Question

Let $\omega$ be a solution of ${x^3} - 1 = 0$ with $\operatorname{Im} \left( \omega \right) > 0$. If $a = 2$ with b and c satisfying

a&b;&c; \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&9&7 \\\ 8&2&7 \\\ 7&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \end{array}} \right]......\left( E \right)$$ Then the value of $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$ is : (A) $ - 2$ (B) $2$ (C) $3$ (D) $ - 3$

Answer 1

The multiplication of two matrices is possible if the no. of columns in matrix A is equal to the no. of rows in matrix B. Here we multiplied the two given matrix and form the equations by comparing the values of both sides.

Complete step-by-step answer:
Since, $a,b$ and $c$ be three real numbers satisfies

a&b;&c; \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&9&7 \\\ 8&2&7 \\\ 7&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \end{array}} \right]$$ So, we get the equations $ a + 8b + 7c = 0 \\\ 9a + 2b + 3c = 0 \\\ 7a + 7b + 7c = 0 \Rightarrow a + b + c = 0 \\\ $ Since $a = 2$, so the equations become $ 2 + 8b + 7c = 0 \Rightarrow 8b + 7c = - 2....(1) \\\ 9\left( 2 \right) + 2b + 3c = 0 \Rightarrow 2b + 3c = - 18....(2) \\\ 2 + b + c = 0 \Rightarrow b + c = - 2....(3) \\\ $ Multiply equation (3) by $7$ and subtract it from (1), we get $8b + 7c - \left( {7b + 7c} \right) = - 2 - \left( { - 14} \right)$ $ \Rightarrow 8b + 7c - 7b - 7c = - 2 + 14 \\\ \Rightarrow b = 12 \\\ $ Substitute the value of $b$ in equation (3), we get $ 12 + c = - 2 \\\ \Rightarrow c = - 2 - 12 \\\ \Rightarrow c = - 14 \\\ $ So, we have $a = 2,b = 12,c = - 14$ Now, $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$ $ = \dfrac{3}{{{\omega ^2}}} + \dfrac{1}{{{\omega ^{12}}}} + \dfrac{3}{{{\omega ^{ - 14}}}}$ $$ = \dfrac{3}{{{\omega ^2}}} + \dfrac{1}{{{\omega ^{12}}}} + 3{\omega ^{14}}$$ Convert the power of $\omega $ in terms of ${\omega ^3}$, because ${\omega ^3} = 1$:- $$ = \dfrac{{3\omega }}{{{\omega ^3}}} + \dfrac{1}{{{{\left( {{\omega ^3}} \right)}^4}}} + 3{\left( {{\omega ^3}} \right)^4} \cdot {\omega ^2}$$ Put ${\omega ^3} = 1$ and simplify the above, $$ = \dfrac{{3\omega }}{1} + \dfrac{1}{{{{\left( 1 \right)}^4}}} + 3{\left( 1 \right)^4} \cdot {\omega ^2}$$ $$ = 3\omega + 1 + 3{\omega ^2}$$ $ = 3\left( {\omega + {\omega ^2}} \right) + 1$ Since $1 + \omega + {\omega ^2} = 0$ $ \Rightarrow \omega + {\omega ^2} = - 1$ $ = 3\left( { - 1} \right) + 1$ $ = - 3 + 1 \\\ = - 2 \\\ $ So, $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$ $ = - 2$ **Hence, option (A) is the correct answer.** **Note:** Remember the formulae regarding $\omega $ i.e., ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$ to solve these types of problems. Also, it is important to reduce the high powers of $\omega $into small powers of $\omega $, as we have done in this question.