Let A be a 2 × 2 matrix with real entries such that A ^ ' = + 1,...

Given,

$A$ be a $2 \times 2$ matrix with real entries such that ${A}^{'} = \alphaA + 1,$

Now let A = \left[\begin{cases}a & b \\ c & d\end{cases}\right]

Now putting the value in ${A}^{'} = \alphaA + I$ we get,

⇒ \left[\begin{cases}a & c \\ b & d\end{cases}\right] = \left[\begin{cases}\alphaa + 1 & \alphab \\ \alphac & \alphad + 1\end{cases}\right]

Now equating both side we get,

$a = \alphaa + 1 ⇒ a = \dfrac{1}{1 - \alpha} . . . . . . . . . \left(i\right)$

$b = \alphac . . . . \left(ii\right)$ and $c = \alphab . . . . . . . . \left(iii\right)$

Now from equation \left(ii\right) \text{ and } \left(iii\right)$we get,</p><p>$c = 0$or$\alpha = \pm 1$(not possible as mentioned in question)</p><p>$∴ c = 0$</p><p>$⇒ c = 0, b = 0$<br />Also$d = \alphad + 1 ⇒ d = \dfrac{1}{1 - \alpha}$</p><p>Also given, \left|{A}^{2} - A\right| = 4$

$⇒ | A | | A - I | = 4$

$⇒ {\left(\dfrac{1}{1 - \alpha}\right)}^{2}{\left(\dfrac{1}{1 - \alpha} - 1\right)}^{2} = 4$

Now let $\dfrac{1}{1 - \alpha} = t$

So, ${t}^{2}{\left(t - 1\right)}^{2} = 4$

$⇒ t\left(t - 1\right) = \pm 2$

$⇒ t = 2\text{or}t = - 1$

$⇒ \alpha = \dfrac{1}{2}, 2$

So, sum of all possible value will be $\dfrac{1}{2} + 2 = \dfrac{5}{2}$