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

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 $A$.

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Hint
To find eigenvalues, solve the characteristic equation.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.