If a,b,c are in G.P. then the value of `|(a,b,a+b),(b,c,b+c),(a+b,b+c,0)|=` (A) 1 (B) -1 (C) `a+b+c` (D) 0

Options

A. 0
B. 1
C. -1
D. None of these

Answer

A

Explanation

Since, a, b, c are in GP.
`rArr b^(2) = ac`
Explanding the determinant we get,
`|{:(" "a," "b,a+b),(" "b," "c,b+c),(a+b,b+c," "0):}|`
`=a|{:(" "c,b+c),(b+c," "0):}|-b|{:(" "b,b+c),(a+b," "0):}|+(a+b)|{:(" "b," "c),(a+b,b+c):}|`
`=-a(b+c)^(2)+b(a+b)(b+c)+(a+b)+(b^(2)+bc-ac-bc)`
`-a(b^(2)+c^(2)+2ac)+b(ab+ac+b^(2)+bc)`
`=-ab^(2)-ac^(2)-2abc+ab^(2)+2abc+b^(2)c" "(because b^(2)=ac)`
`=- ac^(2)+b^(2)c=-ac^(2)+ac.c.`
`=-ac^(2)+ac^(2)=0`