Alpha matrix

|A|=a^2-4 ,then use |A^3|=|A|^3 and solve the equation

Let us consider the form $A = S\Lambda S^{-1}$ , where $\Lambda$ and $S$ are the eigenvalue & eigenvector matrices respectively. Cubing the matrix $A$ yields $A^{3} = S \Lambda^{3} S^{-1}$ , and its determinant can be written as:

$|A^{3}| = |S \Lambda^{3} S^{-1}| = |SS^{-1}| \cdot |\Lambda^{3}| = |I| \cdot |\Lambda^{3}| = |\Lambda^{3}|$ .

If $det(A - \lambda I) = 0$ , then $(\alpha - \lambda)^2 - 2^2 = 0 \Rightarrow \lambda = \alpha \pm 2$ . Substitution of these values into the $\Lambda^{3}$ matrix yields:

$|\Lambda^{3}| = \begin{vmatrix} (\alpha + 2)^{3} & 0 \\ 0 & (\alpha - 2)^{3} \end{vmatrix} = [(\alpha + 2)(\alpha - 2)]^3 = (\alpha^2 - 4)^3 = 125$ ;

or $\alpha^2 - 4 = 5 \Rightarrow \alpha^2 = 9 \Rightarrow \alpha = \boxed{\pm 3}.$