# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:58:12 PM

Title: Q6 TUT 0801
Post by: Victor Ivrii on November 17, 2018, 03:58:12 PM
Find the general solution of the given system of equations:
$$\mathbf{x}'= \begin{pmatrix} 1 &1 &1\\ 2 &1 &-1\\ -8 &-5 &-3 \end{pmatrix}\mathbf{x}.$$
Title: Re: Q6 TUT 0801
Post by: Qinger Zhang on November 17, 2018, 04:27:28 PM
Here is my answer.
Title: Re: Q6 TUT 0801
Post by: Ming Tang on November 17, 2018, 04:28:20 PM
Here is my solution.
Title: Re: Q6 TUT 0801
Post by: Qing Zong on November 17, 2018, 04:34:04 PM
This is my answer
Title: Re: Q6 TUT 0801
Post by: Xue Wang on November 17, 2018, 10:10:07 PM
Here is my solution.
Title: Re: Q6 TUT 0801
Post by: Jiexuan Wei on November 18, 2018, 11:07:12 AM
Here is my solution  :)
Title: Re: Q6 TUT 0801
Post by: Pengyun Li on November 18, 2018, 02:43:08 PM
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 $X(t) = c_1 e^{-t} \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right) + c_2 e^{2t} \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right) + c_3 e^{-2t} \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right)$

Title: Re: Q6 TUT 0801
Post by: Victor Ivrii on November 25, 2018, 09:39:39 AM
I gave credits to the first student who posted readable solution. Scan, do not make pictures. No colour, no grayscale, just black-white