MAT244--2018F > Quiz-6

Q6 TUT 0701

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Victor Ivrii:
Find the general solution of the given system of equations:
$$\mathbf{x}'=
\begin{pmatrix}
3 &2 &4\\
2 &0 &2\\
4 &2 &3
\end{pmatrix}\mathbf{x}.$$

Guanyao Liang:
answer

Qinger Zhang:
Here is my answer.

cindy_wen:
here is my solution

Tzu-Ching Yen:
det($M - rI$) gives
$ (3-r)(r^2 - 3r - 4) - 2(-2r -2) + 4(4 + 4r) = (r+1)((3-r)(r-4)-4 + 16) = -(r+1)(r^2-7r -8) = -(r+1)^2(r-8) $Set this to be zero, $r = -1, 8$

Let $r = 8$, matrix is
$
    M=
   \left[ {\begin{array}{ccc}
    -5 & 2 & 4 \\
    2 & -8 & 2 \\
    4 & 2 & -5 \\
   \end{array} } \right]
$By inspection solution is $x_1 = [2, 1, 2]$

Let $r = -1$
$
    M=
   \left[ {\begin{array}{ccc}
    4 & 2 & 4 \\
    2 & 1 & 2 \\
    4 & 2 & 4 \\
   \end{array} } \right]
 $
gives equation $r = -2s - 2t$ where solution is $[r, s, t]$. Hence two solutions are $x_2 = [1, -2, 0]$ and $x_3 = [0, -2, 1]$

General solution is therefore
$x = c_1e^{8t}x_1 + e^{-t}(c_2x_2 + c_3x_3)$

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