--- title: "Determinant identity proof" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:determinant-identity-proof-14" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-proof-14 site: BCT Engineering --- # Determinant identity proof Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2067 Asar - 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-14 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Using the properties of determinants prove: [a²+1 ba ca da; ab b²+1 cb db; ac bc c²+1 dc; ad bd cd d²+1] (determinant) = a² + b² + c² + d² + 1. ## Short answer Multiplying rows by a,b,c,d and factoring columns leads to determinant |a^2+1 a^2 a^2 a^2; b^2 b^2+1 b^2 b^2; c^2 c^2 c^2+1 c^2; d^2 d^2 d^2 d^2+1|. Adding all rows to the first extracts factor T = 1+a^2+b^2+c^2+d^2; column subtraction yields identity, so Δ = T. ## 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. vvᵀ − I B. I + vvᵀ with v = (a,b,c,d)ᵀ C. a diagonal matrix D. a skew-symmetric matrix Answer: B. Entry (i,j) is vᵢvⱼ plus 1 on the diagonal, so I + vvᵀ. (Key point: Structure) 2. Multiplying R₁..R₄ by a,b,c,d multiplies Δ by: A. abcd B. a+b+c+d C. 1 D. a²b²c²d² Answer: A. Each row scaling contributes its factor, total abcd. (Key point: Multiplying rows) 3. After R₁ → R₁+R₂+R₃+R₄, each entry of the new first row is: A. a²+b²+c²+d² B. 1 C. 0 D. a²+b²+c²+d²+1 Answer: D. Each column sums to T = a²+b²+c²+d²+1. (Key point: Row sum step) 4. After taking T out and clearing with column ops, the remaining determinant is: A. 0 B. T C. 1, since it is triangular with unit diagonal D. abcd Answer: C. A lower triangular matrix with unit diagonal has determinant 1, so Δ = T. 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