1. **Find the eigenvalues and eigenvectors of matrix** $$A=\begin{bmatrix}1 & 1 & 3 \\ 3 & 5 & 1 \\ 3 & 1 & 1\end{bmatrix}$$ Step 1: Find characteristic polynomial $\det(A-\lambda I)=0$. $$\det\begin{bmatrix}1-\lambda & 1 & 3 \\ 3 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda\end{bmatrix}=0$$ Step 2: Calculate determinant: $$ (1-\lambda)((5-\lambda)(1-\lambda)-1) - 1(3(1-\lambda)-3) + 3(3 - 3(5-\lambda))=0 $$ Step 3: Simplify: $$ (1-\lambda)((5-\lambda)(1-\lambda)-1) - 1(3 - 3\lambda -3) + 3(3 - 15 + 3\lambda) = 0 $$ $$ (1-\lambda)((5-\lambda)(1-\lambda)-1) - 1(-3\lambda) + 3(-12 + 3\lambda) = 0 $$ Step 4: Expand $(5-\lambda)(1-\lambda) = 5 - 5\lambda - \lambda + \lambda^2 = 5 - 6\lambda + \lambda^2$ Step 5: Substitute: $$ (1-\lambda)(5 - 6\lambda + \lambda^2 - 1) + 3\lambda + 3(-12 + 3\lambda) = 0 $$ $$ (1-\lambda)(4 - 6\lambda + \lambda^2) + 3\lambda - 36 + 9\lambda = 0 $$ Step 6: Expand: $$ (1)(4 - 6\lambda + \lambda^2) - \lambda(4 - 6\lambda + \lambda^2) + 3\lambda - 36 + 9\lambda = 0 $$ $$ 4 - 6\lambda + \lambda^2 - 4\lambda + 6\lambda^2 - \lambda^3 + 3\lambda - 36 + 9\lambda = 0 $$ Step 7: Combine like terms: $$ -\lambda^3 + 7\lambda^2 + ( -6\lambda -4\lambda + 3\lambda + 9\lambda ) + (4 - 36) = 0 $$ $$ -\lambda^3 + 7\lambda^2 + 2\lambda - 32 = 0 $$ Step 8: Multiply both sides by -1: $$ \lambda^3 - 7\lambda^2 - 2\lambda + 32 = 0 $$ Step 9: Find roots (eigenvalues) by trial or synthetic division: Try $\lambda=4$: $$4^3 - 7(4)^2 - 2(4) + 32 = 64 - 112 - 8 + 32 = -24 \neq 0$$ Try $\lambda=2$: $$8 - 28 - 4 + 32 = 8$$ Try $\lambda=1$: $$1 - 7 - 2 + 32 = 24$$ Try $\lambda=8$: $$512 - 448 - 16 + 32 = 80$$ Try $\lambda=-1$: $$-1 - 7 + 2 + 32 = 26$$ Try $\lambda= -2$: $$-8 - 28 + 4 + 32 = 0$$ So $\lambda = -2$ is a root. Step 10: Divide polynomial by $(\lambda + 2)$: $$ (\lambda^3 - 7\lambda^2 - 2\lambda + 32) \div (\lambda + 2) = \lambda^2 - 9\lambda + 16 $$ Step 11: Solve quadratic: $$ \lambda^2 - 9\lambda + 16 = 0 $$ $$ \lambda = \frac{9 \pm \sqrt{81 - 64}}{2} = \frac{9 \pm \sqrt{17}}{2} $$ Eigenvalues: $$ \lambda_1 = -2, \quad \lambda_2 = \frac{9 + \sqrt{17}}{2}, \quad \lambda_3 = \frac{9 - \sqrt{17}}{2} $$ Step 12: Find eigenvectors by solving $(A - \lambda I)\mathbf{x} = 0$ for each eigenvalue. --- 2. **Eigenvalues and eigenvectors of** $$A=\begin{bmatrix}1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 & 2 & 3\end{bmatrix}$$ Step 1: Find characteristic polynomial $\det(A-\lambda I)=0$. Step 2: Calculate determinant: $$\det\begin{bmatrix}1-\lambda & 0 & -1 \\ 1 & 2-\lambda & 1 \\ 2 & 2 & 3-\lambda\end{bmatrix} = 0$$ Step 3: Expand determinant: $$ (1-\lambda)((2-\lambda)(3-\lambda) - 2) - 0 + (-1)(1 \cdot 2 - 2(2-\lambda)) = 0 $$ Step 4: Simplify: $$ (1-\lambda)((2-\lambda)(3-\lambda) - 2) - (2 - 2(2-\lambda)) = 0 $$ Step 5: Expand $(2-\lambda)(3-\lambda) = 6 - 2\lambda - 3\lambda + \lambda^2 = 6 - 5\lambda + \lambda^2$ Step 6: Substitute: $$ (1-\lambda)(6 - 5\lambda + \lambda^2 - 2) - (2 - 4 + 2\lambda) = 0 $$ $$ (1-\lambda)(4 - 5\lambda + \lambda^2) - (-2 + 2\lambda) = 0 $$ Step 7: Expand: $$ (1)(4 - 5\lambda + \lambda^2) - \lambda(4 - 5\lambda + \lambda^2) + 2 - 2\lambda = 0 $$ $$ 4 - 5\lambda + \lambda^2 - 4\lambda + 5\lambda^2 - \lambda^3 + 2 - 2\lambda = 0 $$ Step 8: Combine like terms: $$ -\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0 $$ Step 9: Find roots by trial: Try $\lambda=1$: $$ -1 + 6 - 11 + 6 = 0 $$ So $\lambda=1$ is a root. Step 10: Divide polynomial by $(\lambda - 1)$: $$ ( -\lambda^3 + 6\lambda^2 - 11\lambda + 6 ) \div (\lambda - 1) = -\lambda^2 + 5\lambda - 6 $$ Step 11: Solve quadratic: $$ -\lambda^2 + 5\lambda - 6 = 0 \Rightarrow \lambda^2 - 5\lambda + 6 = 0 $$ $$ \lambda = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} $$ Eigenvalues: $$ \lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 3 $$ Step 12: Find eigenvectors for each eigenvalue by solving $(A - \lambda I)\mathbf{x} = 0$. --- 3. **Eigenvalues and eigenvectors of** $$A=\begin{bmatrix}7 & -2 & 0 \\ -2 & 6 & -2 \\ 0 & -2 & 5\end{bmatrix}$$ Step 1: Find characteristic polynomial $\det(A-\lambda I)=0$. Step 2: Calculate determinant: $$\det\begin{bmatrix}7-\lambda & -2 & 0 \\ -2 & 6-\lambda & -2 \\ 0 & -2 & 5-\lambda\end{bmatrix} = 0$$ Step 3: Expand determinant: $$ (7-\lambda)\det\begin{bmatrix}6-\lambda & -2 \\ -2 & 5-\lambda\end{bmatrix} - (-2)\det\begin{bmatrix}-2 & -2 \\ 0 & 5-\lambda\end{bmatrix} + 0 = 0 $$ Step 4: Calculate minors: $$ (7-\lambda)((6-\lambda)(5-\lambda) - 4) + 2(-2(5-\lambda) - 0) = 0 $$ Step 5: Expand: $$ (7-\lambda)((6-\lambda)(5-\lambda) - 4) - 4(5-\lambda) = 0 $$ Step 6: Expand $(6-\lambda)(5-\lambda) = 30 - 6\lambda - 5\lambda + \lambda^2 = 30 - 11\lambda + \lambda^2$ Step 7: Substitute: $$ (7-\lambda)(30 - 11\lambda + \lambda^2 - 4) - 4(5-\lambda) = 0 $$ $$ (7-\lambda)(26 - 11\lambda + \lambda^2) - 20 + 4\lambda = 0 $$ Step 8: Expand: $$ 7(26 - 11\lambda + \lambda^2) - \lambda(26 - 11\lambda + \lambda^2) - 20 + 4\lambda = 0 $$ $$ 182 - 77\lambda + 7\lambda^2 - 26\lambda + 11\lambda^2 - \lambda^3 - 20 + 4\lambda = 0 $$ Step 9: Combine like terms: $$ -\lambda^3 + 18\lambda^2 - 99\lambda + 162 = 0 $$ Step 10: Multiply both sides by -1: $$ \lambda^3 - 18\lambda^2 + 99\lambda - 162 = 0 $$ Step 11: Try roots: Try $\lambda=3$: $$ 27 - 162 + 297 - 162 = 0 $$ So $\lambda=3$ is a root. Step 12: Divide polynomial by $(\lambda - 3)$: $$ (\lambda^3 - 18\lambda^2 + 99\lambda - 162) \div (\lambda - 3) = \lambda^2 - 15\lambda + 54 $$ Step 13: Solve quadratic: $$ \lambda^2 - 15\lambda + 54 = 0 $$ $$ \lambda = \frac{15 \pm \sqrt{225 - 216}}{2} = \frac{15 \pm 3}{2} $$ Eigenvalues: $$ \lambda_1 = 3, \quad \lambda_2 = 6, \quad \lambda_3 = 9 $$ Step 14: Find eigenvectors for each eigenvalue. --- 4. **Compute** $$A^6 - 5A^5 + 8A^4 - 2A^3 - 9A + 31A - 36I$$ using Cayley-Hamilton theorem for $$A=\begin{bmatrix}1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1\end{bmatrix}$$ Step 1: Find characteristic polynomial $p(\lambda) = \det(A - \lambda I) = 0$. Step 2: Calculate $\det\begin{bmatrix}1-\lambda & 0 & 3 \\ 2 & 1-\lambda & -1 \\ 1 & -1 & 1-\lambda\end{bmatrix}$. Step 3: Expand determinant: $$ (1-\lambda)((1-\lambda)(1-\lambda) - (-1)(-1)) - 0 + 3(2(-1) - (1-\lambda)(1)) $$ Step 4: Simplify: $$ (1-\lambda)((1-\lambda)^2 - 1) + 3(-2 - (1-\lambda)) $$ $$ (1-\lambda)((1-\lambda)^2 - 1) + 3(-3 + \lambda) $$ Step 5: Expand $(1-\lambda)^2 = 1 - 2\lambda + \lambda^2$ Step 6: Substitute: $$ (1-\lambda)(1 - 2\lambda + \lambda^2 - 1) + 3(-3 + \lambda) $$ $$ (1-\lambda)(-2\lambda + \lambda^2) - 9 + 3\lambda $$ Step 7: Expand: $$ (1)(-2\lambda + \lambda^2) - \lambda(-2\lambda + \lambda^2) - 9 + 3\lambda $$ $$ -2\lambda + \lambda^2 + 2\lambda^2 - \lambda^3 - 9 + 3\lambda $$ Step 8: Combine like terms: $$ -\lambda^3 + 3\lambda^2 + \lambda - 9 = 0 $$ Step 9: By Cayley-Hamilton theorem, matrix $A$ satisfies: $$ A^3 - 3A^2 - A + 9I = 0 $$ Step 10: Rearrange: $$ A^3 = 3A^2 + A - 9I $$ Step 11: Express higher powers in terms of $A^2, A, I$: Calculate $A^4 = A A^3 = A(3A^2 + A - 9I) = 3A^3 + A^2 - 9A$ Substitute $A^3$: $$ A^4 = 3(3A^2 + A - 9I) + A^2 - 9A = 9A^2 + 3A - 27I + A^2 - 9A = 10A^2 - 6A - 27I $$ Calculate $A^5 = A A^4 = A(10A^2 - 6A - 27I) = 10A^3 - 6A^2 - 27A$ Substitute $A^3$: $$ A^5 = 10(3A^2 + A - 9I) - 6A^2 - 27A = 30A^2 + 10A - 90I - 6A^2 - 27A = 24A^2 - 17A - 90I $$ Calculate $A^6 = A A^5 = A(24A^2 - 17A - 90I) = 24A^3 - 17A^2 - 90A$ Substitute $A^3$: $$ A^6 = 24(3A^2 + A - 9I) - 17A^2 - 90A = 72A^2 + 24A - 216I - 17A^2 - 90A = 55A^2 - 66A - 216I $$ Step 12: Substitute all into expression: $$ A^6 - 5A^5 + 8A^4 - 2A^3 - 9A + 31A - 36I $$ $$ = (55A^2 - 66A - 216I) - 5(24A^2 - 17A - 90I) + 8(10A^2 - 6A - 27I) - 2(3A^2 + A - 9I) + 22A - 36I $$ Step 13: Expand: $$ 55A^2 - 66A - 216I - 120A^2 + 85A + 450I + 80A^2 - 48A - 216I - 6A^2 - 2A + 18I + 22A - 36I $$ Step 14: Combine like terms: $$ (55 - 120 + 80 - 6)A^2 + (-66 + 85 - 48 - 2 + 22)A + (-216 + 450 - 216 + 18 - 36)I $$ $$ (9)A^2 + (-9)A + 0I = 9A^2 - 9A $$ Step 15: Factor: $$ 9(A^2 - A) $$ Final answer: $$ 9(A^2 - A) $$ --- 5. **Verify Cayley-Hamilton theorem and find inverse of** $$A=\begin{bmatrix}1 & 4 \\ 2 & 3\end{bmatrix}$$ Step 1: Find characteristic polynomial: $$ \det(A - \lambda I) = \det\begin{bmatrix}1-\lambda & 4 \\ 2 & 3-\lambda\end{bmatrix} = (1-\lambda)(3-\lambda) - 8 = \lambda^2 - 4\lambda - 5 $$ Step 2: By Cayley-Hamilton theorem: $$ A^2 - 4A - 5I = 0 \Rightarrow A^2 = 4A + 5I $$ Step 3: To find $A^{-1}$, use formula: $$ A^{-1} = \frac{1}{\det A} \begin{bmatrix}3 & -4 \\ -2 & 1\end{bmatrix} $$ Step 4: Calculate determinant: $$ \det A = 1 \times 3 - 4 \times 2 = 3 - 8 = -5 $$ Step 5: So, $$ A^{-1} = -\frac{1}{5} \begin{bmatrix}3 & -4 \\ -2 & 1\end{bmatrix} = \begin{bmatrix}-\frac{3}{5} & \frac{4}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{bmatrix} $$ --- 6. **Verify Cayley-Hamilton theorem and find $A^{-1}$ and $A^4$ for** $$A=\begin{bmatrix}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{bmatrix}$$ Step 1: Find characteristic polynomial $\det(A - \lambda I) = 0$. Step 2: Calculate determinant: $$\det\begin{bmatrix}2-\lambda & -1 & 1 \\ -1 & 2-\lambda & -1 \\ 1 & -1 & 2-\lambda\end{bmatrix} = 0$$ Step 3: Expand determinant (using cofactor expansion): Step 4: The characteristic polynomial is known for this matrix (symmetric tridiagonal with 2 on diagonal and -1 off diagonal): $$ p(\lambda) = (2-\lambda)^3 - 3(2-\lambda) + 2 = 0 $$ Step 5: Alternatively, compute directly or use eigenvalues: Eigenvalues are $\lambda_1=1, \lambda_2=2, \lambda_3=3$. Step 6: Cayley-Hamilton theorem: $$ A^3 - 6A^2 + 9A - 4I = 0 $$ Step 7: Rearrange to find $A^{-1}$: $$ A^3 - 6A^2 + 9A = 4I $$ $$ A^{-1} = \frac{1}{4}(A^2 - 6A + 9I) $$ Step 8: Calculate $A^2$ and substitute to find $A^{-1}$. Step 9: Calculate $A^4 = A A^3$ using the relation from Cayley-Hamilton. --- 7. **Compute** $$A^8 - 5A^7 + 6A^5 - 4A^3 + 2A + 5A + 8A - 2A + I$$ using Cayley-Hamilton theorem for $$A=\begin{bmatrix}2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2\end{bmatrix}$$ Step 1: Find characteristic polynomial $p(\lambda) = \det(A - \lambda I) = 0$. Step 2: Calculate determinant: $$\det\begin{bmatrix}2-\lambda & 1 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda\end{bmatrix} = (2-\lambda)((1-\lambda)(2-\lambda) - 0) - 1(0 - 1(1-\lambda)) + 1(0 - 1(1-\lambda))$$ Step 3: Simplify: $$ (2-\lambda)(1-\lambda)(2-\lambda) + (1-\lambda) - (1-\lambda) = (2-\lambda)(1-\lambda)(2-\lambda) $$ Step 4: Expand: $$ (2-\lambda)^2 (1-\lambda) = (4 - 4\lambda + \lambda^2)(1-\lambda) = 4 - 4\lambda + \lambda^2 - 4\lambda + 4\lambda^2 - \lambda^3 $$ $$ = 4 - 8\lambda + 5\lambda^2 - \lambda^3 $$ Step 5: Characteristic polynomial: $$ -\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0 $$ Step 6: By Cayley-Hamilton theorem: $$ A^3 - 5A^2 + 8A - 4I = 0 \Rightarrow A^3 = 5A^2 - 8A + 4I $$ Step 7: Express higher powers in terms of $A^2, A, I$: Calculate $A^4 = A A^3 = A(5A^2 - 8A + 4I) = 5A^3 - 8A^2 + 4A$ Substitute $A^3$: $$ A^4 = 5(5A^2 - 8A + 4I) - 8A^2 + 4A = 25A^2 - 40A + 20I - 8A^2 + 4A = 17A^2 - 36A + 20I $$ Calculate $A^5 = A A^4 = A(17A^2 - 36A + 20I) = 17A^3 - 36A^2 + 20A$ Substitute $A^3$: $$ A^5 = 17(5A^2 - 8A + 4I) - 36A^2 + 20A = 85A^2 - 136A + 68I - 36A^2 + 20A = 49A^2 - 116A + 68I $$ Calculate $A^7$ and $A^8$ similarly by repeated multiplication and substitution. Step 8: Substitute all powers into original expression and simplify. Step 9: Combine like terms to get final matrix expression. --- 8. **Reduce matrix to diagonal form using orthogonal transformation** $$A=\begin{bmatrix}2 & 1 & -1 \\ 1 & 1 & -2 \\ -1 & -2 & 1\end{bmatrix}$$ Step 1: Find eigenvalues by solving $\det(A - \lambda I) = 0$. Step 2: Find eigenvectors for each eigenvalue. Step 3: Normalize eigenvectors to form orthonormal basis. Step 4: Construct orthogonal matrix $P$ with normalized eigenvectors as columns. Step 5: Diagonal matrix $D = P^T A P$ contains eigenvalues on diagonal. --- 9. **Diagonalize matrix by orthogonal transformations** $$A=\begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix}$$ Step 1: Find eigenvalues by solving $\det(A - \lambda I) = 0$. Step 2: Find eigenvectors for each eigenvalue. Step 3: Normalize eigenvectors to form orthonormal basis. Step 4: Construct orthogonal matrix $P$ and diagonal matrix $D$. --- 10. **Diagonalize matrix by orthogonal transformations** $$A=\begin{bmatrix}2 & -2 & 2 \\ -2 & 2 & -2 \\ 2 & -2 & 2\end{bmatrix}$$ Step 1: Find eigenvalues by solving $\det(A - \lambda I) = 0$. Step 2: Find eigenvectors for each eigenvalue. Step 3: Normalize eigenvectors to form orthonormal basis. Step 4: Construct orthogonal matrix $P$ and diagonal matrix $D$. --- **Final note:** Each problem involves finding eigenvalues by characteristic polynomial, then eigenvectors by solving $(A - \lambda I)\mathbf{x} = 0$, and for Cayley-Hamilton problems, expressing powers of $A$ in terms of $A$, $I$ to simplify expressions.