If \( a,b \) and \( c \) are non-zero real number such that \( \begin{vmatrix}1+a & 1 & 1\\1 & 1+b & 1\\1 & 1 & 1+c\end{vmatrix}=0 \), then value of \( \displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −1
Since a, b, c are non-zero, a, b, c can be taken common from rows 1, 2, 3:
\( \displaystyle \Delta=abc\begin{vmatrix}\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a}\\\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b}\\\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{vmatrix} \)
Apply \( R_1\to R_1+R_2+R_3 \) and write \( S=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \); every entry of the new row 1 is S:
\( \displaystyle \Delta=abc\,S\begin{vmatrix}1 & 1 & 1\\\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b}\\\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{vmatrix} \)
Apply \( C_2\to C_2-C_1 \) and \( C_3\to C_3-C_1 \):
\( \displaystyle \Delta=abc\,S\begin{vmatrix}1 & 0 & 0\\\frac{1}{b} & 1 & 0\\\frac{1}{c} & 0 & 1\end{vmatrix} \)
\( \displaystyle =abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \)
Since \( \Delta=0 \) and \( abc\ne 0 \),
\( \displaystyle 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \)
\( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1 \)
Shortcut: Try values: with a = b = c = −3 the determinant has rows (−2, 1, 1), (1, −2, 1), (1, 1, −2), whose sum is the zero row, so \( \Delta=0 \). Then \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1 \).
Tip: The value of this determinant, \( abc+ab+bc+ca \), is a standard result worth remembering.