The value of the determinant `Delta = |(1 + a_(1) b_(1),1 + a_(1) b_(2),1 + a_(1) b_(3)),(1 + a_(2) b_(1),1 + a_(2) b_(2),1 + a_(2) b_(3)),(1 + a_(3) b_(1) ,1 + a_(3) b_(2),1 + a_(3) b_(3))|`, is

Options

A. `a_(1) a_(2) a_(3) + b_(1) b_(2) b_(3)`
B. `(a_(1) a_(2) a_(3)) (b_(1) b_(2) b_(3))`
C. `a_(1) a_(2) b_(1) b_(2) + a_(2) a_(3) b_(2) b_(3) + a_(3) a_(1) b_(3) b_(1)`
D. none of these

Answer

D

Explanation

We have,
`Delta = |(1,1 + a_(1) b_(2),1 + a_(1) b_(3)),(1,1 + a_(2) b_(2),1 + a_(2) b_(3)),(1,1 + a_(3) b_(2),1 + a_(3) b_(3))| + |(a_(1) b_(1),1 + a_(1) b_(2),1 + a_(1) b_(3)),(a_(2) b_(1) ,1 + a_(2) b_(2),1 + a_(2) b_(3)),(a_(3)b_(1) ,1 + a_(3) b_(2),1 + a_(3) b_(3))|`
Appplying `C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)` in first determinant and `C_(2) rarr C_(2) - ((b_(2))/(b_(1))) C_(1), C_(3) rarr C_(3) - (b_(3))/(b_(1)) C_(1)` in second determinant, we get
`Delta = |(1,a_(1) b_(2),a_(1) b_(3)),(1,a_(2) b_(2),a_(2) b_(3)),(1,a_(3) b_(2),a_(3) b_(3))| + |(a_(1) b_(1),1,1),(a_(2) b_(1),1,1),(a_(3) b_(1),1,1)|`
`rArr Delta = b_(1) b_(2) |(1,a_(1),a_(1)),(1,a_(2),a_(2)),(1,a_(3),a_(3))| + 0`
`rArr Delta = 0 + 0 = 0`