Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}\frac{ 1 }{ 10 }&\frac{ 1 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 3 }{ 10 }&\frac{ 1 }{ 5 }&\frac{ 3 }{ 10 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 5 }\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+\dfrac{ 1 }{ 2 }\lambda^2+\dfrac{ 23 }{ 50 }\lambda+\dfrac{ 1 }{ 25 } $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}\frac{ 1 }{ 10 }&\frac{ 1 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 3 }{ 10 }&\frac{ 1 }{ 5 }&\frac{ 3 }{ 10 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 5 }\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}\frac{ 1 }{ 10 } - \lambda &\frac{ 1 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 3 }{ 10 }&\frac{ 1 }{ 5 } - \lambda &\frac{ 3 }{ 10 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 5 } - \lambda \end{matrix} \right| = \\ &= (-\lambda+\frac{ 1 }{ 10 }) (-\lambda+\frac{ 1 }{ 5 }) (-\lambda+\frac{ 1 }{ 5 }) + \frac{ 1 }{ 5 } \cdot \frac{ 3 }{ 10 } \cdot \frac{ 3 }{ 5 } + \frac{ 1 }{ 2 } \cdot \frac{ 3 }{ 10 } \cdot \frac{ 3 }{ 5 } - (-\lambda+\frac{ 1 }{ 10 }) \cdot \frac{ 3 }{ 10 } \cdot \frac{ 3 }{ 5 } - \frac{ 3 }{ 10 } \cdot \frac{ 1 }{ 5 } \cdot (-\lambda+\frac{ 1 }{ 5 }) - \frac{ 3 }{ 5 } \cdot (-\lambda+\frac{ 1 }{ 5 }) \cdot \frac{ 1 }{ 2 } = \\ & = -\lambda^3+\frac{ 1 }{ 2 }\lambda^2-\frac{ 2 }{ 25 }\lambda+\frac{ 1 }{ 250 } + \frac{ 9 }{ 250 } + \frac{ 9 }{ 100 } - (-\frac{ 9 }{ 50 }\lambda+\frac{ 9 }{ 500 }) - (-\frac{ 3 }{ 50 }\lambda+\frac{ 3 }{ 250 }) - (-\frac{ 3 }{ 10 }\lambda+\frac{ 3 }{ 50 }) = \\ & = -\lambda^3+\frac{ 1 }{ 2 }\lambda^2+\frac{ 23 }{ 50 }\lambda+\frac{ 1 }{ 25 } \end{aligned} $$This page was created using
Characteristic polynomial calculator