Let `|(x,2,x),(x^2,x,6),(x,x,6)|=A x^4+B x^3+C x^2+D x+Edot` Then the value of `5A+4B+3C+2D+E` is equal to a. zero b. `-16` c. `11` d. `-11`

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`