Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}3&2&-1\\3&8&-3\\3&6&-1\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+10\lambda^2-28\lambda+24 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}3&2&-1\\3&8&-3\\3&6&-1\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}3 - \lambda &2&-1\\3&8 - \lambda &-3\\3&6&-1 - \lambda \end{matrix} \right| = \\ &= (-\lambda+3) (-\lambda+8) (-\lambda-1) + 2 \cdot (-3) \cdot 3 + (-1) \cdot 3 \cdot 6 - (-\lambda+3) \cdot (-3) \cdot 6 - 3 \cdot 2 \cdot (-\lambda-1) - 3 \cdot (-\lambda+8) \cdot (-1) = \\ & = -\lambda^3+10\lambda^2-13\lambda-24 + (-18) + (-18) - (18\lambda-54) - (-6\lambda-6) - (3\lambda-24) = \\ & = -\lambda^3+10\lambda^2-28\lambda+24 \end{aligned} $$