1. Stating the problem: We want to diagonalize matrix $$A=\begin{bmatrix}1 & -6 & 4 \\ 0 & 4 & 2 \\ 0 & -6 & 3\end{bmatrix}$$. 2. Find the eigenvalues by solving $$\det(A-\lambda I)=0$$: $$A-\lambda I=\begin{bmatrix}1-\lambda & -6 & 4 \\ 0 & 4-\lambda & 2 \\ 0 & -6 & 3-\lambda\end{bmatrix}$$ 3. Compute the determinant: $$\det(A-\lambda I)=(1-\lambda)\cdot\det\begin{bmatrix}4-\lambda & 2 \\ -6 & 3-\lambda\end{bmatrix}-0+0$$ $$=(1-\lambda)((4-\lambda)(3-\lambda)-2(-6))$$ $$=(1-\lambda)((4-\lambda)(3-\lambda)+12)$$ 4. Expand $$(4-\lambda)(3-\lambda) = 12 - 4\lambda - 3\lambda + \lambda^2 = \lambda^2 -7\lambda + 12$$ 5. Substitute back: $$\det(A-\lambda I) = (1-\lambda)(\lambda^2 -7\lambda + 12 + 12) = (1-\lambda)(\lambda^2 -7\lambda + 24)$$ 6. Solve the quadratic equation $$\lambda^2 - 7\lambda + 24 = 0$$: The discriminant $$\Delta = (-7)^2 - 4 \cdot 1 \cdot 24 = 49 - 96 = -47 < 0$$, so the roots are complex: $$\lambda = \frac{7 \pm \sqrt{-47}}{2} = \frac{7 \pm i\sqrt{47}}{2}$$ 7. Therefore, eigenvalues are: $$\lambda_1 = 1$$ $$\lambda_2 = \frac{7 + i\sqrt{47}}{2}$$ $$\lambda_3 = \frac{7 - i\sqrt{47}}{2}$$ 8. Find eigenvectors corresponding to each eigenvalue. For $$\lambda_1=1$$, solve $$(A - I)\mathbf{v} = \mathbf{0}$$: $$A - I = \begin{bmatrix}0 & -6 & 4 \\ 0 & 3 & 2 \\ 0 & -6 & 2\end{bmatrix}$$ This reduces to eigenvector equation. Let $$\mathbf{v} = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$. From first row: $$0\cdot x - 6y + 4z = 0 \Rightarrow -6y + 4z = 0 \Rightarrow 3y = 2z$$. From second row: $$0\cdot x + 3y + 2z = 0 \Rightarrow 3y + 2z = 0$$. From these two equations: - From first: $$3y = 2z$$ - From second: $$3y = -2z$$ This implies $$2z = -2z \Rightarrow 4z=0 \Rightarrow z=0$$. Then from $$3y=2z=0$$ we get $$y=0$$. No restriction on $$x$$, so eigenvectors are of form $$\begin{bmatrix}x \\ 0 \\ 0\end{bmatrix}$$, i.e., scalar multiples of $$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$. 9. For complex eigenvalues, eigenvectors will be complex; diagonalization is in complex field. Summary: Matrix $$A$$ has eigenvalues $$1, \frac{7+i\sqrt{47}}{2}, \frac{7 - i\sqrt{47}}{2}$$ and is diagonalizable over the complex field with eigenvectors corresponding to each eigenvalue. Final diagonal matrix $$D=\begin{bmatrix}1 & 0 & 0 \\ 0 & \frac{7 + i\sqrt{47}}{2} & 0 \\ 0 & 0 & \frac{7 - i\sqrt{47}}{2}\end{bmatrix}$$.