Chapter 5 - Eigenvalues and Eigenvectors - 5.2 Exercises - Page 282: 14

$-28+25\lambda+4\lambda^2-\lambda^3$

$A=\begin{bmatrix} 5-\lambda&-2&3\\0&1-\lambda&0\\6&7&-2-\lambda \end{bmatrix}$ $det(A)=(1-\lambda)\left|\begin{matrix} 5-\lambda&3\\6&-2-\lambda \end{matrix}\right|=$ $=(1-\lambda)\big((5-\lambda)(-2-\lambda)-18\big)=0$ $-28+25\lambda+4\lambda^2-\lambda^3$