# Find $a,b,c$ so that $\begin{bmatrix} 0 & 1& 0 \\ 0 & 0 & 1\\ a & b & c \end{bmatrix} $ has the characteristic polynomial $-\lambda^3+4\lambda^2+5\lambda+6=0$

## Answer

\[0=\det \begin{bmatrix} -\lambda & 1& 0 \\ 0& -\lambda & 1 \\ a & b & c-\lambda \end{bmatrix} \]

\[=-\lambda \det \begin{bmatrix} -\lambda & 1 \\ b & c-\lambda \end{bmatrix}-1 \det \begin{bmatrix} 0 & 1 \\ a & c-\lambda \end{bmatrix} \]

\[=-\lambda [\lambda (\lambda-c)-b]-[-a]=-\lambda^3+c\lambda^2+b\lambda+a\]

\[=-\lambda^3+4\lambda^2+5\lambda+6.\]

Hence

\[a=6, b=5, c=4.\]

Daniel90

436

The answer is accepted.

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