1. **Problem:** Given matrix $$A=\begin{bmatrix}a & a & 0 \\ a & 1 & 0 \\ 0 & 1 & 2\end{bmatrix}$$ where $a$ satisfies $x^2 - x + 1 = 0$, find $$\det \left( 3A^3 A' A^{-1} \right)^{-1}$$. 2. **Step 1:** Solve for $a$ from the quadratic equation: $$x^2 - x + 1 = 0$$ The discriminant is $\Delta = (-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3 < 0$, so $a$ is complex and roots are $$a = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}$$ 3. **Step 2:** Calculate $\det(A)$: $$\det(A) = a \times \det \begin{bmatrix}1 & 0 \\ 1 & 2\end{bmatrix} - a \times \det \begin{bmatrix}a & 0 \\ 0 & 2\end{bmatrix} + 0 = a(1 \times 2 - 0) - a(a \times 2 - 0) = 2a - 2a^2$$ 4. **Step 3:** Use the relation from the quadratic equation: $$a^2 = a - 1$$ Substitute into determinant: $$\det(A) = 2a - 2(a - 1) = 2a - 2a + 2 = 2$$ 5. **Step 4:** Use properties of determinants: $$\det(3A^3 A' A^{-1}) = \det(3I) \det(A^3) \det(A') \det(A^{-1})$$ Since $3I$ is scalar multiplication by 3 on a $3 \times 3$ matrix, its determinant is $3^3 = 27$. 6. **Step 5:** Recall: - $\det(A^3) = (\det A)^3 = 2^3 = 8$ - $\det(A') = \det(A)$ since transpose does not change determinant, so $= 2$ - $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{2}$ 7. **Step 6:** Calculate determinant: $$\det(3A^3 A' A^{-1}) = 27 \times 8 \times 2 \times \frac{1}{2} = 27 \times 8 = 216$$ 8. **Step 7:** Finally, $$\det \left( 3A^3 A' A^{-1} \right)^{-1} = \frac{1}{216}$$ **Answer:** $\boxed{\frac{1}{216}}$