A) 0
B) abc
C) - abc
D) None of these
Correct Answer: B
Solution :
\[k=1,\,|A|\,=\,0\] Applying \[5A=\left[ \begin{matrix} 15 & -25 \\ -20 & 10 \\ \end{matrix} \right]\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\] \[\left| \,\begin{matrix} 1+a & -a & -a \\ 1 & b & 0 \\ 1 & 0 & c \\ \end{matrix}\, \right|\] On expanding w.r.t. \[{{R}_{3}}\], \[ab+bc+ca+abc=\lambda \] ??.(i) Given, \[{{a}^{-1}}+{{b}^{-1}}+{{c}^{-1}}=0\] Þ \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\] Þ \[ab+bc+ca=0\] Þ \[\lambda =abc\], (From equation (i)).
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