[SOLVED] Elementary Linear Algebra, Systems of Linear Equations

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August 29th 2009, 04:45 PM
Alterah

[SOLVED] Elementary Linear Algebra, Systems of Linear Equations

Hello, I am currently stuck on one problem...and am not sure about my solution on another. I believe I am restricted to elementary operations as well. Those are:
1. Interchange two equations
2. Multiply an equation by a nonzero constant
3. Add a multiple of an equation (equation 1) to another equation (equation 2) and replace the second equation (say equation 2) with the result.

Now that I have listed what I must work off of, the first problem I am having is with this system...

$(\cos\theta)x + (\sin\theta)y = 1$
$(-\sin\theta)x + (\cos\theta)y = 1$

I need to solve this system for x and y. I do know that my answer will more than likely be in terms of sines and cosines.

The second problem I am uncertain about is this:

kx + y = 4
2x - 3y = -12

I need to find the value(s) of k such that the system has an infinite number of solutions. I believe my answer of k = -2/3 is accurate and as far as I can tell with k being -2/3 the system has an infinite number of solutions. However, I am uncertain if there are other values of k which would give me the same answer. I apologize if this should be in a different forum, but as my class is Elementary Linear Algebra I figured this location was appropriate. Thanks for all your help.
August 29th 2009, 05:43 PM
Alterah

Ok, I think I found a solution to the first problem...I get the following:

$x = \cos\theta - \sin\theta$
$y = \cos\theta + \sin\theta$

Now, I wound up doing it the long way by adding -1 times the first equation to the second equation. Finding an expression for y in terms of x and plugging back in. I did it this way, as opposed to simply setting the two equations equal to each other because I wasn't really certain how my teacher wants a problem like this done, guess that's a question for him. Can anyone verify my solution?

Thanks!
August 29th 2009, 05:47 PM
mr fantastic

Quote:

Originally Posted by Alterah

Hello, I am currently stuck on one problem...and am not sure about my solution on another. I believe I am restricted to elementary operations as well. Those are:
1. Interchange two equations
2. Multiply an equation by a nonzero constant
3. Add a multiple of an equation (equation 1) to another equation (equation 2) and replace the second equation (say equation 2) with the result.

Now that I have listed what I must work off of, the first problem I am having is with this system...

$(\cos\theta)x + (\sin\theta)y = 1$ .... (1)
$(-\sin\theta)x + (\cos\theta)y = 1$ .... (2)

I need to solve this system for x and y. I do know that my answer will more than likely be in terms of sines and cosines.
[snip]

Multiply equation (1) by $\sin \theta$ and equation (2) by $\cos \theta$ and then add the two resulting equations together.

Quote:

Originally Posted by Alterah

[snip]
The second problem I am uncertain about is this:

kx + y = 4
2x - 3y = -12

I need to find the value(s) of k such that the system has an infinite number of solutions. I believe my answer of k = -2/3 is accurate and as far as I can tell with k being -2/3 the system has an infinite number of solutions. However, I am uncertain if there are other values of k which would give me the same answer. I apologize if this should be in a different forum, but as my class is Elementary Linear Algebra I figured this location was appropriate. Thanks for all your help.

Your answer of k = -2/3 is correct. I cannot explain to you why it's the only value without seeing your working (although I'm sure if yu look at your working the reason will be obvious).
August 29th 2009, 06:27 PM
Alterah

For this system:
$x = \cos\theta - \sin\theta$
$y = \cos\theta + \sin\theta$

I don't think I can multiply through by $\cos\theta\ or \sin\theta$ because those aren't necessarily constants. And therefore that goes against the elementary rules I need to follow for this assignment at least. So, in the end I would up subtracting the first equation from the second one (add -1 * first equation to second one) and then solved for y in the second one. Then proceeded to plug the answer for y into the first one. Did you get the solution I got for that system?

Also, my work for the second set:
1. kx + y = 4
2. 2x - 3y = -12

Went like this:
Multiply equation 1 by 3 and replace equation 1 with the result (3*eq1 ->eq1) . Then add equation 1 to equation 2 and replace equation 2 with the result. So the system will then look like this:
1. 3kx + 3y = 12
2. (3k + 2)x = 0

Equation 2 is true when either x = 0 or (3k + 2) = 0. Since x = 0 is trivial I solved 3k + 2 = 0 for k and got k = -2/3.

With that we can put in any value for x and there will be a corresponding value for y that will make the system true. For example (3,6) or (6, 8). Since we know a system can have only 1 unique solution, no solution, or infinitely many solutions, the fact that we can find two points which satisfy the system when k = -2/3 means it has an infinite number of solutions. I guess that ought to be good enough, There wouldn't be another value for k to make it true.
August 29th 2009, 07:54 PM
mr fantastic

Quote:

Originally Posted by Alterah

For this system:
$x = \cos\theta - \sin\theta$
$y = \cos\theta + \sin\theta$

I don't think I can multiply through by $\cos\theta\ or \sin\theta$ because those aren't necessarily constants.

Mr F says: The unknowns you're solving for are x and y. So yes, you can.

And therefore that goes against the elementary rules I need to follow for this assignment at least. So, in the end I would up subtracting the first equation from the second one (add -1 * first equation to second one) and then solved for y in the second one. Then proceeded to plug the answer for y into the first one. Did you get the solution I got for that system?

Also, my work for the second set:
1. kx + y = 4
2. 2x - 3y = -12

Went like this:
Multiply equation 1 by 3 and replace equation 1 with the result (3*eq1 ->eq1) . Then add equation 1 to equation 2 and replace equation 2 with the result. So the system will then look like this:
1. 3kx + 3y = 12
2. (3k + 2)x = 0

Equation 2 is true when either x = 0 or (3k + 2) = 0. Since x = 0 is trivial I solved 3k + 2 = 0 for k and got k = -2/3.

With that we can put in any value for x and there will be a corresponding value for y that will make the system true. For example (3,6) or (6, 8). Since we know a system can have only 1 unique solution, no solution, or infinitely many solutions, the fact that we can find two points which satisfy the system when k = -2/3 means it has an infinite number of solutions. I guess that ought to be good enough, There wouldn't be another value for k to make it true. Mr F says: I guess so.

..
August 29th 2009, 11:35 PM
Alterah

Thank you for your help. So do you agree with my solution for my first problem or are you getting a different answer?
The system is actually:
$(\cos\theta)x + (\sin\theta)y = 1$
$(-\sin\theta)x + (\cos\theta)y = 1$

The solution I got was:
$x = \cos\theta - \sin\theta$
$y = \cos\theta + \sin\theta$
August 30th 2009, 03:03 AM
mr fantastic

Quote:

Originally Posted by Alterah

Thank you for your help. So do you agree with my solution for my first problem or are you getting a different answer?
The system is actually:
$(\cos\theta)x + (\sin\theta)y = 1$
$(-\sin\theta)x + (\cos\theta)y = 1$

The solution I got was:
$x = \cos\theta - \sin\theta$
$y = \cos\theta + \sin\theta$

Correct. (But why trust me ..... Verification by substitution should be simple enough).
August 30th 2009, 10:03 AM
Alterah

True enough. Thanks for your assistance.