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प्रश्न
Evaluate the following determinant:
\[\begin{vmatrix}1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9\end{vmatrix}\]
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उत्तर
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′ = C2 − 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]
C3 − 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
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