Find the eigenualue and eigen uectors of \( A=\left[\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right] \).

To find the eigenvalues of the matrix \( A=\left[\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right] \), we start by calculating the characteristic polynomial. This involves finding the determinant of \( A - \lambda I \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. 1. Set up the equation: \[ A - \lambda I = \left[\begin{array}{cc}1 - \lambda & -2 \\ 1 & 3 - \lambda\end{array}\right]. \] 2. Calculate the determinant: \[ \det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (-2)(1) = \lambda^2 - 4\lambda + 5. \] 3. Set the determinant to zero and solve for \( \lambda \): \[ \lambda^2 - 4\lambda + 5 = 0. \] Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i. \] The eigenvalues are \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \). Next, let's find the eigenvectors corresponding to each eigenvalue: For \( \lambda_1 = 2 + i \): 1. Substitute \( \lambda_1 \) into \( A - \lambda I \): \[ A - (2 + i)I = \left[\begin{array}{cc}-1 - i & -2 \\ 1 & 1 - i\end{array}\right]. \] 2. Solve the system of equations: From the first row: \((-1 - i)x_1 - 2x_2 = 0\) Thus, \(x_1 = \frac{2x_2}{-1 - i}\). For simplicity, let's set \( x_2 = 1 \): \[ x_1 = \frac{2}{-1 - i} = \frac{2(-1 + i)}{(-1)^2 + 1^2} = -1 + i. \] So, an eigenvector corresponding to \( \lambda_1 \) is: \[ v_1 = \begin{pmatrix}-1 + i \\ 1\end{pmatrix}. \] For \( \lambda_2 = 2 - i \): 1. Substitute \( \lambda_2 \) into \( A - \lambda I \): \[ A - (2 - i)I = \left[\begin{array}{cc}-1 + i & -2 \\ 1 & 1 + i\end{array}\right]. \] 2. Solve similarly to get the eigenvector: Using \( x_2 = 1 \): \[ x_1 = \frac{2}{-1 + i} = \frac{2(-1 - i)}{(-1)^2 + 1^2} = -1 - i. \] So, an eigenvector corresponding to \( \lambda_2 \) is: \[ v_2 = \begin{pmatrix}-1 - i \\ 1\end{pmatrix}. \] Thus, the eigenvalues are \( 2 + i \) and \( 2 - i \), with corresponding eigenvectors \( \begin{pmatrix}-1 + i \\ 1\end{pmatrix} \) and \( \begin{pmatrix}-1 - i \\ 1\end{pmatrix} \), respectively.