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Determinant identity proof

Short answer

Multiply rows by a, b, c, d, split last column into two determinants. The first yields abcd times Vandermonde, the second yields −abcd times Vandermonde, sum zero, so the determinant equals zero.

Using the properties, evaluate the determinant [1 a a² a³+bcd; 1 b b² b³+cda; 1 c c² c³+abd; 1 d d² d³+abc].

Let

Δ = | 1  a  a²  a³+bcd |
    | 1  b  b²  b³+cda |
    | 1  c  c²  c³+abd |
    | 1  d  d²  d³+abc |

Step 1: Multiply R₁, R₂, R₃, R₄ by a, b, c, d. The determinant becomes abcd·Δ, and the last column entries become a⁴ + abcd, b⁴ + abcd, c⁴ + abcd, d⁴ + abcd:

abcd·Δ = | a  a²  a³  a⁴+abcd |
         | b  b²  b³  b⁴+abcd |
         | c  c²  c³  c⁴+abcd |
         | d  d²  d³  d⁴+abcd |

Step 2: Split the last column (additive property of determinants):

abcd·Δ = | a a² a³ a⁴ |         | a a² a³ 1 |
         | b b² b³ b⁴ |  + abcd | b b² b³ 1 |
         | c c² c³ c⁴ |         | c c² c³ 1 |
         | d d² d³ d⁴ |         | d d² d³ 1 |

Step 3: First determinant. Take a, b, c, d common from the rows. It equals abcd·V, where V is the Vandermonde determinant

V = | 1 a a² a³ |
    | 1 b b² b³ |  = Π (xⱼ − xᵢ) over i < j
    | 1 c c² c³ |
    | 1 d d² d³ |

Step 4: Second determinant. Move the column of ones to the front. This needs three adjacent column interchanges, so the sign is (−1)³ = −1; the determinant equals −V.

Step 5: Combine.

abcd·Δ = abcd·V + abcd·(−V) = 0

Since abcd ≠ 0 in general (and both sides are polynomials in a, b, c, d), Δ = 0.

Answer: Δ = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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