Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}1&-2&2\\-2&1&-2\\2&-2&1\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+3\lambda^2+9\lambda+5 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}1&-2&2\\-2&1&-2\\2&-2&1\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}1 - \lambda &-2&2\\-2&1 - \lambda &-2\\2&-2&1 - \lambda \end{matrix} \right| = \\ &= (-\lambda+1) (-\lambda+1) (-\lambda+1) + (-2) \cdot (-2) \cdot 2 + 2 \cdot (-2) \cdot (-2) - (-\lambda+1) \cdot (-2) \cdot (-2) - (-2) \cdot (-2) \cdot (-\lambda+1) - 2 \cdot (-\lambda+1) \cdot 2 = \\ & = -\lambda^3+3\lambda^2-3\lambda+1 + 8 + 8 - (-4\lambda+4) - (-4\lambda+4) - (-4\lambda+4) = \\ & = -\lambda^3+3\lambda^2+9\lambda+5 \end{aligned} $$