Determinant identity proof

The matrix has the form I + vvᵀ with v = (a, b, c, d)ᵀ. Let

Δ = | a²+1  ab    ac    ad   |
    | ab    b²+1  bc    bd   |
    | ac    bc    c²+1  cd   |
    | ad    bd    cd    d²+1 |

(The statement writes the first row as a²+1, ba, ca, da, which is the same.)

Step 1: Multiply R₁, R₂, R₃, R₄ by a, b, c, d respectively. The determinant is multiplied by abcd:

abcd·Δ = | a(a²+1)  a²b     a²c     a²d    |
         | ab²      b(b²+1) b²c     b²d    |
         | ac²      bc²     c(c²+1) c²d    |
         | ad²      bd²     cd²     d(d²+1)|

Step 2: Take a, b, c, d common from C₁, C₂, C₃, C₄. This gives another factor abcd, which cancels:

Δ = | a²+1  a²    a²    a²   |
    | b²    b²+1  b²    b²   |
    | c²    c²    c²+1  c²   |
    | d²    d²    d²    d²+1 |

Step 3: R₁ → R₁ + R₂ + R₃ + R₄. Every column sums to a²+b²+c²+d²+1 = T, so the first row becomes (T, T, T, T). Take T common:

Δ = T | 1    1     1     1   |
      | b²   b²+1  b²    b²  |
      | c²   c²    c²+1  c²  |
      | d²   d²    d²    d²+1|

Step 4: C₂ → C₂ − C₁, C₃ → C₃ − C₁, C₄ → C₄ − C₁.

Δ = T | 1   0  0  0 |
      | b²  1  0  0 |
      | c²  0  1  0 |
      | d²  0  0  1 |

This is lower triangular with unit diagonal, so its determinant is 1.

Δ = T = 1 + a² + b² + c² + d². ∎