Let A be a 2 times 2 matrix with real entries | vedclass

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(C) Given $A^{T} = \alpha A + I$. Taking transpose on both sides,$A = \alpha A^{T} + I$. Substituting $A^{T}$ in the second equation: $A = \alpha(\alp

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$\frac{5}{2}$

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(C) Given $A^{T} = \alpha A + I$. Taking transpose on both sides,$A = \alpha A^{T} + I$. Substituting $A^{T}$ in the second equation: $A = \alpha(\alpha A + I) + I = \alpha^2 A + (\alpha + 1)I$. Rearranging gives $A(1 - \alpha^2) = (\alpha + 1)I$. Since $\alpha 
eq -1$,we can divide by $(1 + \alpha)$ to get $A(1 - \alpha) = I$,so $A = \frac{1}{1 - \alpha}I$. Then $\det(A) = \frac{1}{(1 - \alpha)^2}$. Also,$A - I = \frac{1}{1 - \alpha}I - I = \frac{1 - (1 - \alpha)}{1 - \alpha}I = \frac{\alpha}{1 - \alpha}I$. Then $\det(A - I) = \left(\frac{\alpha}{1 - \alpha}ight)^2$. Given $\det(A^2 - A) = \det(A)\det(A - I) = 4$. Substituting the values: $\frac{1}{(1 - \alpha)^2} \cdot \frac{\alpha^2}{(1 - \alpha)^2} = 4$. $\frac{\alpha^2}{(1 - \alpha)^4} = 4 \Rightarrow \left(\frac{\alpha}{(1 - \alpha)^2}ight)^2 = 2^2$. This implies $\frac{\alpha}{(1 - \alpha)^2} = 2$ or $\frac{\alpha}{(1 - \alpha)^2} = -2$. Case $1$: $\alpha = 2(1 - 2\alpha + \alpha^2) \Rightarrow 2\alpha^2 - 5\alpha + 2 = 0$. The roots are $\alpha = 2$ and $\alpha = 1/2$. Case $2$: $\alpha = -2(1 - 2\alpha + \alpha^2) \Rightarrow 2\alpha^2 - 3\alpha + 2 = 0$. The discriminant $D = 9 - 16 = -7 The possible values of $\alpha$ are $2$ and $1/2$. The sum is $2 + 1/2 = 5/2$.

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

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