Determinant identity: −(b − a)⁴

Let

Δ = | a b a a |
    | a b b b |
    | b b b a |
    | a a b a |

and put d = b − a.

Step 1: Make zeros in the first column

Apply R₂ → R₂ − R₁, R₃ → R₃ − R₁, R₄ → R₄ − R₁. Then R₂ − R₁ = (0, 0, d, d), R₃ − R₁ = (d, 0, d, 0) and R₄ − R₁ = (0, −d, d, 0):

Δ = | a   b    a    a  |
    | 0   0    d    d  |
    | d   0    d    0  |
    | 0  −d    d    0  |

Step 2: Take d common from R₂, R₃, R₄

Δ = d³ | a  b  a  a |
       | 0  0  1  1 |
       | 1  0  1  0 |
       | 0 −1  1  0 |

Step 3: C₃ → C₃ − C₄

Δ = d³ | a  b  0  a |
       | 0  0  0  1 |
       | 1  0  1  0 |
       | 0 −1  1  0 |

Step 4: Expand along R₂

The only non-zero entry is 1 at position (2, 4), with sign (−1)^(2+4) = +1:

Δ = d³ · | a  b  0 |
         | 1  0  1 |
         | 0 −1  1 |

Expanding this 3 × 3 determinant along R₁:

a(0·1 − 1·(−1)) − b(1·1 − 0) + 0 = a − b = −d

So Δ = d³(−d) = −d⁴ = −(b − a)⁴. ∎