Let's solve **Problem 4** step by step: Given matrix: \( A = \begin{bmatrix} 10 & -11 \\ -30 & 41 \end{bmatrix} \) --- ### Part (a): Find all eigenvalues of \(A\) The eigenvalues \(\lambda\) of \(A\) satisfy \[ \det(A - \lambda I) = 0 \] Let's compute \(A - \lambda I\): \[ A - \lambda I = \begin{bmatrix} 10-\lambda & -11 \\ -30 & 41-\lambda \end{bmatrix} \] So, \[ \det(A - \lambda I) = (10-\lambda)(41-\lambda) - (-11)(-30) \] \[ = (10-\lambda)(41-\lambda) - 330 \] Now expand \((10-\lambda)(41-\lambda)\): \[ = 10 \times 41 - 10\lambda - 41\lambda + \lambda^2 - 330 \] \[ = 410 - 10\lambda - 41\lambda + \lambda^2 - 330 \] \[ = \lambda^2 - 51\lambda + 410 - 330 \] \[ = \lambda^2 - 51\lambda + 80 \] Set this equal to zero: \[ \lambda^2 - 51\lambda + 80 = 0 \] Solve this quadratic equation for \(\lambda\): Use the quadratic formula: \[ \lambda = \frac{51 \pm \sqrt{(-51)^2 - 4 \cdot 1 \cdot 80}}{2} \] \[ = \frac{51 \pm \sqrt{2601 - 320}}{2} \] \[ = \frac{51 \pm \sqrt{2281}}{2} \] \(\sqrt{2281}\) is not a perfect square, but let's see if it can be simplified: - \(47^2 = 2209\), - \(48^2 = 2304\). So \(\sqrt{2281}\) is between 47 and 48. #### **Final Answer (a):** **Eigenvalues of \(A\):** \[ \boxed{ \lambda_{1,2} = \frac{51 \pm \sqrt{2281}}{2} } \] --- ### Part (b): For each eigenvalue, find the set of eigenvectors Let’s denote a generic eigenvalue as \(\lambda\). The eigenvector \(\mathbf{v}\) satisfies \[ (A - \lambda I)\mathbf{v} = 0 \] That is, \[ \begin{bmatrix} 10-\lambda & -11 \\ -30 & 41-\lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This gives two equations: 1. \((10-\lambda)x - 11y = 0\) 2. \(-30x + (41-\lambda)y = 0\) Let’s solve the first equation for \(x\): \[ (10-\lambda)x - 11y = 0 \implies x = \frac{11}{10-\lambda} y \] So every eigenvector is of the form \[ \mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix} = y \begin{bmatrix} \dfrac{11}{10-\lambda} \\ 1 \end{bmatrix} \] So for each eigenvalue \(\lambda\), the eigenvectors are all scalar multiples of \[ \boxed{ \mathbf{v}_{\lambda} = \begin{bmatrix} \frac{11}{10-\lambda} \\ 1 \end{bmatrix} } \] where \(\lambda = \frac{51 \pm \sqrt{2281}}{2}\). **Alternatively, you can use the second equation to get the same result.** --- ### Part (c): Diagonalization A matrix is diagonalizable if it has enough linearly independent eigenvectors (here, 2 for a \(2 \times 2\) matrix). Since our characteristic polynomial has two distinct roots, \(A\) is diagonalizable. #### Let's construct the diagonalization: - Let \(\lambda_1 = \frac{51 + \sqrt{2281}}{2}\) - Let \(\lambda_2 = \frac{51 - \sqrt{2281}}{2}\) The corresponding eigenvectors are: - For \(\lambda_1\): \[ \mathbf{v}_1 = \begin{bmatrix} \dfrac{11}{10 - \lambda_1} \\ 1 \end{bmatrix} \] - For \(\lambda_2\): \[ \mathbf{v}_2 = \begin{bmatrix} \dfrac{11}{10 - \lambda_2} \\ 1 \end{bmatrix} \] Let \(P = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{bmatrix}\) be the matrix with eigenvectors as columns. Then, \[ D = P^{-1}AP = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \] #### **Final Answer (c):** \[ \boxed{ A \text{ is diagonalizable.} } \] \[ \boxed{ P = \begin{bmatrix} \dfrac{11}{10-\lambda_1} & \dfrac{11}{10-\lambda_2} \\ 1 & 1 \end{bmatrix} } \] where \(\lambda_1 = \frac{51 + \sqrt{2281}}{2}\), \(\lambda_2 = \frac{51 - \sqrt{2281}}{2}\). \[ \boxed{ D = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} } \] --- ### **Summary Table** | Part | Answer | |------|--------| | (a) | \(\lambda_{1,2} = \dfrac{51 \pm \sqrt{2281}}{2}\) | | (b) | Eigenvectors: \(\mathbf{v}_\lambda = \begin{bmatrix} \dfrac{11}{10-\lambda} \\ 1 \end{bmatrix}\) | | (c) | Diagonalizable. \(P = \begin{bmatrix} \dfrac{11}{10-\lambda_1} & \dfrac{11}{10-\lambda_2} \\ 1 & 1 \end{bmatrix}\), \(D = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}\) | If you want explicit decimal values, let me know!