Question

`|((1+alpha)^2,(1+ 2alpha)^2, (1+3alpha)^2),((2+alpha)^2,(2+ 2alpha)^2, (2+3alpha)^2),((3+alpha)^2,(3+ 2alpha)^2, (3+3alpha)^2)| =-648alpha` FInd the value of `alpha`
A. `-4`
B. 9
C. `-9`
D. 4

Answer 1

Correct Answer - B::C
We have,
`|((1 + alpha)^(2),(1 + 2 alpha^(2)),(1 + 3 alpha)^(2)),((2 + alpha)^(2),(2 + 2 alpha)^(2),(2 + 3 alpha)^(2)),((3 + alpha)^(2),(3 + 2 alpha)^(2),(3 + 3 alpha)^(2))| = -648 alpha`
`rArr|((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(2(4 + 2 alpha),2(4 + 4 alpha),2(4 + 6 alpha))| = -648 alpha " " ["Applying " R_(2) - R_(2) - R_(1), R_(3) rarr R_(3) - R_(1)]`
`rArr 2 |((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(4 + 2 alpha,4 + 4 alpha,4 + 6 alpha)| = -648 alpha` [Taking 2 common from `R_(3)`]
`rArr 2|((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(1,1,1)| = -648 alpha " " [ "Applying " R_(3) rarr R_(3) - R_(2)]`
`rArr 2 |((1 + alpha)^(2),alpha (2 + 3 alpha),2alpha (2 + 4alpha)),(3 + 2 alpha,2 alpha,4 alpha),(1,0,0)| = -648 alpha " " ["Applying " C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)]`
`rArr 2|(alpha (2 + 3 alpha),2 alpha (2 + 4 alpha)),(2 alpha,4 alpha)| = -648 alpha` [Expanding along `R_(3)`]
`rArr 4 alpha^(2) |(2 + 3 alpha,2 + 4 alpha),(2,2)| = -648 alpha`
`rArr 4 alpha^(2) (4 + 6 alpha - 4 - 8 alpha) = -648 alpha`
`rArr - 8 alpha^(3) = -648 alpha`
`rArr - 8 alpha (al[ha^(2) - 81) = 0`
`rArr alpha = 0, alpha +- 9`