Question

Задание во вложении.

Answer 1

$det\left[\begin{array}{ccc}0&0&4- \alpha \\0&6- \alpha &0\\4- \alpha &0&0\end{array}\right] = - (6- \alpha )(4- \alpha )^2$

$\alpha _1 = 6 \\ \alpha _2 = \alpha _3 = 4$