Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}-9&4&4\\-8&3&4\\-16&8&7\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+\lambda^2+5\lambda+3 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}-9&4&4\\-8&3&4\\-16&8&7\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}-9 - \lambda &4&4\\-8&3 - \lambda &4\\-16&8&7 - \lambda \end{matrix} \right| = \\ &= (-\lambda-9) (-\lambda+3) (-\lambda+7) + 4 \cdot 4 \cdot (-16) + 4 \cdot (-8) \cdot 8 - (-\lambda-9) \cdot 4 \cdot 8 - (-8) \cdot 4 \cdot (-\lambda+7) - (-16) \cdot (-\lambda+3) \cdot 4 = \\ & = -\lambda^3+\lambda^2+69\lambda-189 + (-256) + (-256) - (-32\lambda-288) - (32\lambda-224) - (64\lambda-192) = \\ & = -\lambda^3+\lambda^2+5\lambda+3 \end{aligned} $$