Question

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

Answer 1

0.5

Answer 2

The problem provides a $2 \times 2$ matrix $A$ with real entries and an equation $A - \alpha A^2 = I$, where $\alpha \in \mathbb{R} - \{-1, 1\}$. We are also given $\det(A^2 - A) = 4$. We need to find twice the sum of all possible values of $\alpha$.

1. Relate the given matrix equation to the characteristic polynomial: The given equation is $A - \alpha A^2 = I$. Rearranging, we get $\alpha A^2 - A + I = 0$. Since $\alpha \in \mathbb{R} - \{-1, 1\}$, $\alpha \neq 0$. We can divide by $\alpha$: $A^2 - \frac{1}{\alpha} A + \frac{1}{\alpha} I = 0$. By the Cayley-Hamilton theorem, a $2 \times 2$ matrix $A$ satisfies its characteristic equation: $A^2 - (\text{tr} A)A + (\det A)I = 0$. Comparing the two equations, we can identify: $\text{tr} A = \frac{1}{\alpha}$ $\det A = \frac{1}{\alpha}$

2. Express $A^2 - A$ in terms of $A$ and $\alpha$: From the given equation $A - \alpha A^2 = I$, we can factor out $A$: $A(I - \alpha A) = I$. This implies that $A$ is invertible, and $A^{-1} = I - \alpha A$. Now consider the expression $A^2 - A$. We can write it as $A(A - I)$. From $A - \alpha A^2 = I$, we can rearrange to get $A - I = \alpha A^2$. Substitute this into $A(A - I)$: $A^2 - A = A(\alpha A^2) = \alpha A^3$.

3. Use the determinant condition: We are given $\det(A^2 - A) = 4$. Substitute the expression from step 2: $\det(\alpha A^3) = 4$. For a $2 \times 2$ matrix $M$ and a scalar $k$, $\det(kM) = k^2 \det(M)$. So, $\det(\alpha A^3) = \alpha^2 \det(A^3)$. Also, $\det(A^3) = (\det A)^3$. Therefore, $\alpha^2 (\det A)^3 = 4$.

4. Substitute $\det A$ and solve for $\alpha$: From step 1, we found $\det A = \frac{1}{\alpha}$. Substitute this into the equation from step 3: $\alpha^2 \left(\frac{1}{\alpha}\right)^3 = 4$ $\alpha^2 \frac{1}{\alpha^3} = 4$ $\frac{1}{\alpha} = 4$ $\alpha = \frac{1}{4}$.

5. Check for validity and final calculation: The obtained value $\alpha = 1/4$ satisfies the condition $\alpha \in \mathbb{R} - \{-1, 1\}$. Since we derived $\det A = 1/\alpha$ and used it, this implies that the characteristic polynomial of $A$ is $x^2 - \frac{1}{\alpha} x + \frac{1}{\alpha} = 0$. For $A$ to be a matrix with real entries, its eigenvalues must be real or complex conjugate pairs. The roots of this quadratic equation are $\lambda = \frac{1 \pm \sqrt{1 - 4\alpha}}{\text{2}\alpha}$. For the eigenvalues to be real, we need $1 - 4\alpha \ge 0$, so $1 \ge 4\alpha$, or $\alpha \le 1/4$. Our value $\alpha = 1/4$ satisfies this condition, as $1 - 4(1/4) = 0$, meaning $A$ has repeated real eigenvalues. There is only one possible value for $\alpha$, which is $1/4$. The sum of all possible values of $\alpha$ is $1/4$. The twice of the sum of all possible values of $\alpha$ is $2 \times \frac{1}{4} = \frac{1}{2}$.

The final answer is $\boxed{0.5}$.