Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}0&\dfrac{ 3 }{ 10 }\\-\dfrac{ 6 }{ 5 }&0\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = \lambda^2+\dfrac{ 9 }{ 25 } $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(\lambda) &= det(A - \lambda I) = det \left( \left[ \begin{matrix}0&\dfrac{ 3 }{ 10 }\\-\dfrac{ 6 }{ 5 }&0\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}0 - \lambda &\dfrac{ 3 }{ 10 }\\-\dfrac{ 6 }{ 5 }&0 - \lambda \end{matrix} \right| = (-\lambda) (-\lambda) - \frac{ 3 }{ 10 } \cdot (-\frac{ 6 }{ 5 }) = \\ &= \lambda^2 - (-\frac{ 9 }{ 25 }) = \lambda^2+\frac{ 9 }{ 25 } \end{aligned} $$This page was created using
Characteristic polynomial calculator