1. **State the problem:** Find the eigenvalues and eigenvectors of the matrix $$A = \begin{bmatrix} 14 & -10 \\ 5 & -1 \end{bmatrix}$$. 2. **Eigenvalues formula:** Eigenvalues $\lambda$ satisfy the characteristic equation $$\det(A - \lambda I) = 0,$$ where $I$ is the identity matrix. 3. **Calculate $A - \lambda I$:** $$A - \lambda I = \begin{bmatrix} 14 - \lambda & -10 \\ 5 & -1 - \lambda \end{bmatrix}$$ 4. **Find the determinant:** $$\det(A - \lambda I) = (14 - \lambda)(-1 - \lambda) - (-10)(5)$$ $$= (14 - \lambda)(-1 - \lambda) + 50$$ 5. **Expand the determinant:** $$= (14)(-1) + 14(-\lambda) - \lambda(-1) - \lambda^2 + 50$$ $$= -14 - 14\lambda + \lambda - \lambda^2 + 50$$ $$= -\lambda^2 - 13\lambda + 36$$ 6. **Set determinant to zero:** $$-\lambda^2 - 13\lambda + 36 = 0$$ Multiply both sides by $-1$: $$\lambda^2 + 13\lambda - 36 = 0$$ 7. **Solve quadratic equation:** Using the quadratic formula $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a=1$, $b=13$, $c=-36$: $$\lambda = \frac{-13 \pm \sqrt{13^2 - 4(1)(-36)}}{2} = \frac{-13 \pm \sqrt{169 + 144}}{2} = \frac{-13 \pm \sqrt{313}}{2}$$ 8. **Eigenvalues:** $$\lambda_1 = \frac{-13 + \sqrt{313}}{2}, \quad \lambda_2 = \frac{-13 - \sqrt{313}}{2}$$ 9. **Find eigenvectors:** For each eigenvalue $\lambda$, solve $(A - \lambda I)\mathbf{v} = 0$. 10. **Eigenvector for $\lambda_1$:** $$\begin{bmatrix} 14 - \lambda_1 & -10 \\ 5 & -1 - \lambda_1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ From the first row: $$(14 - \lambda_1)x - 10y = 0 \implies y = \frac{14 - \lambda_1}{10} x$$ Choose $x=1$, then $$y = \frac{14 - \lambda_1}{10}$$ 11. **Eigenvector for $\lambda_2$:** Similarly, $$(14 - \lambda_2)x - 10y = 0 \implies y = \frac{14 - \lambda_2}{10} x$$ Choose $x=1$, then $$y = \frac{14 - \lambda_2}{10}$$ **Final answer:** Eigenvalues: $$\lambda_1 = \frac{-13 + \sqrt{313}}{2}, \quad \lambda_2 = \frac{-13 - \sqrt{313}}{2}$$ Eigenvectors: $$\mathbf{v}_1 = \begin{bmatrix} 1 \\ \frac{14 - \lambda_1}{10} \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ \frac{14 - \lambda_2}{10} \end{bmatrix}$$