Question

If $a ^{-1}+ b ^{-1}+ c ^{-1}=0$ such that
$\begin{vmatrix}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{vmatrix}=\lambda$ then the value of $\lambda$ is :

• A. 0
• B. - abc
• C. abc
• D. None of these

Answer 1

Answer: (C)

Given: $a ^{-1}+ b ^{-1}+ c ^{-1}=0$
and $\begin{vmatrix}1+ a & 1 & 1 \\ 1 & 1+ b & 1 \\ 1 & 1 & 1+ c \end{vmatrix}=\lambda$,
$\Rightarrow abc\begin{vmatrix}\frac{1+a}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1+b}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1+c}{c}\end{vmatrix}=\lambda$
$\Rightarrow a b c\begin{vmatrix}\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{vmatrix}=\lambda$
$\left( R _{1} \rightarrow R _{1}+ R _{2}+ R _{3}\right)$
$\Rightarrow abc \begin{vmatrix}\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+1 & \frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+1 & \frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+1 \\ \frac{1}{ b } & \frac{1}{ b }+1 & \frac{1}{ b } \\ \frac{1}{ c } & \frac{1}{ c } & \frac{1}{ c }+1\end{vmatrix}=\lambda$
$\Rightarrow abc (\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+1)\begin{vmatrix}1 & 1 & 1 \\ \frac{1}{ b } & \frac{1}{ b }+1 & \frac{1}{ b } \\ \frac{1}{ c } & \frac{1}{ c } & \frac{1}{ c }+1\end{vmatrix}=\lambda$
$\Rightarrow a b c\begin{vmatrix}1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{vmatrix}=\lambda$
$R _{2} \rightarrow \frac{1}{ b } R _{1}- R _{2}, R _{3} \rightarrow \frac{1}{ c } R _{1}- R _{3}$
$\Rightarrow abc \begin{vmatrix}1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{vmatrix}=\lambda$
$ \Rightarrow abc =\lambda$