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March 21st, 2015, 06:07 PM   #1
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11.3, homogeneous systems, linear al

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 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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