Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (b)/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 ${\textstyle \color {blue}\lambda _{2}=-3,\lambda _{3}=5.}$