Находим характеристический многочлен матрицы
![A-\lambda I](https://tex.z-dn.net/?f=A-%5Clambda+I)
:
![(2-\lambda)(3+\lambda)(2-\lambda)+(-5(2+\lambda)+3)+2(0-(3+\lambda))= \\ =(4-\lambda^2)(3+\lambda)-13-7\lambda=\lambda^3+3\lambda^2+3\lambda+1=(\lambda+1)^3 \\ c(\lambda)=(\lambda+1)^3](https://tex.z-dn.net/?f=%282-%5Clambda%29%283%2B%5Clambda%29%282-%5Clambda%29%2B%28-5%282%2B%5Clambda%29%2B3%29%2B2%280-%283%2B%5Clambda%29%29%3D+%5C%5C%0A%3D%284-%5Clambda%5E2%29%283%2B%5Clambda%29-13-7%5Clambda%3D%5Clambda%5E3%2B3%5Clambda%5E2%2B3%5Clambda%2B1%3D%28%5Clambda%2B1%29%5E3+%5C%5C%0Ac%28%5Clambda%29%3D%28%5Clambda%2B1%29%5E3)
Собственные значения:
![det(A-\lambda |)=0 \Rightarrow \ (\lambda+1)^3=0 \Rightarrow \ \lambda_{1,2,3}=-1](https://tex.z-dn.net/?f=det%28A-%5Clambda+%7C%29%3D0+%5CRightarrow+%5C+%28%5Clambda%2B1%29%5E3%3D0+%5CRightarrow+%5C+%5Clambda_%7B1%2C2%2C3%7D%3D-1)
Собственные векторы:
![\left(\left.\begin{array}[t]{ccc} 3 & -1 & 2\\ 5 & -2 & 3\\ -1 & 0 & -1 \end{array}\right|\begin{array}[t]{c} 0\\ 0\\ 0 \end{array}\right) \rightarrow \left(\left.\begin{array}[t]{ccc} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{array}\right|\begin{array}[t]{c} 0\\ 0\\ 0 \end{array}\right) ](https://tex.z-dn.net/?f=%5Cleft%28%5Cleft.%5Cbegin%7Barray%7D%5Bt%5D%7Bccc%7D%0A3+%26+-1+%26+2%5C%5C%0A5+%26+-2+%26+3%5C%5C%0A-1+%26+0+%26+-1%0A%5Cend%7Barray%7D%5Cright%7C%5Cbegin%7Barray%7D%5Bt%5D%7Bc%7D%0A0%5C%5C%0A0%5C%5C%0A0%0A%5Cend%7Barray%7D%5Cright%29%0A+%5Crightarrow+%5Cleft%28%5Cleft.%5Cbegin%7Barray%7D%5Bt%5D%7Bccc%7D%0A1+%26+0+%26+1%5C%5C%0A0+%26+1+%26+1%5C%5C%0A0+%26+0+%26+0%0A%5Cend%7Barray%7D%5Cright%7C%5Cbegin%7Barray%7D%5Bt%5D%7Bc%7D%0A0%5C%5C%0A0%5C%5C%0A0%0A%5Cend%7Barray%7D%5Cright%29%0A+)
![V_{-1}= Span\left\{ \begin{pmatrix}-1\\ -1\\ 1 \end{pmatrix}\right\} ](https://tex.z-dn.net/?f=V_%7B-1%7D%3D+Span%5Cleft%5C%7B+%5Cbegin%7Bpmatrix%7D-1%5C%5C%0A-1%5C%5C%0A1%0A%5Cend%7Bpmatrix%7D%5Cright%5C%7D+%0A+)
Собственный вектор указан как базис собственного подпространства
![V_{-1}](https://tex.z-dn.net/?f=V_%7B-1%7D)
4(3y-0,6)=3(4t-0,8)
12y-12y=-2,4+2,4
0y=0
То есть под y может подразумеваться любое значение
Из двух шаров можно решить- двумя способами
а из трёх шаров- шестью способами