If $M=\begin{pmatrix} 0 & 1 & -1 \\ -1 & 0 & \alpha \\ 2 & -\alpha & 0 \end{pmatrix} $ and $Mx=b$, for some $x$,then find $x^TMx$.
My approach:
If $Mx=b$ has a unique solution $x$ then, $\text{Rank(M)}=3$ Thus, there are three linearly independent columns in $M$. Thus if $\alpha \neq 0$, $c_1(0,-1,2)+c_2(1, 0 ,-\alpha)+c_3(-1, \alpha ,0)=(0, 0 ,0)$ has only $c_1,c_2,c_3 =0$ as solutions. But if $\alpha=0$, then the second and third row are linearly dependent, so the $\text{rank}(M)=2$ which is not possible. But then I am not getting any specific value of $\alpha$ in $M$. Am I going wrong somewhere?