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January 19th, 2015, 05:09 PM  #1 
Member Joined: Jan 2015 From: Colorado Posts: 32 Thanks: 0  Need to satisfy constants, so each of the systems have a solution
Hi, I am really having trouble on this problem. "What conditions must the constants, the b's, satisfy so that each of these systems has a solution? Hint: Apply Gauss's Method and see what happens to the right side. x  3y + b1 3x + y = b2 x + 7y = b3 2x + 4y = b4 So what I have done, is put it into a matrix like so: 1 3  b1 3 1  b2 1 7  b3 2 4  b4 So what I am confused with, is if you have to do Guass's method, and get it into row echelon form (right? that's his method?) then how do you do that, the first row can stay the same, the second row has to have 0 #  b2 and then the last two rows must have 0 0  b3/b4. Right? I am having trouble getting there. 
January 19th, 2015, 05:33 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
$2r_2 + 2r_3  4r_4$ gives $0 \, 0  2b_2 + 2b_3  4b_4$. Thus we have $2b_2 + 2b_3  4b_4 = 0$ as one condition that the $b_i$ must satisfy. $r_2  3r_1$ gives $0 \, 10  b_2  3b_1$. We can also have $r_1 + r_3  r_4$ which gives $0 \, 0  b_1 + b_3  b_4$ and thus $b_1 + b_3  b_4 = 0$ is another condition for the $b_i$. 
January 19th, 2015, 05:38 PM  #3  
Member Joined: Jan 2015 From: Colorado Posts: 32 Thanks: 0  Quote:
I don't think I am following which row you are manipulating here...  
January 19th, 2015, 05:51 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
$r_i$ is row $i$.

January 19th, 2015, 05:54 PM  #5 
Member Joined: Jan 2015 From: Colorado Posts: 32 Thanks: 0  
January 19th, 2015, 05:58 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Oh, I wasn't thinking about actually doing it all, just giving you some examples of manipulations that you can do.

January 19th, 2015, 06:02 PM  #7  
Member Joined: Jan 2015 From: Colorado Posts: 32 Thanks: 0  Quote:
So am I right though, about how I need to make the last two rows exactly 0 0  b3 and 0 0  b4 in order to complete this problem?  
January 20th, 2015, 02:59 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
No. They'll both be 0 0  f(b1,b2,b3,b4) where f is some linear function.

January 20th, 2015, 06:15 AM  #9  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
Your first object must be to get all zeros, in the first column, below the first row. To do that, since the top row has "1" and the other "3", "1", and "2", you can 1) subtract 3 times the first row from the second 2) subtract the first row from the third 3) subtract 2 times the first row from the fourth That will give $\displaystyle \begin(bmatrix}1 & 3 & b_1 \\ 0 & 10 & b2 3b_1 \\ 0 & 10 & b_3 b_1 \\ 0 & 0 & b4 2b_1\end{bmatrix}$ Can you continue from there? Last edited by Country Boy; January 20th, 2015 at 06:19 AM.  

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