Let `|(x,2,x),(x^(2),x,6),(x,x,6)| = ax^(4) + bx^(3) + cx^(2) + dx + e` Then, the value of `5a + 4b + 3c + 2d + e` is equal to

Options

A. zero
B. -16
C. 16
D. -11

Answer

D

Explanation

Let the given determinant be equal to `Delta (x). ` Then
`5A +4B +3C + 2D + E =Delta (1) +Delta (1)`
Now `Delta (1)=0` as `R_(2)` and `R_(3)` are identical .
`Delta (x) = |{:(1,,0,,1),(x^(2),,x,,6),(x,,x,,6):}|+|{:(x,,2,,x),(2x,,1,,0),(x,,x,,6):}|+|{:(x,,2,,x),(x^(2),,x,,6),(1,,1,,0):}|`
`Delta (1) =|{:(1,,2,,1),(2,,1,,0),(1,,1,,6):}|+|{:(x,,2 ,,x),(x^(2),,x,,6),(1,,1,,0):}|`
`Delta'(1) = |{:(1,,2,,1),(2,,1,,0),(1,,1,,6):}|+|{:(1,,2,,1),(1,,1,,6),(1,,1,,0):}|`
`=-17 +(12 +1-1-6)=-11`