Question

Evaluate the following determinant:

\[\begin{vmatrix}1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9\end{vmatrix}\]

Answer 1

D = \[\begin{vmatrix}1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9\end{vmatrix}\]

Let’s label the columns:

`C_1 = [(1),(3),(9),(27)], C_2 = [(3),(9),(27),(1)], C_3 = [(9),(27),(1),(3)], C_4 = [(27),(1),(3),(9)]`

C₂′​ = C_2​ − 3C₁​, C₃′ ​= C₃​ − 9C₁​, C₄′ ​= C₄​ − 27C₁

​Compute each:

C₂​ − 3C₁​ = [3, 9, 27, 1] − 3[1, 3, 9, 27] = [0, 0, 0, 1 − 81 = −80]

C_3​ − 9C₁ ​= [9, 27, 1, 3] − 9[1, 3, 9, 27] = [0, 0, −80, 3 − 243 = −240]

C₄​ − 27C₁​ = [27, 1, 3, 9] − 27[1, 3, 9, 27] = [0, −80, −240, 9 − 729 = −720]

`D = |(1,0,0,0),(3,0,0,-80),(9,0,-80,-240),(27,-80,-240,-720)|`

`D = 1. |(0,0,-80),(0,-80,-240),(-80,-240,-720)|`

evaluate the 3×3 determinant:

`|(0,0,-80),(0,-80,-240),(-80,-240,-720)|`

Only the third element is nonzero:

`= (-1)^(1+3). (-80). |(0,-80),(-80,-240)|`

= −80 ⋅ (0 ⋅ −240 − (−80) (−80)) = −80 ⋅ (0 − 6400) = −80 ⋅ (−6400) = 512000

D = 1 ⋅ 512000

= 512000​