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January 19th, 2015, 05:09 PM   #1
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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.
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January 19th, 2015, 05:33 PM   #2
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$2r_2 + 2r_3 - 4r_4$ 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 $b_i$.
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January 19th, 2015, 05:38 PM   #3
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Quote:
Originally Posted by v8archie View Post
$2r_2 + 2r_3 - 4r_4$ 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 $b_i$.

I don't think I am following which row you are manipulating here...
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January 19th, 2015, 05:51 PM   #4
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$r_i$ is row $i$.
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January 19th, 2015, 05:54 PM   #5
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$r_i$ is row $i$.
right, so your first manipulation, what row were you manipulating and leaving?

for example

2x + 2y = 1
x + y = 2

and I manipulate r1 - r2, I am going to manipulate and leave row 2 as x + y = 1
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January 19th, 2015, 05:58 PM   #6
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Oh, I wasn't thinking about actually doing it all, just giving you some examples of manipulations that you can do.
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January 19th, 2015, 06:02 PM   #7
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Originally Posted by v8archie View Post
Oh, I wasn't thinking about actually doing it all, just giving you some examples of manipulations that you can do.
Oh! Okay, sorry I was just a little confused.
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?
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January 20th, 2015, 02:59 AM   #8
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No. They'll both be 0 0 | f(b1,b2,b3,b4) where f is some linear function.
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January 20th, 2015, 06:15 AM   #9
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Quote:
Originally Posted by mophiejoe View Post
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.
You have 4 equations (so a matrix with four rows) and only two unknowns. In order that this have a solution, "Gaussian elimination" must the last two rows to all zeros.

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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