Question

# If a, b, c all are non-zero and unequal and $$\begin{vmatrix}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{vmatrix}=0$$, then $$1+\displaystyle \frac {1}{a}+\frac {1}{b}+\frac {1}{c}$$ is equal to

A
abc
B
1abc
C
1a+1b+1c
D
0

Solution

## The correct option is D $$0$$$$a\neq b\neq c\neq 0$$$$\begin{vmatrix}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{vmatrix}=0$$$$abc\begin{vmatrix}\dfrac {1}{a}+1 & \dfrac {1}{a} & \dfrac {1}{a}\\ \dfrac {1}{b} & \dfrac {1}{b}+1 & \dfrac {1}{b} \\ \dfrac {1}{c} & \dfrac {1}{c} & \dfrac {1}{c}+1\end{vmatrix}=0$$$$R_1\rightarrow R_1+R_2+R_3$$$$(abc)\left (1+\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}\right )\begin{vmatrix}1 & 1 & 1 \\ \dfrac {1}{b} & \dfrac {1}{b}-1 & \dfrac {1}{b}\\ \dfrac {1}{c} & \dfrac {1}{c} & \dfrac {1}{c}+1\end{vmatrix}=0$$$$C_1\rightarrow C_1-C_2; C_2\rightarrow C_2-C_3$$$$(abc)\left( 1+\dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right) \begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & \dfrac { 1 }{ b } \\ 0 & -1 & \dfrac { 1 }{ c } +1 \end{vmatrix}=0$$$$(abc)\left (1+\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}\right )=0$$$$\because abc\neq 0$$ $$\Rightarrow$$ $$1+\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=0$$Ans: DMathematics

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