1. **Problem Statement:** Diagonalise the matrix $$A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$$ by an orthogonal transformation. 2. **Key Concept:** To diagonalise a symmetric matrix by an orthogonal transformation, we find an orthogonal matrix $$P$$ such that $$P^T A P = D$$, where $$D$$ is a diagonal matrix containing the eigenvalues of $$A$$. 3. **Step 1: Verify Symmetry** Matrix $$A$$ is symmetric since $$A = A^T$$. 4. **Step 2: Find Eigenvalues** Solve the characteristic equation $$\det(A - \lambda I) = 0$$: $$\det\begin{bmatrix}6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda \end{bmatrix} = 0$$ 5. **Calculate determinant:** $$ (6-\lambda) \begin{vmatrix}3-\lambda & -1 \\ -1 & 3-\lambda \end{vmatrix} - (-2) \begin{vmatrix}-2 & -1 \\ 2 & 3-\lambda \end{vmatrix} + 2 \begin{vmatrix}-2 & 3-\lambda \\ 2 & -1 \end{vmatrix} = 0 $$ 6. **Compute minors:** $$ (6-\lambda)((3-\lambda)^2 - (-1)(-1)) + 2((-2)(3-\lambda) - (-1)(2)) + 2((-2)(-1) - (3-\lambda)(2)) = 0 $$ 7. **Simplify:** $$ (6-\lambda)((3-\lambda)^2 - 1) + 2(-2(3-\lambda) + 2) + 2(2 - 2(3-\lambda)) = 0 $$ 8. **Expand:** $$ (6-\lambda)((3-\lambda)^2 - 1) + 2(-6 + 2\lambda + 2) + 2(2 - 6 + 2\lambda) = 0 $$ 9. **Simplify further:** $$ (6-\lambda)((3-\lambda)^2 - 1) + 2(-4 + 2\lambda) + 2(-4 + 2\lambda) = 0 $$ 10. **Calculate $$(3-\lambda)^2 - 1$$:** $$ (3-\lambda)^2 - 1 = (9 - 6\lambda + \lambda^2) - 1 = 8 - 6\lambda + \lambda^2 $$ 11. **Substitute:** $$ (6-\lambda)(8 - 6\lambda + \lambda^2) + 4(-4 + 2\lambda) = 0 $$ 12. **Expand:** $$ (6)(8 - 6\lambda + \lambda^2) - \lambda(8 - 6\lambda + \lambda^2) + 4(-4 + 2\lambda) = 0 $$ 13. **Calculate:** $$ 48 - 36\lambda + 6\lambda^2 - 8\lambda + 6\lambda^2 - \lambda^3 - 16 + 8\lambda = 0 $$ 14. **Combine like terms:** $$ -\lambda^3 + 12\lambda^2 - 36\lambda + 32 = 0 $$ 15. **Multiply both sides by -1:** $$ \lambda^3 - 12\lambda^2 + 36\lambda - 32 = 0 $$ 16. **Find roots:** Try possible rational roots (factors of 32): 1, 2, 4, 8, 16, 32. 17. **Test $$\lambda=2$$:** $$ 2^3 - 12(2)^2 + 36(2) - 32 = 8 - 48 + 72 - 32 = 0 $$ So, $$\lambda=2$$ is a root. 18. **Divide polynomial by $$(\lambda - 2)$$:** $$ \lambda^3 - 12\lambda^2 + 36\lambda - 32 = (\lambda - 2)(\lambda^2 - 10\lambda + 16) $$ 19. **Solve quadratic:** $$ \lambda^2 - 10\lambda + 16 = 0 $$ 20. **Use quadratic formula:** $$ \lambda = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} $$ 21. **Eigenvalues:** $$ \lambda_1 = 2, \quad \lambda_2 = 8, \quad \lambda_3 = 2 $$ 22. **Step 3: Find eigenvectors for each eigenvalue** 23. **For $$\lambda=2$$:** Solve $$ (A - 2I)\mathbf{x} = 0 $$ 24. **Matrix:** $$ \begin{bmatrix}4 & -2 & 2 \\ -2 & 1 & -1 \\ 2 & -1 & 1 \end{bmatrix} \mathbf{x} = 0 $$ 25. **Find basis vectors for eigenspace corresponding to $$\lambda=2$$.** 26. **For $$\lambda=8$$:** Solve $$ (A - 8I)\mathbf{x} = 0 $$ 27. **Matrix:** $$ \begin{bmatrix}-2 & -2 & 2 \\ -2 & -5 & -1 \\ 2 & -1 & -5 \end{bmatrix} \mathbf{x} = 0 $$ 28. **Find eigenvector for $$\lambda=8$$.** 29. **Step 4: Orthogonalise eigenvectors if needed and normalise to get orthonormal basis.** 30. **Step 5: Form orthogonal matrix $$P$$ with eigenvectors as columns.** 31. **Step 6: Compute $$D = P^T A P$$, which will be diagonal with eigenvalues on the diagonal.** **Final answer:** The matrix $$A$$ is diagonalised by an orthogonal matrix $$P$$ whose columns are the normalised eigenvectors corresponding to eigenvalues $$2, 2, 8$$, and the diagonal matrix is $$D = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{bmatrix}$$.