MAT244--2018F > Quiz-6

Q6 TUT 0701

(1/2) > >>

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:

Qinger Zhang:

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