Let's solve **part (a)** step by step. Matrix: \[ A = \begin{bmatrix} 2 & -3 & 6 \\ 0 & 3 & -4 \\ 0 & 2 & -3 \end{bmatrix} \] --- ### **Step 1: Find the Eigenvalues** Find the characteristic equation: \[ \det(A - \lambda I) = 0 \] So, \[ \begin{vmatrix} 2 - \lambda & -3 & 6 \\ 0 & 3 - \lambda & -4 \\ 0 & 2 & -3 - \lambda \end{vmatrix} = 0 \] Expand the determinant: \[ (2-\lambda)\begin{vmatrix}3-\lambda & -4 \\ 2 & -3-\lambda\end{vmatrix} - 0 + 0 \] \[ = (2-\lambda)\left[(3-\lambda)(-3-\lambda) - (-4)(2)\right] \] \[ = (2-\lambda)\left[-(3-\lambda)(3+\lambda) + 8\right] \] \[ = (2-\lambda)\left[-(9 - \lambda^2) + 8\right] \] \[ = (2-\lambda)\left[-9 + \lambda^2 + 8\right] \] \[ = (2-\lambda)(\lambda^2 - 1) \] Set to zero: \[ (2-\lambda)(\lambda^2 - 1) = 0 \] \[ \therefore \lambda = 2, \; 1, \; -1 \] **Explanation:** We calculated the determinant of \(A - \lambda I\) to obtain the characteristic polynomial and solved for the eigenvalues. --- ### **Step 2: Find the Eigenvectors** #### For \(\lambda_1 = 2\): Solve \((A - 2I)\vec{x} = 0\): \[ A - 2I = \begin{bmatrix} 0 & -3 & 6 \\ 0 & 1 & -4 \\ 0 & 2 & -5 \end{bmatrix} \] Set up equations: - Row 1: \(-3x_2 + 6x_3 = 0\) - Row 2: \(x_2 - 4x_3 = 0\) - Row 3: \(2x_2 - 5x_3 = 0\) From Row 2: \(x_2 = 4x_3\) Plug into Row 1: \(-3(4x_3) + 6x_3 = -12x_3 + 6x_3 = -6x_3 = 0 \implies x_3 = 0\) So, \(x_2 = 0\). \(x_1\) is free. Thus, eigenvector is any scalar multiple of \(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\). --- #### For \(\lambda_2 = 1\): \[ A - I = \begin{bmatrix} 1 & -3 & 6 \\ 0 & 2 & -4 \\ 0 & 2 & -4 \end{bmatrix} \] Rows 2 and 3 are the same, so focus on Rows 1 and 2: - Row 1: \(x_1 - 3x_2 + 6x_3 = 0\) - Row 2: \(2x_2 - 4x_3 = 0 \implies x_2 = 2x_3\) Substitute into Row 1: \(x_1 - 3(2x_3) + 6x_3 = x_1 - 6x_3 + 6x_3 = x_1 = 0\) So \(x_1 = 0, x_2 = 2x_3, x_3\) free. Eigenvector: any scalar multiple of \(\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}\). --- #### For \(\lambda_3 = -1\): \[ A + I = \begin{bmatrix} 3 & -3 & 6 \\ 0 & 4 & -4 \\ 0 & 2 & -2 \end{bmatrix} \] Row 2: \(4x_2 - 4x_3 = 0 \implies x_2 = x_3\) Row 1: \(3x_1 - 3x_2 + 6x_3 = 0 \implies 3x_1 - 3x_2 + 6x_3 = 0 \implies 3x_1 + 3x_3 = 0 \implies x_1 = -x_3\) So eigenvector: any multiple of \(\begin{bmatrix}-1 \\ 1 \\ 1\end{bmatrix}\). **Explanation:** We solved \((A-\lambda I)\vec{x} = 0\) for each eigenvalue to obtain the corresponding eigenvectors. --- ### **Step 3: Check Linear Independence** The three eigenvectors are: - \(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\) - \(\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}\) - \(\begin{bmatrix}-1 \\ 1 \\ 1\end{bmatrix}\) You can verify their linear independence by checking if the determinant of the matrix formed by placing them as columns is non-zero. \[ \begin{vmatrix} 1 & 0 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{vmatrix} \] Expand: \[ 1 \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} - 0 + 0 = 1(2\cdot1 - 1\cdot1) = 1(2 - 1) = 1 \] Since the determinant is non-zero, the eigenvectors are linearly independent. **Explanation:** We checked the determinant of the matrix formed by eigenvectors and found it to be non-zero, which means they are linearly independent. --- ## **Final Answer (for part a):** **Eigenvalues:** \(\boxed{2, 1, -1}\) **Associated eigenvectors:** - For \(\lambda = 2\): \(\boxed{\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}}\) - For \(\lambda = 1\): \(\boxed{\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}}\) - For \(\lambda = -1\): \(\boxed{\begin{bmatrix}-1 \\ 1 \\ 1\end{bmatrix}}\) **There is a set of 3 linearly independent eigenvectors.** --- Would you like to see the solution for (b) or (c) as well?