1. Key Concepts and Formulas

1. Matrix Transpose Properties: For matrices $M$ and $N$, and a scalar $k$:
   • $(M^T)^T = M$ (Transpose of a transpose is the original matrix).
   • $(kM + N)^T = kM^T + N^T$ (Transpose distributes over scalar multiplication and addition).
   • $I^T = I$ (The identity matrix is symmetric).
2. Determinant Properties for $n \times n$ matrices: For matrices $M$ and $N$, and a scalar $k$:
   • $\det(MN) = \det(M) \det(N)$ (Determinant of a product is the product of determinants).
   • $\det(kM) = k^n \det(M)$ (Scalar multiplication scales the determinant by $k^n$). For a $2 \times 2$ matrix, $n=2$, so $\det(kM) = k^2 \det(M)$.
   • $\det(I) = 1$.
3. Quadratic Equation Roots: For $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$. For real roots, the discriminant $D = b^2 - 4ac$ must be non-negative.

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2. Step-by-Step Solution

Step 1: Determine the structure of Matrix $A$

We are given the relation: $A^T = \alpha A + I \quad (*)$ where $A$ is a $2 \times 2$ matrix with real entries, and $\alpha \in \mathbb{R}-\{-1,1\}$. Our first goal is to express $A$ in a simpler form.

• Take the transpose of both sides of equation $(*)$: Applying the transpose operation to both sides helps us create a second equation involving $A$ and $A^T$, which can then be used to eliminate $A^T$. $(A^T)^T = (\alpha A + I)^T$
• Apply transpose properties: Using $(A^T)^T = A$, $(kA + B)^T = kA^T + B^T$, and $I^T = I$: $A = \alpha A^T + I \quad (**)$
• Substitute $A^T$ from $(*)$ into $(**)$: This eliminates $A^T$ from the equation, allowing us to solve for $A$ in terms of $\alpha$ and $I$. $A = \alpha (\alpha A + I) + I$
• Simplify the expression: $A = \alpha^2 A + \alpha I + I$ $A = \alpha^2 A + (\alpha + 1)I$
• Rearrange to solve for $A$: Gather terms involving $A$ on one side: $A - \alpha^2 A = (\alpha + 1)I$ Factor out $A$: $A(1 - \alpha^2) = (\alpha + 1)I$
• Isolate $A$: We can divide by $(1 - \alpha^2)$ because the problem states $\alpha \in \mathbb{R}-\{-1,1\}$, which means $\alpha \neq 1$ and $\alpha \neq -1$. Therefore, $1 - \alpha^2 = (1 - \alpha)(1 + \alpha) \neq 0$. $A = \frac{(\alpha + 1)}{(1 - \alpha)(1 + \alpha)} I$ Since $\alpha \neq -1$, we can cancel the $(\alpha + 1)$ terms: $A = \frac{1}{1 - \alpha} I \quad (1)$ This shows that $A$ is a scalar multiple of the identity matrix. This significantly simplifies subsequent calculations.

Step 2: Calculate $\det(A)$ and $\det(A-I)$

We are given $\det(A^2 - A) = 4$. To use this, we will factor $A^2 - A$ as $A(A-I)$ and then calculate $\det(A)$ and $\det(A-I)$ separately.

• Calculate $\det(A)$: Since $A = \frac{1}{1 - \alpha} I$ and $A$ is a $2 \times 2$ matrix, we use the property $\det(kI) = k^2$ (for $n=2$). Here, $k = \frac{1}{1 - \alpha}$. $\det(A) = \det\left(\frac{1}{1 - \alpha} I\right) = \left(\frac{1}{1 - \alpha}\right)^2 = \frac{1}{(1 - \alpha)^2}$
• Calculate $A - I$: Using equation $(1)$: $A - I = \frac{1}{1 - \alpha} I - I = \left(\frac{1}{1 - \alpha} - 1\right) I$ Simplify the scalar term: $\frac{1}{1 - \alpha} - 1 = \frac{1 - (1 - \alpha)}{1 - \alpha} = \frac{\alpha}{1 - \alpha}$ So, $A - I = \frac{\alpha}{1 - \alpha} I$
• Calculate $\det(A - I)$: Again, using $\det(kI) = k^2$: $\det(A - I) = \det\left(\frac{\alpha}{1 - \alpha} I\right) = \left(\frac{\alpha}{1 - \alpha}\right)^2 = \frac{\alpha^2}{(1 - \alpha)^2}$

Step 3: Use the given determinant condition to form an equation for $\alpha$

We are given $\det(A^2 - A) = 4$.

• Factor the expression inside the determinant: $A^2 - A = A(A - I)$
• Apply the determinant property $\det(MN) = \det(M)\det(N)$: $\det(A(A - I)) = \det(A) \det(A - I) = 4$
• Substitute the calculated determinants: $\left(\frac{1}{(1 - \alpha)^2}\right) \left(\frac{\alpha^2}{(1 - \alpha)^2}\right) = 4$
• Simplify the equation: $\frac{\alpha^2}{(1 - \alpha)^4} = 4$

Step 4: Solve for $\alpha$

To solve for $\alpha$, we take the square root of both sides. Remember that $\sqrt{x^2} = |x|$ and $\sqrt{y^4} = y^2$ (since $y^2$ is always non-negative). $\sqrt{\frac{\alpha^2}{(1 - \alpha)^4}} = \sqrt{4}$ $\frac{|\alpha|}{(1 - \alpha)^2} = 2$ This equation implies two cases due to the absolute value:

Case 1: $\alpha \ge 0 \implies \frac{\alpha}{(1 - \alpha)^2} = 2$ $2(1 - \alpha)^2 = \alpha$ Expand the quadratic term: $2(1 - 2\alpha + \alpha^2) = \alpha$ $2 - 4\alpha + 2\alpha^2 = \alpha$ Rearrange into a standard quadratic equation $ax^2 + bx + c = 0$: $2\alpha^2 - 5\alpha + 2 = 0$ To find the real solutions for $\alpha$, we calculate the discriminant $D = b^2 - 4ac$: $D = (-5)^2 - 4(2)(2) = 25 - 16 = 9$. Since $D = 9 > 0$, there are two distinct real roots. The roots are $\alpha = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}$. So, $\alpha_1 = \frac{5 + 3}{4} = \frac{8}{4} = 2$ and $\alpha_2 = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}$. Both $2$ and $1/2$ are non-negative and satisfy the condition $\alpha \in \mathbb{R}-\{-1,1\}$. So both are valid possible values for $\alpha$.

Case 2: $\alpha < 0 \implies \frac{-\alpha}{(1 - \alpha)^2} = 2$ $-2(1 - \alpha)^2 = \alpha$ $-2(1 - 2\alpha + \alpha^2) = \alpha$ $-2 + 4\alpha - 2\alpha^2 = \alpha$ Rearrange into a standard quadratic equation: $2\alpha^2 - 3\alpha + 2 = 0$ Calculate the discriminant $D = b^2 - 4ac$: $D = (-3)^2 - 4(2)(2) = 9 - 16 = -7$. Since $D = -7 < 0$, there are no real roots for $\alpha$ in this case. The roots are complex, and we are only interested in $\alpha \in \mathbb{R}$.

Step 5: Sum of all possible values of $\alpha$

From Case 1, the possible real values for $\alpha$ are $2$ and $1/2$. Case 2 yielded no real values. The sum of all possible values of $\alpha$ is $2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$.

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3. Common Mistakes & Tips

• Absolute Value: When solving $\sqrt{x^2} = |x|$, remember to consider both positive and negative cases for the expression inside the absolute value. Failing to do so can lead to missing valid solutions.
• Matrix vs. Scalar Algebra: Always be mindful of the rules of matrix algebra. For instance, $\det(A+B) \neq \det(A)+\det(B)$ generally, but $\det(AB) = \det(A)\det(B)$ is crucial.
• Conditions on $\alpha$: The condition $\alpha \in \mathbb{R}-\{-1,1\}$ is vital. It justifies division by $1-\alpha^2$ and filters out non-real solutions for $\alpha$.

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4. Summary

We began by utilizing matrix transpose properties to simplify the given relation $A^T = \alpha A + I$. This revealed that matrix $A$ must be a scalar multiple of the identity matrix, specifically $A = \frac{1}{1-\alpha}I$. We then used this simplified form of $A$ to calculate $\det(A)$ and $\det(A-I)$. Substituting these into the given condition $\det(A^2 - A) = 4$ led to an equation involving $\alpha$. Solving this equation, which included handling an absolute value, resulted in two quadratic equations. One quadratic yielded two real solutions for $\alpha$ (2 and 1/2), both satisfying the problem's constraints. The other quadratic yielded no real solutions. The sum of these two valid real values of $\alpha$ is $2 + 1/2 = 5/2$.

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5. Final Answer

The sum of all possible values of $\alpha$ is $\frac{5}{2}$. This corresponds to option (D). The final answer is $\boxed{\frac{5}{2}}$