1. **Problem:** Find the eigenvalues of matrix $$A = \begin{bmatrix}1 & 2 \\ 5 & 4\end{bmatrix}$$. 2. **Formula:** Eigenvalues $$\lambda$$ satisfy $$\det(A - \lambda I) = 0$$ where $$I$$ is the identity matrix. 3. **Work:** $$A - \lambda I = \begin{bmatrix}1-\lambda & 2 \\ 5 & 4-\lambda\end{bmatrix}$$ Calculate determinant: $$\det(A - \lambda I) = (1-\lambda)(4-\lambda) - (2)(5) = (1-\lambda)(4-\lambda) - 10$$ Expand: $$= 4 - \lambda - 4\lambda + \lambda^2 - 10 = \lambda^2 - 5\lambda - 6$$ 4. **Solve characteristic equation:** $$\lambda^2 - 5\lambda - 6 = 0$$ Factor or use quadratic formula: $$\lambda = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm \sqrt{49}}{2} = \frac{5 \pm 7}{2}$$ 5. **Eigenvalues:** $$\lambda_1 = \frac{5 + 7}{2} = 6$$ $$\lambda_2 = \frac{5 - 7}{2} = -1$$ **Final answer:** The eigenvalues of matrix $$A$$ are $$6$$ and $$-1$$.