Question

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

• A. 0
• B. \frac{3}{2}
• C. 2
• D. \frac{5}{2}

Answer 1

Answer: (D)

\frac{5}{2}

Answer 2

The relation $A^T = \alpha A + I$ for a 2x2 matrix $A$ with $\alpha \neq \pm 1$ implies that $A$ must be a scalar matrix, $A=kI$, where $k = \frac{1}{1-\alpha}$. The condition det($A^2 - A$) = 4 then translates to $(k^2-k)^2 = 4$. This yields $k^2-k=2$ or $k^2-k=-2$. The equation $k^2-k=-2$ has no real solutions for $k$. The equation $k^2-k=2$ yields $k=2$ and $k=-1$. Substituting $k = \frac{1}{1-\alpha}$ back, we get $\alpha = 1/2$ and $\alpha = 2$. The sum of these possible values of $\alpha$ is $1/2 + 2 = 5/2$.