Sure! Let's tackle part (a) in two steps, as you requested. --- ## (a) Find the characteristic polynomial and eigenvalues of matrix \( A \). ### **Step 1: Find the characteristic polynomial of \( A \).** The characteristic polynomial of a matrix \( A \) is defined as \( \det(A - \lambda I) \). Given: \[ A = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 4 & 1 \\ 1 & -1 & 2 \end{bmatrix} \] Subtract \( \lambda \) from the diagonal entries: \[ A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 1 \\ -1 & 4-\lambda & 1 \\ 1 & -1 & 2-\lambda \end{bmatrix} \] Now, compute the determinant: \[ \begin{align*} \det(A - \lambda I) &= \begin{vmatrix} 2-\lambda & 1 & 1 \\ -1 & 4-\lambda & 1 \\ 1 & -1 & 2-\lambda \end{vmatrix} \\ &= (2-\lambda)\begin{vmatrix}4-\lambda & 1 \\ -1 & 2-\lambda \end{vmatrix} - 1\begin{vmatrix}-1 & 1 \\ 1 & 2-\lambda\end{vmatrix} + 1\begin{vmatrix}-1 & 4-\lambda \\ 1 & -1\end{vmatrix} \end{align*} \] Calculate each minor: - \( (4-\lambda)(2-\lambda) - (1)(-1) = (4-\lambda)(2-\lambda) + 1 \) - \( (-1)(2-\lambda) - (1)(1) = -(2-\lambda) - 1 = -2 + \lambda - 1 = \lambda - 3 \) - \( (-1)(-1) - (4-\lambda)(1) = 1 - (4-\lambda) = \lambda - 3 \) Plug in the values: \[ \begin{align*} \det(A - \lambda I) &= (2-\lambda)[(4-\lambda)(2-\lambda) + 1] - 1[\lambda-3] + 1[\lambda-3] \\ &= (2-\lambda)[(4-\lambda)(2-\lambda) + 1] \\ \end{align*} \] Now, expand \( (4-\lambda)(2-\lambda) = 8 - 4\lambda - 2\lambda + \lambda^2 = 8 - 6\lambda + \lambda^2 \) So, \[ (4-\lambda)(2-\lambda) + 1 = (8 - 6\lambda + \lambda^2) + 1 = 9 - 6\lambda + \lambda^2 \] Now, multiply: \[ \det(A - \lambda I) = (2-\lambda)(9 - 6\lambda + \lambda^2) \] Expand: \[ (2-\lambda)(\lambda^2 - 6\lambda + 9) = 2(\lambda^2 - 6\lambda + 9) - \lambda(\lambda^2 - 6\lambda + 9) \] \[ = 2\lambda^2 - 12\lambda + 18 - (\lambda^3 - 6\lambda^2 + 9\lambda) \] \[ = 2\lambda^2 - 12\lambda + 18 - \lambda^3 + 6\lambda^2 - 9\lambda \] \[ = -\lambda^3 + 8\lambda^2 - 21\lambda + 18 \] **Explanation:** We subtracted \(\lambda\) from the diagonal entries of \(A\), calculated the determinant, and expanded to obtain the characteristic polynomial. **Characteristic Polynomial:** \[ \boxed{-\lambda^3 + 8\lambda^2 - 21\lambda + 18} \] --- ### **Step 2: Find the eigenvalues of \( A \).** The eigenvalues are the roots of the characteristic polynomial: \[ -\lambda^3 + 8\lambda^2 - 21\lambda + 18 = 0 \] Or, \[ \lambda^3 - 8\lambda^2 + 21\lambda - 18 = 0 \] Let’s try rational roots (factors of 18): \( \pm1, \pm2, \pm3, \pm6, \pm9, \pm18 \). Try \(\lambda = 1\): \( 1 - 8 + 21 - 18 = -4 \) (not zero) Try \(\lambda = 2\): \( 8 - 32 + 42 - 18 = (8 - 32) + (42 - 18) = -24 + 24 = 0 \) So, \(\lambda = 2\) is a root. Now, perform synthetic or polynomial division to factor out \((\lambda - 2)\): Divide \( \lambda^3 - 8\lambda^2 + 21\lambda - 18 \) by \( (\lambda - 2) \): \[ \lambda^3 - 8\lambda^2 + 21\lambda - 18 = (\lambda - 2)(\lambda^2 - 6\lambda + 9) \] Factor \( \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 \) So, characteristic equation factors as: \[ (\lambda - 2)(\lambda - 3)^2 = 0 \] So, eigenvalues are: \[ \boxed{2, 3, 3} \] **Explanation:** We solved the cubic polynomial and factored it to find the eigenvalues. --- ## **Final Answers for (a):** - **Characteristic polynomial:** \( \boxed{-\lambda^3 + 8\lambda^2 - 21\lambda + 18} \) - **Eigenvalues:** \( \boxed{2, 3, 3} \) --- Would you like to proceed to (b)?