Determinant identity proof

Let k = 1 + a² + b² and

Δ = | 1+a²−b²    2ab       −2b     |
    |  2ab     1−a²+b²      2a     |
    |  2b       −2a      1−a²−b²   |

Step 1: R₁ → R₁ + b·R₃ and R₂ → R₂ − a·R₃.

Row 1:

• (1 + a² − b²) + 2b² = 1 + a² + b² = k
• 2ab + b(−2a) = 0
• −2b + b(1 − a² − b²) = −b − a²b − b³ = −bk

Row 2:

• 2ab − a(2b) = 0
• (1 − a² + b²) − a(−2a) = 1 + a² + b² = k
• 2a − a(1 − a² − b²) = a + a³ + ab² = ak

Δ = | k    0    −bk |
    | 0    k     ak |
    | 2b  −2a  1−a²−b² |

Step 2: Take k common from R₁ and from R₂ (factor k²):

Δ = k² | 1    0   −b |
       | 0    1    a |
       | 2b  −2a  1−a²−b² |

Step 3: Expand along R₁:

= 1·[1·(1−a²−b²) − a·(−2a)] − 0 + (−b)·[0·(−2a) − 1·2b]
= (1 − a² − b² + 2a²) + 2b²
= 1 + a² + b² = k

Therefore Δ = k² · k = k³ = (1 + a² + b²)³. ∎