Q6 TUT 0801

[Victor Ivrii]

Find the general solution of the given system of equations:
$$\mathbf{x}'=
\begin{pmatrix}
1 &1 &1\\
2 &1 &-1\\
-8 &-5 &-3
\end{pmatrix}\mathbf{x}.$$

[Qinger Zhang]

Here is my answer.

[Ming Tang]

Here is my solution.

[Qing Zong]

This is my answer

[Xue Wang]

Here is my solution.

[Jiexuan Wei]

Here is my solution 

[Pengyun Li]

Firstly, we need to find the eigenvalues and eigenvectors.

$det(A-\lambda I) = det\left|\begin{matrix}1-\lambda & 1 & 1 \\ 2 & 1-\lambda & -1 \\ -8 & -5 & -3-\lambda\end{matrix}\right| = (\lambda + 2)(\lambda - 2)(\lambda + 1)$
 
Thus, the eigenvalues are $\lambda_1 = -2, \lambda_2 = -1, \lambda_3 = 2$

When $\lambda_1 = -2, (A - \lambda I) = (A + 2I) = \left(\begin{matrix}3 & 1 & 1 \\ 2 & 3 & -1 \\ -8 & -5 & -1\end{matrix}\right)$

\begin{equation*}
   \begin{pmatrix}
  3 & 1 & 1\\
  2 & 3 & -1\\
  -8 & -5 & -1
  \end{pmatrix}
  \begin{pmatrix}
  x_1\\
  x_2\\
  x_3
  \end{pmatrix}=0
  \end{equation*}
   Let x_1 = t
 \begin{equation*}
\begin{pmatrix}
  x_1\\
  x_2\\
  x_3
  \end{pmatrix}=t
 \begin{pmatrix}
  -4\\
  5\\
  7
  \end{pmatrix}
\end{equation*}

Therefore, the corresponding eigenvector $\vec{v_1} = \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right) $

Similarly, when $\lambda_2 = -1, \lambda_3 = 2$, we can get $\vec{v_2} = \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right)
 and \ \vec{v_3} = \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right)$ respectively.

Therefore, the general solution $X(t) = c_1 e^{-t} \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right) + c_2 e^{2t} \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right) + c_3 e^{-2t} \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right)$

[Victor Ivrii]

I gave credits to the first student who posted readable solution. Scan, do not make pictures. No colour, no grayscale, just black-white