1. **State the problem:** Find the eigenvectors of the matrix $$A = \begin{pmatrix}1 & 1 \\ 3 & -1\end{pmatrix}$$ using the transformation method. 2. **Find the eigenvalues:** Solve the characteristic equation $$\det(A - \lambda I) = 0$$ where $$I$$ is the identity matrix. $$A - \lambda I = \begin{pmatrix}1 - \lambda & 1 \\ 3 & -1 - \lambda\end{pmatrix}$$ Calculate the determinant: $$\det(A - \lambda I) = (1 - \lambda)(-1 - \lambda) - 3 \times 1 = (1 - \lambda)(-1 - \lambda) - 3$$ Expand: $$= -(1 - \lambda)(1 + \lambda) - 3 = -(1 - \lambda^2) - 3 = -1 + \lambda^2 - 3 = \lambda^2 - 4$$ Set equal to zero: $$\lambda^2 - 4 = 0$$ Solve for $$\lambda$$: $$\lambda = \pm 2$$ 3. **Find eigenvectors for $$\lambda = 2$$:** Solve $$ (A - 2I)\mathbf{x} = 0 $$ $$A - 2I = \begin{pmatrix}1 - 2 & 1 \\ 3 & -1 - 2\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 3 & -3\end{pmatrix}$$ Set up the system: $$-x_1 + x_2 = 0$$ $$3x_1 - 3x_2 = 0$$ From the first equation: $$x_2 = x_1$$ Eigenvector corresponding to $$\lambda = 2$$ is any scalar multiple of: $$\mathbf{v}_1 = \begin{pmatrix}1 \\ 1\end{pmatrix}$$ 4. **Find eigenvectors for $$\lambda = -2$$:** Solve $$ (A + 2I)\mathbf{x} = 0 $$ $$A + 2I = \begin{pmatrix}1 + 2 & 1 \\ 3 & -1 + 2\end{pmatrix} = \begin{pmatrix}3 & 1 \\ 3 & 1\end{pmatrix}$$ Set up the system: $$3x_1 + x_2 = 0$$ $$3x_1 + x_2 = 0$$ From the equation: $$x_2 = -3x_1$$ Eigenvector corresponding to $$\lambda = -2$$ is any scalar multiple of: $$\mathbf{v}_2 = \begin{pmatrix}1 \\ -3\end{pmatrix}$$ **Final answer:** Eigenvalues: $$\lambda_1 = 2$$ with eigenvector $$\mathbf{v}_1 = \begin{pmatrix}1 \\ 1\end{pmatrix}$$ Eigenvalues: $$\lambda_2 = -2$$ with eigenvector $$\mathbf{v}_2 = \begin{pmatrix}1 \\ -3\end{pmatrix}$$