 Answer: D. |N| = 32s^3 and Δ = |N|/16 = 2s^3 = 2(ab+bc+ca)^3. 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