1. **Problem Statement:** Find the modal matrix and diagonalize the matrix $$A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{bmatrix}$$. 2. **Step 1: Find the eigenvalues of $$A$$.** The eigenvalues $$\lambda$$ satisfy $$\det(A - \lambda I) = 0$$, where $$I$$ is the identity matrix. 3. **Calculate $$A - \lambda I$$:** $$A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{bmatrix}$$ 4. **Find the determinant:** $$\det(A - \lambda I) = (1-\lambda)((5-\lambda)(1-\lambda) - 1 \cdot 1) - 1(1(1-\lambda) - 1 \cdot 3) + 3(1 \cdot 1 - (5-\lambda) \cdot 3)$$ 5. **Simplify the determinant:** Calculate each term: - $$(5-\lambda)(1-\lambda) - 1 = (5 - 5\lambda - \lambda + \lambda^2) - 1 = 4 - 6\lambda + \lambda^2$$ - $$1(1-\lambda) - 3 = 1 - \lambda - 3 = -2 - \lambda$$ - $$1 - 3(5-\lambda) = 1 - 15 + 3\lambda = -14 + 3\lambda$$ So, $$\det(A - \lambda I) = (1-\lambda)(4 - 6\lambda + \lambda^2) - 1(-2 - \lambda) + 3(-14 + 3\lambda)$$ 6. **Expand and simplify:** $$ (1-\lambda)(4 - 6\lambda + \lambda^2) = 4 - 6\lambda + \lambda^2 - 4\lambda + 6\lambda^2 - \lambda^3 = 4 - 10\lambda + 7\lambda^2 - \lambda^3 $$ Add the other terms: $$ -1(-2 - \lambda) = 2 + \lambda $$ $$ 3(-14 + 3\lambda) = -42 + 9\lambda $$ Sum all: $$4 - 10\lambda + 7\lambda^2 - \lambda^3 + 2 + \lambda - 42 + 9\lambda = (4 + 2 - 42) + (-10\lambda + \lambda + 9\lambda) + 7\lambda^2 - \lambda^3 = -36 + 0\lambda + 7\lambda^2 - \lambda^3$$ So, $$\det(A - \lambda I) = -36 + 7\lambda^2 - \lambda^3 = -\lambda^3 + 7\lambda^2 - 36$$ 7. **Rewrite characteristic polynomial:** $$-\lambda^3 + 7\lambda^2 - 36 = 0 \implies \lambda^3 - 7\lambda^2 + 36 = 0$$ 8. **Find roots of $$\lambda^3 - 7\lambda^2 + 36 = 0$$.** Try rational roots: factors of 36 are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36. Test $$\lambda=3$$: $$3^3 - 7(3)^2 + 36 = 27 - 63 + 36 = 0$$ So, $$\lambda=3$$ is a root. 9. **Factor polynomial:** Divide by $$\lambda - 3$$: $$\lambda^3 - 7\lambda^2 + 36 = (\lambda - 3)(\lambda^2 - 4\lambda - 12)$$ 10. **Solve quadratic:** $$\lambda^2 - 4\lambda - 12 = 0$$ Use quadratic formula: $$\lambda = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}$$ So, $$\lambda = 6$$ or $$\lambda = -2$$ 11. **Eigenvalues:** $$\lambda_1 = 3, \quad \lambda_2 = 6, \quad \lambda_3 = -2$$ 12. **Step 2: Find eigenvectors for each eigenvalue.** Solve $$ (A - \lambda I)\mathbf{v} = 0 $$ for each $$\lambda$$. 13. **Eigenvector for $$\lambda=3$$:** $$A - 3I = \begin{bmatrix} -2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & -2 \end{bmatrix}$$ Solve: $$-2x + y + 3z = 0$$ $$x + 2y + z = 0$$ $$3x + y - 2z = 0$$ From first equation: $$y = 2x - 3z$$ Substitute into second: $$x + 2(2x - 3z) + z = 0 \implies x + 4x - 6z + z = 0 \implies 5x - 5z = 0 \implies x = z$$ Then, $$y = 2x - 3z = 2z - 3z = -z$$ Eigenvector: $$\mathbf{v}_1 = z \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$ 14. **Eigenvector for $$\lambda=6$$:** $$A - 6I = \begin{bmatrix} -5 & 1 & 3 \\ 1 & -1 & 1 \\ 3 & 1 & -5 \end{bmatrix}$$ Solve: $$-5x + y + 3z = 0$$ $$x - y + z = 0$$ $$3x + y - 5z = 0$$ From second: $$y = x + z$$ Substitute into first: $$-5x + (x + z) + 3z = 0 \implies -5x + x + z + 3z = 0 \implies -4x + 4z = 0 \implies z = x$$ Then, $$y = x + z = x + x = 2x$$ Eigenvector: $$\mathbf{v}_2 = x \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$$ 15. **Eigenvector for $$\lambda=-2$$:** $$A + 2I = \begin{bmatrix} 3 & 1 & 3 \\ 1 & 7 & 1 \\ 3 & 1 & 3 \end{bmatrix}$$ Solve: $$3x + y + 3z = 0$$ $$x + 7y + z = 0$$ $$3x + y + 3z = 0$$ First and third are the same. From first: $$y = -3x - 3z$$ Substitute into second: $$x + 7(-3x - 3z) + z = 0 \implies x - 21x - 21z + z = 0 \implies -20x - 20z = 0 \implies x = -z$$ Then, $$y = -3(-z) - 3z = 3z - 3z = 0$$ Eigenvector: $$\mathbf{v}_3 = z \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$ 16. **Step 3: Form the modal matrix $$P$$ and diagonal matrix $$D$$:** $$P = \begin{bmatrix} 1 & 1 & -1 \\ -1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ 17. **Step 4: Verify diagonalization:** $$A = P D P^{-1}$$ **Final answer:** The modal matrix is $$P = \begin{bmatrix} 1 & 1 & -1 \\ -1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}$$ and the diagonal matrix is $$D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$.