Determinant identity proof

Let

Δ = | x 1 y 1 |
    | 1 y 1 x |
    | 1 x 1 y |
    | y 1 x 1 |

and s = x + y.

Step 1: R₁ → R₁ + R₄ and R₂ → R₂ + R₃.

• R₁ + R₄ = (x+y, 2, x+y, 2) = (s, 2, s, 2)
• R₂ + R₃ = (2, x+y, 2, x+y) = (2, s, 2, s)

Step 2: R₄ → R₄ − ½R₁ and R₃ → R₃ − ½R₂ (using the new R₁, R₂).

• R₄ − ½(R₁+R₄) = ½(R₄ − R₁) = ½(y−x, 0, x−y, 0) = ((y−x)/2)(1, 0, −1, 0)
• R₃ − ½(R₂+R₃) = ½(R₃ − R₂) = ½(0, x−y, 0, y−x) = ((x−y)/2)(0, 1, 0, −1)

These operations do not change the value of the determinant. Take the constants (x−y)/2 and (y−x)/2 out of R₃ and R₄; their product is −(x−y)²/4:

Δ = −((x−y)²/4) | s  2  s  2 |
                 | 2  s  2  s |
                 | 0  1  0 −1 |
                 | 1  0 −1  0 |

Step 3: C₃ → C₃ + C₁, C₄ → C₄ + C₂.

Δ = −((x−y)²/4) | s  2  2s  4  |
                 | 2  s  4   2s |
                 | 0  1  0   0  |
                 | 1  0  0   0  |

Step 4: Evaluate. Expand along R₄: its only non-zero entry is 1 at (4, 1), with sign (−1)⁵ = −1. The remaining 3 × 3 minor is [2 2s 4; s 4 2s; 1 0 0]; expand it along its last row (entry 1 at (3, 1), sign +):

det = (−1) · | 2s  4 | = −(4s² − 16) = −4(s² − 4)
             | 4   2s|

Therefore

Δ = −((x−y)²/4) · (−4)(s² − 4) = (x−y)² (s² − 4) = (x−y)² (s − 2)(s + 2)

Δ = (x + y + 2)(x − y)²(x + y − 2). ∎

(Check: x = 3, y = 1 gives rows (3,1,1,1), (1,1,1,3), (1,3,1,1), (1,1,3,1), whose determinant is 48 = 6 · 2² · 2 ✓.)