A) Zero
B) One
C) Non-zero
D) [a b c]
Correct Answer: C
Solution :
\[{{\mathbf{a}}^{-1}}=\frac{\mathbf{b}\times \mathbf{c}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]},\] \[{{\mathbf{c}}^{-1}}=\frac{\mathbf{a}\times \mathbf{b}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]},\] \[{{\mathbf{b}}^{-1}}=\frac{\mathbf{c}\times \mathbf{a}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}\] \[\Rightarrow [{{\mathbf{a}}^{-1}}\,{{\mathbf{b}}^{-1}}\,{{\mathbf{c}}^{-1}}]=\frac{(\mathbf{b}\times \mathbf{c})}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}.\left( \frac{(\mathbf{c}\times \mathbf{a})}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}\times \frac{(\mathbf{a}\times \mathbf{b})}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]} \right)\] \[=\frac{\mathbf{b}\times \mathbf{c}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}.\left[ \frac{\mathbf{a}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]} \right]=\frac{1}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}\ne 0\].
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