Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}-6&28&21\\4&-15&-12\\-8&\frac{ 321 }{ 10 }&25\end{matrix} \right) $$The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+4\lambda^2-\dfrac{ 31 }{ 5 }\lambda+\dfrac{ 16 }{ 5 } $$The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}-6&28&21\\4&-15&-12\\-8&\frac{ 321 }{ 10 }&25\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}-6 - \lambda &28&21\\4&-15 - \lambda &-12\\-8&\frac{ 321 }{ 10 }&25 - \lambda \end{matrix} \right| = \\ &= (-\lambda-6) (-\lambda-15) (-\lambda+25) + 28 \cdot (-12) \cdot (-8) + 21 \cdot 4 \cdot \frac{ 321 }{ 10 } - (-\lambda-6) \cdot (-12) \cdot \frac{ 321 }{ 10 } - 4 \cdot 28 \cdot (-\lambda+25) - (-8) \cdot (-\lambda-15) \cdot 21 = \\ & = -\lambda^3+4\lambda^2+435\lambda+2250 + 2688 + \frac{ 13482 }{ 5 } - (\frac{ 1926 }{ 5 }\lambda+\frac{ 11556 }{ 5 }) - (-112\lambda+2800) - (168\lambda+2520) = \\ & = -\lambda^3+4\lambda^2-\frac{ 31 }{ 5 }\lambda+\frac{ 16 }{ 5 } \end{aligned} $$