Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (b)

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MATH152 April 2015

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Question B 2 (b)
Let $A$ be the matrix ${\begin{bmatrix}4&-1&7\\0&3&0\\1&2&-2\end{bmatrix}}$ (b) Find all the other eigenvalues of $A$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
To find eigenvalues, solve the characteristic equation.

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Solution 1
In order to find other eigenvalues, we solve the characteristic equation $\operatorname {det} (A-\lambda I)=0,$ i.e., $\operatorname {det} {\begin{bmatrix}4-\lambda &-1&7\\0&3-\lambda &0\\1&2&-2-\lambda \end{bmatrix}}=(3-\lambda )\cdot \operatorname {det} {\begin{bmatrix}4-\lambda &7\\1&-2-\lambda \end{bmatrix}}=0.$ Thus we have $(3-\lambda )[(4-\lambda )(-2-\lambda )-7]=0.$ After simplification, we obtain ${\textstyle (3-\lambda )(\lambda +3)(\lambda -5)=0}$ ; we know of the eigenvalue ${\textstyle \lambda _{1}=3}$ from part (a), and the others are ${\textstyle \color {blue}\lambda _{2}=-3,\lambda _{3}=5.}$

Solution 2
Let $\lambda _{1},\lambda _{2},\lambda _{3}$ denote the three roots of the characteristic polynomial of $A$ (counted with multiplicity). We recall from part (a) that $\lambda _{1}=3$ is one of the roots. Moreover, $\operatorname {tr} (A)=\sum _{i=1}^{3}\lambda _{i}$ and $\det(A)=\prod _{i=1}^{3}\lambda _{i}$ . Thus $3+\lambda _{2}+\lambda _{3}=\operatorname {tr} {\begin{bmatrix}4&-1&7\\0&3&0\\1&2&-2\end{bmatrix}}=4+3+-2=5$ and $3\lambda _{2}\lambda _{3}=\det {\begin{bmatrix}4&-1&7\\0&3&0\\1&2&-2\end{bmatrix}}=3\cdot \det {\begin{bmatrix}4&7\\1&-2\end{bmatrix}}=3\cdot -15.$ Hence $\lambda _{2}+\lambda _{3}=2$ and $\lambda _{2}\lambda _{3}=-15$ . We find that $\color {blue}\lambda _{2}=-3,\lambda _{3}=5$ , as in the first solution.