For a real number $\alpha$, if the system $\left[\begin{array}{ccc}1 & \alpha & \alpha^{2} \ \alpha

Question

For a real number $\alpha$, if the system $\left[\begin{array}{ccc}1 & \alpha & \alpha^{2} \ \alpha & 1 & \alpha \ \alpha^{2} & \alpha & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{c}1 \ -1 \ 1\end{array}ight]$ of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^{2}=$

Professor PC Thomas Classes · Answer

(1) : For $A X=b$ to have infinitely many solutions, a necessary (but NOT sufficient) condition is $\operatorname{det} A=0$, for if $\operatorname{det} A$ $
eq 0$, then $A^{-1}$ exists and we get $X=A^{-1} b$ to be a unique solution.\ $\operatorname{det} A=0 \Rightarrow\left|\begin{array}{ccc}1 & \alpha & \alpha^{2} \ \alpha & 1 & \alpha \ \alpha^{2} & \alpha & 1\end{array}ight|=0$\ Applying $R_{1} ightarrow R_{1}+R_{2}+R_{3}$, we get\ $\left|\begin{array}{ccc}1+\alpha+\alpha^{2} & 2 \alpha+1 & \alpha^{2}+\alpha+1 \ \alpha & 1 & \alpha \ \alpha^{2} & \alpha & 1\end{array}ight|=0$\ Applying $C_{3} ightarrow C_{3}-C_{1}$, we get\ $\left|\begin{array}{ccc}1+\alpha+\alpha^{2} & 2 \alpha+1 & 0 \ \alpha & 1 & 0 \ \alpha^{2} & \alpha & 1-\alpha^{2}\end{array}ight|=0$\ $\Rightarrow \quad\left(1-\alpha^{2}ight)\left(1+\alpha+\alpha^{2}-2 \alpha^{2}-\alphaight)=0$\ $\Rightarrow \quad\left(1-\alpha^{2}ight)\left(1-\alpha^{2}ight)=0$\ $	herefore \quad \alpha^{2}=1 \Rightarrow \alpha=1,-1$\ For $\alpha=1$, the equation becomes\ $x+y+z=1$\ $x+y+z=-1$\ $x+y+z=1$\ which gives no solution.\ For $\alpha=-1$, the equation becomes\ $x-y+z=1$\ $-x+y-z=-1$\ $x-y+z=1$\ which are all the same and hence infinitely many solutions.\ Now, $1+\alpha+\alpha^{2}=1-1+1=1$\