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