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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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