Search any question & find its solution
Question:
Answered & Verified by Expert
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 what is $\lambda$ equal to?
Options:
then what is $\lambda$ equal to?
Solution:
2850 Upvotes
Verified Answer
The correct answer is:
$a b c$
Let $a^{-1}+b^{-1}+c^{-1}=0$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow \quad \frac{b c+a c+a b}{a b c}=0$
$\Rightarrow a b+b c+c a=0$ ...(1)
Consider $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$
$\Rightarrow(1+a)[(1+b)(1+c)-1]-(1+c-1)+(1-1-b)=\lambda$
$\Rightarrow(1+a)(c+b+b c)-c-b=\lambda$
$\Rightarrow \mathrm{bc}+\mathrm{ac}+\mathrm{ab}+\mathrm{abc}=\lambda$
$\Rightarrow$ abc $=\lambda \quad$ (using (1))
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow \quad \frac{b c+a c+a b}{a b c}=0$
$\Rightarrow a b+b c+c a=0$ ...(1)
Consider $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$
$\Rightarrow(1+a)[(1+b)(1+c)-1]-(1+c-1)+(1-1-b)=\lambda$
$\Rightarrow(1+a)(c+b+b c)-c-b=\lambda$
$\Rightarrow \mathrm{bc}+\mathrm{ac}+\mathrm{ab}+\mathrm{abc}=\lambda$
$\Rightarrow$ abc $=\lambda \quad$ (using (1))
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.