The coefficient of x in the determinant $|\begin{array}{lll} (1 + x)^{a_{1} b_{1}} (1 + x)^{a_{1} b_{2}} (1 + x)^{a_{1} b_{3}} \\ (1 + x)^{a_{2} b_{1}} (1 + x)^{a_{2} b_{2}} (1 + x)^{a_{2} b_{3}} \\ (1 + x)^{a_{3} b_{1}} (1 + x)^{a_{3} b_{2}} (1 + x)^{a_{3} b_{3}} \end{array}|$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$|\begin{array}{lll} (1 + x)^{a_{1} b_{1}} (1 + x)^{a_{1} b_{2}} (1 + x)^{a_{1} b_{3}} \\ (1 + x)^{a_{2} b_{2}} (1 + x)^{a_{2} b_{2}} (1 + x)^{a_{2} b_{3}} \\ (1 + x)^{a_{3} b_{1}} (1 + x)^{a_{3} b_{2}} (1 + x)^{a_{3} b_{3}} \end{array}| = A + Bx + {Cx}^{2} + \dots . . . . .$
Differentiating both sides w.r.t.x and then put x=0

$= |\begin{array}{ccc} a_{1} b_{1} (1 + x)^{a_{1} b_{1} - 1} & a_{1} b_{2} (1 + x)^{a_{1} b_{2} - 1} & a_{1} b_{3} (1 + x)^{a_{1} b_{3} - 1} \\ (1 + x)^{a_{2} b_{1}} & (1 + x)^{a_{2} b_{2}} & (1 + x)^{a_{2} b_{3}} \\ (1 + x)^{a_{3} b_{1}} & (1 + x)^{a_{3} b_{2}} & (1 + x)^{a_{3} b_{3}} \end{array}| + |\begin{array}{ccc} (1 + x)^{a_{1} b_{1}} & (1 + x)^{a_{1} b_{2}} & (1 + x)^{a_{1} b_{3}} \\ a_{2} b_{1} (1 + x)^{a_{2} b_{1} - 1} & a_{2} b_{2} (1 + x)^{a_{2} b_{2} - 1} & a_{2} b_{3} (1 + x)^{a_{2} b_{3} - 1} \\ (1 + x)^{a_{3} b_{1}} & (1 + x)^{a_{3} b_{2}} & (1 + x)^{a_{3} b_{3}} \end{array}| + |\begin{array}{ccc} (1 + x)^{a_{1} b_{1}} & (1 + x)^{a_{1} b_{2}} & (1 + x)^{a_{1} b_{3}} \\ (1 + x)^{a_{2} b_{1}} & (1 + x)^{a_{2} b_{2}} & (1 + x)^{a_{2} b_{3}} \\ b_{3} b_{1} (1 + x)^{a_{3} b_{1} - 1} & a_{3} b_{2} (1 + x)^{a_{3} b_{2} - 1} & a_{3} b_{3} (1 + x)^{a_{3} b_{3} - 1} \end{array}| = |\begin{array}{ccc} a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}| + |\begin{array}{ccc} 1 & 1 & 1 \\ a_{2} b_{1} & a_{2} b_{2} & a_{3} b_{3} \\ 1 & 1 & 1 \end{array}| + |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ a_{3} b_{1} & a_{3} b_{2} & a_{3} b_{3} \end{array}|$
we get 0+0+0=B