1. **State the problem:** Find the eigenvalues of the matrix $$A = \begin{pmatrix} 2 & 4 \\ 1 & 5 \end{pmatrix}$$. 2. **Formula and explanation:** Eigenvalues $$\lambda$$ satisfy the characteristic equation $$\det(A - \lambda I) = 0$$, where $$I$$ is the identity matrix. 3. **Set up the characteristic matrix:** $$A - \lambda I = \begin{pmatrix} 2 - \lambda & 4 \\ 1 & 5 - \lambda \end{pmatrix}$$ 4. **Calculate the determinant:** $$\det(A - \lambda I) = (2 - \lambda)(5 - \lambda) - 4 \times 1$$ 5. **Expand the determinant:** $$= (2 - \lambda)(5 - \lambda) - 4 = 10 - 2\lambda - 5\lambda + \lambda^2 - 4 = \lambda^2 - 7\lambda + 6$$ 6. **Solve the quadratic equation:** $$\lambda^2 - 7\lambda + 6 = 0$$ 7. **Factor the quadratic:** $$\lambda^2 - 7\lambda + 6 = (\lambda - 6)(\lambda - 1) = 0$$ 8. **Find eigenvalues:** $$\lambda = 6 \quad \text{or} \quad \lambda = 1$$ **Final answer:** The eigenvalues of matrix $$A$$ are $$6$$ and $$1$$.