Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}3&\dfrac{ 5 }{ 2 }\\-2&4\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = \lambda^2-7\lambda+17 $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(\lambda) &= det(A - \lambda I) = det \left( \left[ \begin{matrix}3&\dfrac{ 5 }{ 2 }\\-2&4\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}3 - \lambda &\dfrac{ 5 }{ 2 }\\-2&4 - \lambda \end{matrix} \right| = (-\lambda+3) (-\lambda+4) - \frac{ 5 }{ 2 } \cdot (-2) = \\ &= \lambda^2-7\lambda+12 - (-5) = \lambda^2-7\lambda+17 \end{aligned} $$This page was created using
Characteristic polynomial calculator