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March 21st, 2015, 06:07 PM  #1 
Newbie Joined: Mar 2015 From: dy/dx = dy/du X du/dx 9. CHAIN Posts: 1 Thanks: 0  11.3, homogeneous systems, linear al gebra 
March 22nd, 2015, 05:17 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
If your question is "How can I look at this and instantly know the answer without doing any work?", I can't help you because I can't do that myself! Do you understand that this matrix equation is equivalent to the three simultaneous equations $\displaystyle x_1 + 3x_2  5x_3 + x_4 + 3x_5 + 2x_6 = 0$ $\displaystyle x_3 + 5x_4 + 2x_6 = 0$ $\displaystyle x_5  x_6 = 0$? Surely you can see that the last equation is the same as $\displaystyle x_5 = x_6$. We can also solve the second equation for $\displaystyle x_3$: $\displaystyle x_3 = x_4 2x_6$ Replacing $\displaystyle x_3$ and $\displaystyle x_5$ in the first equation by those, $\displaystyle x_1 + 3x_2 5(x_4  2x_6) + x_4 + 3(x_6) + 2x_6 = x_1 + 3x_2 + 6x_4 + 15x_6 = 0$ So $\displaystyle x_1= 3x_2  6x_4  15x_6 = 0$ and now all can be written in terms of $\displaystyle x_2$, $\displaystyle x_4$, and $\displaystyle x_6$. Last edited by skipjack; March 22nd, 2015 at 06:43 PM. 

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