Solution
= Solution
Expand the <determinant> along the first row:
$$
\begin{aligned}
\det A_\alpha={}&(\alpha+1)\left[\alpha-(2-\alpha)(\alpha+3)\right]\\
&-(1-\alpha)\left[2-(2-\alpha)^2\right]+2\left[2(\alpha+3)-\alpha(2-\alpha)\right]\\
={}&(\alpha+1)(\alpha^2+2\alpha-6)-(1-\alpha)(-\alpha^2+4\alpha-2)+2(\alpha^2+6)\\
={}&\boxed{10\alpha^2-10\alpha+8.}
\end{aligned}
$$
The cubic terms cancel in the expansion.