Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}4&2&0\\-1&1&0\\1&1&2\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+7\lambda^2-16\lambda+12 $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}4&2&0\\-1&1&0\\1&1&2\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}4 - \lambda &2&0\\-1&1 - \lambda &0\\1&1&2 - \lambda \end{matrix} \right| = \\ &= (-\lambda+4) (-\lambda+1) (-\lambda+2) + 2 \cdot 0 \cdot 1 + 0 \cdot (-1) \cdot 1 - (-\lambda+4) \cdot 0 \cdot 1 - (-1) \cdot 2 \cdot (-\lambda+2) - 1 \cdot (-\lambda+1) \cdot 0 = \\ & = -\lambda^3+7\lambda^2-14\lambda+8 + 0 + 0 - 0 - (2\lambda-4) - 0 = \\ & = -\lambda^3+7\lambda^2-16\lambda+12 \end{aligned} $$