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
Lets compute the characteristic equation
\[0=\det \begin{bmatrix} -\lambda & 1& 0 \\ 0& -\lambda & 1 \\ a & b & c-\lambda \end{bmatrix} \]
\[=-\lambda \det \begin{bmatrix} -\lambda & 1 \\ b & c-\lambda \end{bmatrix}-1 \det \begin{bmatrix} 0 & 1 \\ a & c-\lambda \end{bmatrix} \]
\[=-\lambda [\lambda (\lambda-c)-b]-[-a]=-\lambda^3+c\lambda^2+b\lambda+a\]
\[=-\lambda^3+4\lambda^2+5\lambda+6.\]
Hence
\[a=6, b=5, c=4.\]
The answer is accepted.
- answered
- 248 views
- $3.00
Related Questions
- Solve $abc=2(a-2)(b-2)(c-2)$ where $a,b $ and $c$ are integers
- Algebra Word Problem 3
- Representation theory question
- Prove that: |x| + |y| ≤ |x + y| + |x − y|.
- Can enough pizza dough be made to cover the surface of the earth?
- Five times the larger of two consecutive odd integers is equal to one more than eight times the smaller. Find the integers.
- Need Upper Bound of an Integral
- Linear Algebra - matrices and vectors