Discussion Forum

1)

Let a, b and c be three real numbers satisfying [a b c]$\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 Let $\omega$  be a solution of   $x^{3}-1=0$  with  Im $(\omega$) >0.If a=2  with b and c satisfying Eq (E). then the value of $\frac{3}{\omega^{\alpha}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}$

Answer:

Option A

Explanation:

 

 Given  ,$[a  b c]_{1  \times 3}$  $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 $\Rightarrow$ $\begin{bmatrix}a+8b+7c  \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

$\Rightarrow$  a+8b+7c=0....(i)

$\Rightarrow$  9a+2b+3c=0 .....(ii)

$\Rightarrow$  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

 $\therefore$  b=6a and c=-7a

 if a=2,b=12 and c=-14

 $\therefore$   $\frac{3}{ \omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}$

 $\Rightarrow$   $\frac{3}{\omega^{2}}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}$

            $= \frac{3}{ \omega^{2}}+1+3 \omega^{2}$

 $=3 \omega+1+ 3 \omega^{2}$

=$1+3(\omega+\omega^{2})$

=1-3=-2