[SOLVED] Elementary Linear Algebra, Systems of Linear Equations

Aug 2009
71
1
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...

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

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.
 
Aug 2009
71
1
Ok, I think I found a solution to the first problem...I get the following:

\(\displaystyle x = \cos\theta - \sin\theta\)
\(\displaystyle y = \cos\theta + \sin\theta\)

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!
 
Dec 2007
16,948
6,768
Zeitgeist
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...

\(\displaystyle (\cos\theta)x + (\sin\theta)y = 1\) .... (1)
\(\displaystyle (-\sin\theta)x + (\cos\theta)y = 1\) .... (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 \(\displaystyle \sin \theta\) and equation (2) by \(\displaystyle \cos \theta\) and then add the two resulting equations together.

[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).
 
  • Like
Reactions: Alterah
Aug 2009
71
1
For this system:
\(\displaystyle x = \cos\theta - \sin\theta\)
\(\displaystyle y = \cos\theta + \sin\theta\)

I don't think I can multiply through by \(\displaystyle \cos\theta\ or \sin\theta \) 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.
 
Dec 2007
16,948
6,768
Zeitgeist
For this system:
\(\displaystyle x = \cos\theta - \sin\theta\)
\(\displaystyle y = \cos\theta + \sin\theta\)

I don't think I can multiply through by \(\displaystyle \cos\theta\ or \sin\theta \) 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.
..
 
Aug 2009
71
1
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:
\(\displaystyle (\cos\theta)x + (\sin\theta)y = 1\)
\(\displaystyle (-\sin\theta)x + (\cos\theta)y = 1 \)

The solution I got was:
\(\displaystyle x = \cos\theta - \sin\theta\)
\(\displaystyle y = \cos\theta + \sin\theta\)
 
Dec 2007
16,948
6,768
Zeitgeist
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:
\(\displaystyle (\cos\theta)x + (\sin\theta)y = 1\)
\(\displaystyle (-\sin\theta)x + (\cos\theta)y = 1 \)

The solution I got was:
\(\displaystyle x = \cos\theta - \sin\theta\)
\(\displaystyle y = \cos\theta + \sin\theta\)
Correct. (But why trust me ..... Verification by substitution should be simple enough).