Let $\mathrm{A}$ be a $2 \times 2$ matrix with real entries such that $\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I}$, where $\alpha \in \mathbb{R}-\{-1,1\}$. If $\operatorname{det}\left(A^{2}-A\right)=4$, then the sum of all possible values of $\alpha$ is equal to :

We have, $A^T=\alpha A+I$, where $A$ is $2 \times 2$ matrix and $\alpha \in R-\{-1,1\}$ <br/><br/>$$ \begin{aligned} \left(A^T\right)^T & =\alpha A^T+I \\\\ A & =\alpha A^T+I \\\\ A & =\alpha(\alpha A+I)+I \left[\because A^T=\alpha A+I\right]\\\\ A & =\alpha^2 A+(\alpha+1) I \\\\ A & \left(1-\alpha^2\right)=(\alpha+1) I \\\\ A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I \end{aligned} $$ <br/><br/>$$ \begin{aligned} A & =\frac{1}{1-\alpha} I ..........(i)\\\\ |A| & =\frac{1}{(1-\alpha)^2} ...........(ii) \end{aligned} $$ <br/><br/>Also, <br/><br/>$$ \begin{aligned} \left|A^2-A\right| & =4 \\\\ |A||A-I| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\\\ \frac{\alpha^2}{(1-\alpha)^4} & =4 \\\\ \frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\\\ 2(1-\alpha)^2 & = \pm \alpha \end{aligned} $$ <br/><br/>If $2(1-\alpha)^2=\alpha$, then $2 \alpha^2-5 \alpha+2=0$ <br/><br/>Sum of value of $\alpha=\frac{5}{2} \quad\left[\because\right.$ Sum of zero $\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]$ <br/><br/>If $2(1-\alpha)^2=-\alpha$, then $2 \alpha^2-3 \alpha+2=0$ <br/><br/>$\therefore$ No real value of $\alpha$ <br/><br/>Hence, sum of all values of $\alpha=\frac{5}{2}$