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If $a^{-1}+b^{-1}+c^{-1}=0$ such that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$, then the value of $\lambda$ is
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Verified Answer
The correct answer is:
$a b c$
$\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$,
$\left|\begin{array}{ccc}1+a & -a & -a \\ 1 & b & 0 \\ 1 & 0 & c\end{array}\right|$
On expanding w.r.t. $R_3$
Given, $a b+b c+c a+a b c=\lambda$ ....(i)
$\begin{aligned} & \text { Given, } a^{-1}+b^{-1}+c^{-1}=0 \\ & \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \Rightarrow a b+b c+c a=0\end{aligned}$
$\Rightarrow \lambda=a b c, \quad$ (From equation (i)).
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$,
$\left|\begin{array}{ccc}1+a & -a & -a \\ 1 & b & 0 \\ 1 & 0 & c\end{array}\right|$
On expanding w.r.t. $R_3$
Given, $a b+b c+c a+a b c=\lambda$ ....(i)
$\begin{aligned} & \text { Given, } a^{-1}+b^{-1}+c^{-1}=0 \\ & \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \Rightarrow a b+b c+c a=0\end{aligned}$
$\Rightarrow \lambda=a b c, \quad$ (From equation (i)).
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