- 2075 Asoj · 5 marks
Determinant identity proof
Short answer
A determinant is an alternating multilinear function assigning to each square matrix a number via permutations. Apply C₁ → C₁+C₂+C₃, then C₂−C₁, C₃−C₁, and take factors; the result equals twice the determinant of the transposed matrix.
Define the determinant as a function and, using its properties, show that [b+c c+a a+b; q+r r+p p+q; y+z z+x x+y] (determinant) = 2 [a p x; b q y; c r z] (determinant).
Determinant as a function
For a square matrix A = [aᵢⱼ] of order n, the determinant is a function det : Mₙ → ℝ (or ℂ) that assigns to A the single number
|A| = Σ sgn(π) · a₁π(1) · a₂π(2) · … · aₙπ(n)
summed over all n! permutations π of {1, …, n}, with sgn(π) = ±1 according as π is even or odd. This function is linear in each row (or column) separately, changes sign when two rows are interchanged, and equals 1 for the identity matrix. These properties are used below.
Proof
Let
Δ = | b+c c+a a+b |
| q+r r+p p+q |
| y+z z+x x+y |
Step 1: C₁ → C₁ + C₂ + C₃. Each entry of the new first column is twice a sum:
Δ = | 2(a+b+c) c+a a+b |
| 2(p+q+r) r+p p+q | = 2 | a+b+c c+a a+b |
| 2(x+y+z) z+x x+y | | p+q+r r+p p+q |
| x+y+z z+x x+y |
Step 2: C₂ → C₂ − C₁, C₃ → C₃ − C₁. Then c + a − (a+b+c) = −b and a + b − (a+b+c) = −c, and similarly for the other rows:
Δ = 2 | a+b+c −b −c |
| p+q+r −q −r |
| x+y+z −y −z |
Step 3: C₁ → C₁ + C₂ + C₃.
Δ = 2 | a −b −c |
| p −q −r |
| x −y −z |
Step 4: Take (−1) out of C₂ and of C₃; the factor is (−1)(−1) = 1:
Δ = 2 | a b c |
| p q r |
| x y z |
Finally, a determinant is unchanged when rows and columns are interchanged (|M| = |Mᵀ|):
Δ = 2 | a p x ; b q y ; c r z |. ∎
✓ Answer checked by Claude Opus 5.5 (high effort)
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