Let $$P = \begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}$$ and $$A = \begin{bmatrix} 2 & 7 & \omega^2 \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1 \end{bmatrix}$$ where $$\omega = \frac{-1+i\sqrt{3}}{2}$$, and $$I_3$$ be the identity matrix of order 3. If the determinant of the matrix $$\left(P^{-1}AP - I_3\right)^2$$ is $$\alpha\omega^2$$, then the value of $$\alpha$$ is equal to ________.

We are given $$\omega = \frac{-1 + i\sqrt{3}}{2}$$, a primitive cube root of unity, so $$\omega^3 = 1$$ and $$1 + \omega + \omega^2 = 0$$.

The matrix $$A = \begin{bmatrix} 2 & 7 & \omega^2 \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega + 1 \end{bmatrix}$$.

Since $$P^{-1}AP$$ is similar to $$A$$, we have $$\det(P^{-1}AP - I_3) = \det(A - I_3)$$, and therefore $$\det((P^{-1}AP - I_3)^2) = [\det(A - I_3)]^2$$.

We compute $$A - I_3 = \begin{bmatrix} 1 & 7 & \omega^2 \\ -1 & -\omega - 1 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$$.

Using the identity $$1 + \omega + \omega^2 = 0$$, we note that $$-\omega - 1 = \omega^2$$. So the matrix becomes $$A - I_3 = \begin{bmatrix} 1 & 7 & \omega^2 \\ -1 & \omega^2 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$$.

Computing the determinant by expanding along the first column: $$\det(A - I_3) = 1 \cdot \begin{vmatrix} \omega^2 & 1 \\ -\omega & -\omega \end{vmatrix} - (-1) \cdot \begin{vmatrix} 7 & \omega^2 \\ -\omega & -\omega \end{vmatrix} + 0$$.

The first minor is $$\omega^2 \cdot (-\omega) - 1 \cdot (-\omega) = -\omega^3 + \omega = -1 + \omega$$.

The second minor is $$7 \cdot (-\omega) - \omega^2 \cdot (-\omega) = -7\omega + \omega^3 = -7\omega + 1$$.

So $$\det(A - I_3) = 1 \cdot (-1 + \omega) + 1 \cdot (-7\omega + 1) = -1 + \omega - 7\omega + 1 = -6\omega$$.

Therefore $$[\det(A - I_3)]^2 = (-6\omega)^2 = 36\omega^2$$.

Since $$\det((P^{-1}AP - I_3)^2) = 36\omega^2 = \alpha\omega^2$$, we get $$\alpha = 36$$.