--- title: "Determinant identity proof" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:determinant-identity-proof-eight" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-proof-eight site: BCT Engineering --- # Determinant identity proof Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2072 Chaitra, 5 marks - Video: [Linear Algebra : Proofs for Determinant Identities](https://www.youtube.com/watch?v=jcb3o4MITgg) — Maths Resource - 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-proof-eight - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Use properties of determinants to prove: [a²+1 ba ca da; ab b²+1 cb db; ac bc c²+1 dc; ad bd cd d²+1] (determinant) = 1 + a² + b² + c² + d². ## Short answer Multiply rows R₁,R₂,R₃,R₄ by a,b,c,d, then take common factors a,b,c,d from columns; after row addition and column subtraction the determinant reduces to 1 + a² + b² + c² + d². ## Answer The matrix has the form I + vvᵀ with v = (a, b, c, d)ᵀ. Let ```text Δ = | a²+1 ab ac ad | | ab b²+1 bc bd | | ac bc c²+1 cd | | ad bd cd d²+1 | ``` (The statement writes the first row as a²+1, ba, ca, da, which is the same.) **Step 1: Multiply R₁, R₂, R₃, R₄ by a, b, c, d respectively.** The determinant is multiplied by abcd: ```text abcd·Δ = | a(a²+1) a²b a²c a²d | | ab² b(b²+1) b²c b²d | | ac² bc² c(c²+1) c²d | | ad² bd² cd² d(d²+1)| ``` **Step 2: Take a, b, c, d common from C₁, C₂, C₃, C₄.** This gives another factor abcd, which cancels: ```text Δ = | a²+1 a² a² a² | | b² b²+1 b² b² | | c² c² c²+1 c² | | d² d² d² d²+1 | ``` **Step 3: R₁ → R₁ + R₂ + R₃ + R₄.** Every column sums to a²+b²+c²+d²+1 = T, so the first row becomes (T, T, T, T). Take T common: ```text Δ = T | 1 1 1 1 | | b² b²+1 b² b² | | c² c² c²+1 c² | | d² d² d² d²+1| ``` **Step 4: C₂ → C₂ − C₁, C₃ → C₃ − C₁, C₄ → C₄ − C₁.** ```text Δ = T | 1 0 0 0 | | b² 1 0 0 | | c² 0 1 0 | | d² 0 0 1 | ``` This is lower triangular with unit diagonal, so its determinant is 1. **Δ = T = 1 + a² + b² + c² + d².** ∎ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The matrix has the form: A. I - vv^T B. vv^T C. I + vv^T D. I + v^Tv Answer: C. With v = (a,b,c,d)^T, entries are delta_ij + v_i v_j. (Key point: Matrix form) 2. Why multiply the rows by a, b, c, d then take a, b, c, d from the columns? A. To change the value of Δ B. To get a^2, b^2, c^2, d^2 patterns while the factors abcd cancel C. To make the matrix triangular at once D. To remove the 1s Answer: B. The factor abcd on both sides cancels, leaving rows built from squares. (Key point: Row and column scaling) 3. R1 -> R1 + R2 + R3 + R4 makes row 1 equal to: A. (T, T, T, T), T = 1 + a^2 + b^2 + c^2 + d^2 B. (1, 1, 1, 1) C. (0, 0, 0, 0) D. (a^2, b^2, c^2, d^2) Answer: A. Each column sums to T. (Key point: Row sum) 4. After column operations the determinant is triangular with unit diagonal, so Δ is: A. 1 B. T = 1 + a^2 + b^2 + c^2 + d^2 C. abcd D. T^2 Answer: B. Δ = T · 1. 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