If \(\begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix}\) \(= 0\), where \(a, b\) and \(c\) are non zero constants, then the value of \(\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −1
Expanding the determinant along the first row:
\( (1+a)\big[(1+b)(1+c) - 1\big] - 1\big[(1+c) - 1\big] + 1\big[1 - (1+b)\big] \)
\(= (1+a)(b + c + bc) - c - b\)
\(= ab + bc + ca + abc\)
Setting it equal to 0 and dividing by \(abc\) (which is not 0):
\( \displaystyle \frac{1}{c} + \frac{1}{a} + \frac{1}{b} + 1 = 0 \)
\( \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -1 \)
Tip: This determinant appears often: it equals \(abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\). It is worth remembering.
Note: NTA dropped this question. For the benefit of students and to give them additional practice, the question has been suitably modified so that it has one clear answer.