1. **State the problem:** Find the characteristic polynomial of the matrix $$A = \begin{bmatrix} 4 & 0 & 10 \\ -3 & 1 & -10 \\ 0 & 1 & -2 \end{bmatrix}$$ using the variable $$\lambda$$. 2. **Formula:** The characteristic polynomial $$p(\lambda)$$ of a matrix $$A$$ is given by $$p(\lambda) = \det(A - \lambda I)$$ where $$I$$ is the identity matrix of the same size as $$A$$. 3. **Set up the matrix $$A - \lambda I$$:** $$$ A - \lambda I = \begin{bmatrix} 4-\lambda & 0 & 10 \\ -3 & 1-\lambda & -10 \\ 0 & 1 & -2-\lambda \end{bmatrix} $$$ 4. **Calculate the determinant:** $$$ \det(A - \lambda I) = (4-\lambda) \cdot \det \begin{bmatrix} 1-\lambda & -10 \\ 1 & -2-\lambda \end{bmatrix} - 0 + 10 \cdot \det \begin{bmatrix} -3 & 1-\lambda \\ 0 & 1 \end{bmatrix} $$$ 5. **Calculate the 2x2 determinants:** $$$ \det \begin{bmatrix} 1-\lambda & -10 \\ 1 & -2-\lambda \end{bmatrix} = (1-\lambda)(-2-\lambda) - (-10)(1) = -(2+\lambda - 2\lambda - \lambda^2) + 10 = -(-\lambda^2 - \lambda - 2) + 10 = \lambda^2 + \lambda + 2 + 10 = \lambda^2 + \lambda + 12 $$$ $$$ \det \begin{bmatrix} -3 & 1-\lambda \\ 0 & 1 \end{bmatrix} = (-3)(1) - 0 = -3 $$$ 6. **Substitute back:** $$$ \det(A - \lambda I) = (4-\lambda)(\lambda^2 + \lambda + 12) + 10(-3) = (4-\lambda)(\lambda^2 + \lambda + 12) - 30 $$$ 7. **Expand:** $$$ (4-\lambda)(\lambda^2 + \lambda + 12) = 4\lambda^2 + 4\lambda + 48 - \lambda^3 - \lambda^2 - 12\lambda = -\lambda^3 + (4\lambda^2 - \lambda^2) + (4\lambda - 12\lambda) + 48 = -\lambda^3 + 3\lambda^2 - 8\lambda + 48 $$$ 8. **Combine all terms:** $$$ \det(A - \lambda I) = -\lambda^3 + 3\lambda^2 - 8\lambda + 48 - 30 = -\lambda^3 + 3\lambda^2 - 8\lambda + 18 $$$ 9. **Final answer:** The characteristic polynomial is $$$ p(\lambda) = -\lambda^3 + 3\lambda^2 - 8\lambda + 18 $$$ You can also write it as $$$ p(\lambda) = -(\lambda^3 - 3\lambda^2 + 8\lambda - 18) $$$ This polynomial can be used to find eigenvalues by solving $$p(\lambda) = 0$$.