Linear system: determ inconclusive, unique, infinity solutions, find infinity s.

eckimz

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Hi. I'm majoring in petroleum engineer, got this linear system in a test. So first thing I did was to make the matrices and try to 0 under the principal diagonal. But I basically stopped at that. I'm not sure if I don't know what I was doing or if I was just actually doing it wrong. I did not try krammer. I know that after I get the final matrice, I should just do free parameter and define the solutions for each question above, like, for unique solution I know I have to get an answer different of 0. I think inconclusive has to be equal to 0 or that is the infinite, I don't know. I just want to see how this works to understand. I got other tests coming that I need to focus and study and I can't stop thinking of this stupid question.

Here's the linear system equations:

1x -2y + 3z = 3
2x -3y -2z = -2
ax + by + cz = 1

Determine:
a) inconclusive solution.
b) unique solution.
c) infinite solutions.
d) find the solutions in C.

I studied a lot to left something in blank lol :(. The linear system was suppose to be the easy question of the test haha.
The rest of the test was just so easy (gauss-seidel method of iterations, independent linear or dependent linear, vector coordinates with basis changing and using canonical basis).

Thanks for your time.
 
Hi. I'm majoring in petroleum engineer, got this linear system in a test. So first thing I did was to make the matrices and try to 0 under the principal diagonal. But I basically stopped at that. I'm not sure if I don't know what I was doing or if I was just actually doing it wrong. I did not try krammer. I know that after I get the final matrice, I should just do free parameter and define the solutions for each question above, like, for unique solution I know I have to get an answer different of 0. I think inconclusive has to be equal to 0 or that is the infinite, I don't know. I just want to see how this works to understand. I got other tests coming that I need to focus and study and I can't stop thinking of this stupid question.

Here's the linear system equations:

1x -2y + 3z = 3
2x -3y -2z = -2
ax + by + cz = 1

Determine:
a) inconclusive solution.
b) unique solution.
c) infinite solutions.
d) find the solutions in C.

I studied a lot to left something in blank lol :(. The linear system was suppose to be the easy question of the test haha.
The rest of the test was just so easy (gauss-seidel method of iterations, independent linear or dependent linear, vector coordinates with basis changing and using canonical basis).

Thanks for your time.

I'm confused. How can you tell what kind of solution it has, when three coefficients are unknown? Did it actually ask you to find conditions on a, b, and c under which each result occurs? This is why we ask for a complete copy of the problem. Also, we ask to see your work; can you show the result of your diagonalization? Also, to make sure we are on the same page, can you show the rules you were taught for recognizing the kind of system? There are several ways to characterize this.
 
I'm confused. How can you tell what kind of solution it has, when three coefficients are unknown? Did it actually ask you to find conditions on a, b, and c under which each result occurs? This is why we ask for a complete copy of the problem. Also, we ask to see your work; can you show the result of your diagonalization? Also, to make sure we are on the same page, can you show the rules you were taught for recognizing the kind of system? There are several ways to characterize this.

Sir, that's all it had on the test, it is complete. I copy exactly how it was. I'm also confused. I thought I was crazy about asking a, b and c at same time. I thought I was doing it wrong or something. Well, I don't have a progress, since I could not do it at all. I knew what to do, but I just couldn't since I had to find a, b and c, how am I suppose to do that using Gaussian elimination. What I did it was:

WhatsApp Image 2018-06-22 at 11.24.12 AM.jpg

After that (in the test), I just put what I would've done after finding the 0 under the principal diagonal (written).
I just don't know how to find x, y and z with a, b, and c. Is it even possible?

Here's how I usually do in the homework (other exercise, none doubts about this, just complement so you can understand me). It doesn't say in my photo, but in this exercise only asks for unique solution.
I do gaussian elimination and set free parameter and find x, y and z. But in this exercise it only gives "a". Look, I did and found the results correctly.


WhatsApp Image 2018-06-22 at 11.16.24 AM.jpg

Btw, I read forum rules, sorry if I did not complain with all, I didn't mean it, thought I had done the post correctly, I'm confused, so it might get little hard to explain better, let me know if I didn't make clear.
 
Sir, that's all it had on the test, it is complete. I copy exactly how it was. I'm also confused. I thought I was crazy about asking a, b and c at same time. I thought I was doing it wrong or something. Well, I don't have a progress, since I could not do it at all. I knew what to do, but I just couldn't since I had to find a, b and c, how am I suppose to do that using Gaussian elimination. What I did it was:

View attachment 9681

After that (in the test), I just put what I would've done after finding the 0 under the principal diagonal (written).
I just don't know how to find x, y and z with a, b, and c. Is it even possible?

Here's how I usually do in the homework (other exercise, none doubts about this, just complement so you can understand me). It doesn't say in my photo, but in this exercise only asks for unique solution.
I do gaussian elimination and set free parameter and find x, y and z. But in this exercise it only gives "a". Look, I did and found the results correctly.


View attachment 9682

Btw, I read forum rules, sorry if I did not comply with all, I didn't mean it, thought I had done the post correctly, I'm confused, so it might get little hard to explain better, let me know if I didn't make clear.

When a question is confusing, helpers need more information; it's not a matter of your doing something wrong. Here, your pictures are a little fuzzy, but I can read enough to see what you are doing.

In the second example, it looks like you were given two unknown parameters, a and b, and solved in terms of those. I would think you would do essentially the same thing for the problem you are asking about, giving the solution in terms of a, b, and c. (It doesn't say to find a, b, and c, right?) That would apply to the unique solution case. Then you would look for situations where the solution would be nonexistent or non-unique.
 
I presume that the question is actually. "Determine values of a, b, and c such that this system of equations has
"a) inconclusive solution.
b) unique solution.
c) infinite solutions.
d) find the solutions in c."

The first thing I would do is try to solve the system! Given
1x -2y + 3z = 3
2x -3y -2z = -2
ax + by + cz = 1

Multiply the third equation by 3 and the first equation c. Subtract the first from the third to get (3a- c)x+ (3b+ 2)y= 3- 3c.

Multiply the second equation by 3 and the first equation by 2. Add the two equations to get 8x- 10y= 0. From that, y= 8x/10= 4x/5.

Setting y= 4x/5 in (3a- c)x+ (3b+ 2)y= 3- 3c gives (3a- c)x+ (3b+ 2)(4x/5)= (3a+12b/5- c+ 2)x= 3- 3c.

That will have a unique solution as long as 3a+ 12b/5- c+ 2 is not equal to 0. There will be NO solution (I suspect that is what is meant by "inconclusive solution") if 3a+ 12b/5- c+ 2= 0 but 3- 3c is not 0.

There will be an infinite number of solutions if both 3a+ 12b/5- c+ 2= 0 and 3- 3c= 0. Of course that last means c= 1 so 3a+ 12b/5- c+ 2= 3a+ 12b/5+ 1= 0. We can write b= -(15a+ 5)/12, c= 1 as the solution for any a.
 
This problem can be solved by converting the original system to a triangular system. The triangular system has the same solution set as the original, but its form makes it much easier to find values of a, b, and c such that there are zero, one, or infinitely many solutions.

In matrix form, the original system of linear equations is:
Code:
[FONT=Courier New][1 -2  3] [x]   [ 3]
[1 -3  2] [y] = [-2]
[a  b  c] [z]   [ 1][/FONT]
To transform the system to triangular form, multiply both sides by:
Code:
[FONT=courier new][ 1  0  0]
[-2  1  0]
[-a  0  1][/FONT]
and then by:
Code:
[FONT=courier new][1     0  0]
[0     1  0]
[0  2a+b  1] [/FONT]
which yields the triangular system:
Code:
[FONT=courier new][1 -2         3] [x]   [       3]
[0  1        -8] [y] = [      -8]
[0  0  13a+8b+c] [z]   [13a+8b+1][/FONT]
Written out, the three equations of the new system are:
Code:
[FONT=courier new]x - 2y + 3z =  3

     y - 8z = -8

(13a+8b+c)z = 13a+8b+1[/FONT]
If (13a+8b+c)z = 13a+8b+1 has no solution, then the system has no solutions.
Setting a=b=1, c=-21, reduces the equation previous to 0z=22, which has no solutions.

If a triangular matrix U has non-zero diagonal entries, then U*x=b has a unique solution.
Therefore, setting a=b=c=1, guarantees the system of equations has a unique solution.

If (13a+8b+c)z = 13a+8b+1 has infinitely many solutions, then the system will have infinitely many solutions,
so we need to find a, b, and c such that 13a+8b+c = 0 and 13a+8b+1 = 0.
One choice is a=0, b=-1/8, c=1, which makes the system of equations:
Code:
[FONT=courier new]
[1 -2  3] [x] = [ 3]
[0  1 -8] [y] = [-8]
[0  0  0] [z] = [ 0][/FONT]
whose solution is x=-13 + 13t, y=-8 + 8t, z = t
 
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