Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 04 (b)

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MATH152 April 2013

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Question B 04 (b)
Let A be the $3\times 3$ matrix given by ${\begin{bmatrix}-7&0&1\\0&3&0\\-2&0&-4\end{bmatrix}}$ It is known that $\lambda _{1}=3$ is an eigenvalue of A. Find the other eigenvalues $\lambda _{2}$ and $\lambda _{3}$ 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
Compute the roots of the characteristic polynomial as given by $\det(A-\lambda I)$

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Solution 1

Let's compute the characteristic polynomial ${\begin{aligned}\det(A-\lambda I)&=\det \left({\begin{bmatrix}-7&0&1\\0&3&0\\-2&0&-4\end{bmatrix}}-{\begin{bmatrix}\lambda &0&0\\0&\lambda &0\\0&0&\lambda \end{bmatrix}}\right)\\&=\det \left({\begin{bmatrix}-7-\lambda &0&1\\0&3-\lambda &0\\-2&0&-4-\lambda \end{bmatrix}}\right)\\&=(-7-\lambda )(3-\lambda )(-4-\lambda )-(1(3-\lambda )(-2))\\&=(3-\lambda )((-7-\lambda )(-4-\lambda )+2)\\&=(3-\lambda )(28+11\lambda +\lambda ^{2}+2)\\&=(3-\lambda )(30+11\lambda +\lambda ^{2})\\&=(3-\lambda )(\lambda +5)(\lambda +6)\end{aligned}}$ Solving for the roots gives the eigenvalues $\lambda =-6,-5,3$

Solution 2

A party trick is to use the fact that the product of eigenvalues equals the determinant $\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=\mathrm {det} (A)$ and that the sum of the eigenvalues equals the sum of the diagonal entries of A $\displaystyle \lambda _{1}+\lambda _{2}+\lambda _{3}=-7+3-4=-8$ The determinant of A is quickly calculated as $\displaystyle \mathrm {det} (A)=(-7)(3)(-4)+0+0-(-2)(3)(1)-0-0=90$ Since $\lambda _{1}=3$ is given, we get the two equations $\lambda _{2}\lambda _{3}=30,\quad {\text{and}}\quad \lambda _{2}+\lambda _{3}=-11$ Hence $\lambda _{2}=-11-\lambda _{3}$ and thus ${\begin{aligned}(-11-\lambda _{3})\lambda _{3}&=30\\\lambda _{3}^{2}+11\lambda _{3}+30&=0\\(\lambda _{3}+5)(\lambda _{3}+6)&=0\end{aligned}}$ Hence the eigenvalues are $\lambda =3,-5,-6$.

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