Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}1&2\\-1&3\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = \lambda^2-4\lambda+5 $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(\lambda) &= det(A - \lambda I) = det \left( \left[ \begin{matrix}1&2\\-1&3\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}1 - \lambda &2\\-1&3 - \lambda \end{matrix} \right| = (-\lambda+1) (-\lambda+3) - 2 \cdot (-1) = \\ &= \lambda^2-4\lambda+3 - (-2) = \lambda^2-4\lambda+5 \end{aligned} $$This page was created using
Characteristic polynomial calculator