Question

If $ \alpha $ and $ \beta $ are non-real numbers satisfying $ {{x}^{3}}-1=0, $ then the value of $ \left| \begin{matrix} \lambda +1 & \alpha & \beta \\ \alpha & \lambda +\beta & 1 \\ \beta & 1 & \lambda +\alpha \\ \end{matrix} \right| $ is

• A. 0
• B. $ {{\lambda }^{3}} $
• C. $ {{x}^{2}}+2bx+4=0 $
• D. $ {{x}^{2}}-bx+1=0 $

Answer 1

Answer: (B)

It is given that $ \alpha $ and $ \beta $ are the non-real roots of the equation $ {{x}^{3}}-1=0 $ . We have, $ {{x}^{3}}-1=0 $ $ \Rightarrow $ $ {{x}^{3}}=1 $ $ \Rightarrow $ $ x=1,\omega ,{{\omega }^{2}} $ Hence, $ \alpha =\omega $ and $ \beta ={{\omega }^{2}} $ Now, $ \left| \begin{matrix} \lambda +1 & \alpha & \beta \\ \alpha & \lambda +\beta & 1 \\ \beta & 1 & \lambda +\alpha \\ \end{matrix} \right| $ $ =\left| \begin{matrix} \lambda +1+\alpha +\beta & \lambda +1+\alpha +\beta & \lambda +1+a+\beta \\ \alpha & \alpha +\beta & 1 \\ \beta & 1 & \lambda +\alpha \\ \end{matrix} \right| $ $ [\because {{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}] $ $ =(\lambda +1+\alpha +\beta )\left| \begin{matrix} 0 & 0 & 1 \\ \alpha -\lambda -\beta & \lambda +\beta -1 & 1 \\ \beta -1 & 1-\lambda -\alpha & \lambda +\alpha \\ \end{matrix} \right| $ $ [\because {{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}] $ $ =(\lambda +1+\alpha +\beta ).1[(\alpha -\lambda -\beta )(1-\lambda -\alpha ) $ $ -(\beta -1)(\lambda +\beta -1)] $ $ =(\lambda +1+\alpha +\beta )(\alpha -{{\alpha }^{2}}+{{\lambda }^{2}}+\alpha \beta -{{\beta }^{2}}+\beta -1) $ $ (\lambda +1+\omega +{{\omega }^{2}})(\omega -{{\omega }^{2}}+{{\lambda }^{2}}+{{\omega }^{3}}-{{\omega }^{4}}+{{\omega }^{2}}-1) $ [putting $ \alpha =\omega $ and $ \beta ={{\omega }^{2}} $ ] $ =\lambda (\omega -{{\omega }^{2}}+{{\lambda }^{2}}+1-\omega +{{\omega }^{2}}-1) $ $ ={{\lambda }^{3}} $