1. **State the problem:** Reduce the matrix $$\begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}$$ to diagonal form. 2. **Method:** To diagonalize a matrix, we find its eigenvalues and eigenvectors, then form a diagonal matrix with eigenvalues on the diagonal. 3. **Find eigenvalues:** Solve $$\det(A - \lambda I) = 0$$ where $$A$$ is the matrix and $$I$$ is the identity matrix. Calculate $$\det\begin{bmatrix}5-\lambda & 3 & 7 \\ 3 & 26-\lambda & 2 \\ 7 & 2 & 10-\lambda\end{bmatrix} = 0$$. 4. **Compute determinant:** $$ (5-\lambda)((26-\lambda)(10-\lambda) - 2 \times 2) - 3(3(10-\lambda) - 2 \times 7) + 7(3 \times 2 - (26-\lambda) \times 7) = 0 $$ Simplify step-by-step: - Calculate $$ (26-\lambda)(10-\lambda) - 4 = (260 - 26\lambda - 10\lambda + \lambda^2) - 4 = \lambda^2 - 36\lambda + 256 $$ - First term: $$ (5-\lambda)(\lambda^2 - 36\lambda + 256) $$ - Second term: $$ -3(3(10-\lambda) - 14) = -3(30 - 3\lambda - 14) = -3(16 - 3\lambda) = -48 + 9\lambda $$ - Third term: $$ 7(6 - 7(26 - \lambda)) = 7(6 - 182 + 7\lambda) = 7(-176 + 7\lambda) = -1232 + 49\lambda $$ 5. **Expand first term:** $$ (5-\lambda)(\lambda^2 - 36\lambda + 256) = 5\lambda^2 - 180\lambda + 1280 - \lambda^3 + 36\lambda^2 - 256\lambda = -\lambda^3 + 41\lambda^2 - 436\lambda + 1280 $$ 6. **Sum all terms:** $$ -\lambda^3 + 41\lambda^2 - 436\lambda + 1280 - 48 + 9\lambda - 1232 + 49\lambda = -\lambda^3 + 41\lambda^2 - 378\lambda + 0 $$ Simplify constants and coefficients: $$ -\lambda^3 + 41\lambda^2 - 378\lambda = 0 $$ 7. **Factor out $$-\lambda$$:** $$ -\lambda(\lambda^2 - 41\lambda + 378) = 0 $$ 8. **Solve quadratic:** $$ \lambda^2 - 41\lambda + 378 = 0 $$ Calculate discriminant: $$ \Delta = 41^2 - 4 \times 378 = 1681 - 1512 = 169 $$ Roots: $$ \lambda = \frac{41 \pm \sqrt{169}}{2} = \frac{41 \pm 13}{2} $$ So, $$ \lambda_1 = \frac{41 + 13}{2} = 27, \quad \lambda_2 = \frac{41 - 13}{2} = 14 $$ 9. **Eigenvalues:** $$\lambda = 0, 14, 27$$. 10. **Find eigenvectors:** For each eigenvalue, solve $$ (A - \lambda I)\mathbf{v} = 0 $$. 11. **Diagonal matrix:** $$ D = \begin{bmatrix}0 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 27\end{bmatrix} $$ This is the diagonal form of the matrix. **Final answer:** The matrix diagonalizes to $$D = \begin{bmatrix}0 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 27\end{bmatrix}$$ with eigenvalues 0, 14, and 27 on the diagonal.