Let a,b,c be such that b(a+c) `ne ` 0. If
`|{:(a,a+1,a-1),(-b,b+1,b-1),(c,c-1,c+1):}|+|{:(a+1,b+1,c-1),(a-1,b-1,c+1),((-1)^(n+2)a,(-1)^(n+1)b,(-1)^(n)c):}|`=0 then the value of n is

Options

A. zero
B. any even integer
C. any odd integer
D. any integer

Answer

C

Explanation

We have,
`|(a,a +1,a -1),(-b,b +1,b -1),(c,c -1,c +1)| + |(a +1,b +1,c -1),(a -1,b -1,c +1),((-1)^(n+2)a,(-1)^(n+1)b,(-1)^(n) c)| = 0`
`rArr |(a,-b,c),(a +1,b +1,c -1),(a -1,b -1,c +1)| + (-1)^(n) |(a +1,b +1,c -1),(a -1,b -1,c +1),(a,-b,c)| = 0` [Using `|A| = |A^(T)|` in first determinant]
`rArr |(a,-b,c),(a +1,b +1,c -1),(a -1,b -1,c +1)| + (-1)^(n+1) |(a,-b,c),(a +1,b +1,c -1),(a -1,b -1,c +1)| = 0` [Applying first `R_(1) hArr R_(3) and " then " R_(2) hArr R_(3)`]
`rArr |(a,-b,c),(a +1,b +1,c -1),(a -1,b -1,c +1)|{1 + (-1)^(n +2)} = 0`
`rArr |(a,-b,c),(1,2b +1,-1),(-1,2b -1,1)| {1 + (-1)^(n+2)} = 0" " ["Applying " R_(2) rarrR_(2) - R_(1), R_(3) rarr R_(3) - R_(1)]`
`rArr |(a,-b,c),(1,2b +1,-1),(0,4b,0)| {1 + (-1)^(n+2)} = 0 " " [("Applying"),(R_(3) rarr R_(3) + R_(2))]`
`rArr -4b (-a -c) {1 + (-1)^(n+2)} = 0`
`rArr 1 + (-1)^(n+2) = 0 rArr` n is any odd integer