JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 21)
Explanation
Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$
$$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$
$$ \Rightarrow a + 8b + 7c = 0$$ ..... (i)
$$ \Rightarrow 9a + 2b + 3c = 0$$ .... (ii)
$$ \Rightarrow a + b + c = 0$$ .... (iii)
On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get
$$7a + c = 0$$ ..... (iv)
Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get
$$6a - b = 0$$ ..... (v)
$$\therefore$$ b = 6a and c = $$-$$ 7a
If a = 2, b = 12 and c = $$-$$14
$$\therefore$$ $${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$$
$$ \Rightarrow {3 \over {{\omega ^2}}} + {1 \over {{\omega ^{12}}}} + {3 \over {{\omega ^{ - 14}}}} = {3 \over {{\omega ^2}}} + 1 + 3{\omega ^2}$$
$$ = 3\omega + 1 + 3{\omega ^2}$$
$$ = 1 + 3(\omega + {\omega ^2})$$
$$ = 1 - 3 = - 2$$
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