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Question: If **a, b, c** are any three vectors and their inverse are \(\mathbf{a}^{- 1},\mathbf{b}^{- 1},\math...

If a, b, c are any three vectors and their inverse are a−1,b−1,c−1\mathbf{a}^{- 1},\mathbf{b}^{- 1},\mathbf{c}^{- 1}and [abc]≠0,\lbrack\mathbf{abc}\rbrack \neq 0, then [a−1b−1c−1]\lbrack\mathbf{a}^{- 1}\mathbf{b}^{- 1}\mathbf{c}^{- 1}\rbrack will be

A

Zero

B

One

C

Non-zero

D

[a b c]

Answer

Non-zero

Explanation

Solution

a−1=b×c[abc],\mathbf{a}^{- 1} = \frac{\mathbf{b} \times \mathbf{c}}{\lbrack\mathbf{abc}\rbrack}, c−1=a×b[abc],\mathbf{c}^{- 1} = \frac{\mathbf{a} \times \mathbf{b}}{\lbrack\mathbf{abc}\rbrack}, b−1=c×a[abc]\mathbf{b}^{- 1} = \frac{\mathbf{c} \times \mathbf{a}}{\lbrack\mathbf{abc}\rbrack}

⇒[a−1b−1c−1]=(b×c)[abc].((c×a)[abc]×(a×b)[abc])\Rightarrow \lbrack\mathbf{a}^{- 1}\mathbf{b}^{- 1}\mathbf{c}^{- 1}\rbrack = \frac{(\mathbf{b} \times \mathbf{c})}{\lbrack\mathbf{abc}\rbrack}.\left( \frac{(\mathbf{c} \times \mathbf{a})}{\lbrack\mathbf{abc}\rbrack} \times \frac{(\mathbf{a} \times \mathbf{b})}{\lbrack\mathbf{abc}\rbrack} \right)

=b×c[abc].[a[abc]]=1[abc]≠0= \frac{\mathbf{b} \times \mathbf{c}}{\lbrack\mathbf{abc}\rbrack}.\left\lbrack \frac{\mathbf{a}}{\lbrack\mathbf{abc}\rbrack} \right\rbrack = \frac{1}{\lbrack\mathbf{abc}\rbrack} \neq 0.