1. **Problem statement:** Find the characteristic polynomial of matrix $ A = \begin{bmatrix}-5 & 10 & -10 \\ -14 & 19 & -17 \\ -14 & 14 & -12 \end{bmatrix} $ using the variable $ \lambda $. 2. **Formula:** The characteristic polynomial of a matrix $ A $ is given by $ p(\lambda) = \det(\lambda I - A) $, where $ I $ is the identity matrix. 3. **Construct $ \lambda I - A $:** $$ \lambda I - A = \begin{bmatrix} \lambda + 5 & -10 & 10 \\ 14 & \lambda - 19 & 17 \\ 14 & -14 & \lambda + 12 \end{bmatrix} $$ 4. **Calculate the determinant:** $$ \det(\lambda I - A) = (\lambda + 5) \begin{vmatrix} \lambda - 19 & 17 \\ -14 & \lambda + 12 \end{vmatrix} - (-10) \begin{vmatrix} 14 & 17 \\ 14 & \lambda + 12 \end{vmatrix} + 10 \begin{vmatrix} 14 & \lambda - 19 \\ 14 & -14 \end{vmatrix} $$ 5. **Calculate each minor:** - First minor: $$ (\lambda - 19)(\lambda + 12) - (17)(-14) = (\lambda - 19)(\lambda + 12) + 238 $$ - Second minor: $$ 14(\lambda + 12) - 14(17) = 14(\lambda + 12 - 17) = 14(\lambda - 5) $$ - Third minor: $$ 14(-14) - 14(\lambda - 19) = -196 - 14\lambda + 266 = 70 - 14\lambda $$ 6. **Substitute back:** $$ \det(\lambda I - A) = (\lambda + 5)[(\lambda - 19)(\lambda + 12) + 238] + 10 \cdot 14(\lambda - 5) + 10(70 - 14\lambda) $$ 7. **Expand and simplify:** - Expand $(\lambda - 19)(\lambda + 12) = \lambda^2 - 7\lambda - 228$ - So inside the bracket: $$ \lambda^2 - 7\lambda - 228 + 238 = \lambda^2 - 7\lambda + 10 $$ - Multiply by $\lambda + 5$: $$ (\lambda + 5)(\lambda^2 - 7\lambda + 10) = \lambda^3 - 7\lambda^2 + 10\lambda + 5\lambda^2 - 35\lambda + 50 = \lambda^3 - 2\lambda^2 - 25\lambda + 50 $$ - Add the other terms: $$ 10 \cdot 14(\lambda - 5) = 140\lambda - 700 $$ $$ 10(70 - 14\lambda) = 700 - 140\lambda $$ - Sum all: $$ \lambda^3 - 2\lambda^2 - 25\lambda + 50 + 140\lambda - 700 + 700 - 140\lambda = \lambda^3 - 2\lambda^2 - 25\lambda + 50 $$ 8. **Final characteristic polynomial:** $$ p(\lambda) = \lambda^3 - 2\lambda^2 - 25\lambda + 50 $$ 9. **Find eigenvalues:** Solve $ p(\lambda) = 0 $: Try rational roots using factors of 50: $ \pm1, \pm2, \pm5, \pm10, \pm25, \pm50 $. - Test $ \lambda = 2 $: $$ 2^3 - 2(2)^2 - 25(2) + 50 = 8 - 8 - 50 + 50 = 0 $$ So $ \lambda = 2 $ is a root. 10. **Factor out $ (\lambda - 2) $:** Divide $ p(\lambda) $ by $ (\lambda - 2) $: $$ \lambda^3 - 2\lambda^2 - 25\lambda + 50 = (\lambda - 2)(\lambda^2 - 25) $$ 11. **Solve quadratic:** $$ \lambda^2 - 25 = 0 \implies \lambda = \pm 5 $$ 12. **Eigenvalues:** $$ \boxed{\lambda = 2, 5, -5} $$