Solve System with Matrices: Finding Unique Solutions for k Values

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The discussion focuses on determining values of k for which two systems of equations either lack a unique solution or possess one. For the first system, it is established that setting the determinant of the coefficient matrix to zero indicates nonuniqueness, specifically when k equals 4. The second system similarly requires finding k values that yield a unique solution, with the consensus that any k other than 4 will result in a unique solution. Participants emphasize the importance of understanding determinants and the conditions for unique solutions in linear systems. The conversation highlights the need for foundational knowledge in matrix theory to effectively tackle these problems.
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I know how to determine the determinant, and the inverse, and how to sovle a system with the inverse of a matrice, but I have no idea what these two questions are talking about:
1. Find the values of k for which the system
kx+2y=1
3x+(k-1)y=1
does not have a unique solution. If k does not have these values, find the unique solution. For each value of k for which no unique solution exists, determine whether or not any solution of the system exists.

what is this question asking me for? What is a unique solution?

2. Find the values of k for which the system
x+2y+2z=1
2x+y+2z=4
3x+3y+kz=5
has a unique solution. Find this unique solution and solve the system for any values of k for which the unique solution does not exist.

again, wtf?
 
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What's the condition for a nonhomogenous system to have unique solution?

Daniel.
 
I do not know what a unique solution is, nor have I ever heard the term nonhomogenous.
 
Then my advice is to study theory first and then get busy (trying to) solve(ing) problems...

Daniel.
 
this is not some independant study topic. This is from a practice booklet the teacher gave us, a question which the tacher recomnded we do. So, my current knowledge should suffice in answering this question.
What I know: -determinants
-inverses
-simple matrix operations
-solving systems with matrix inverses
-there being no solution if a system forms a singular matrix
 
The IV-th item of your "list" is the key.

Yes,setting the determinant of coefficients to zero will ensure nonuniqueness of solutions.

Daniel.
 
For the 2nd question, if nonuniqueness equates to there being no solution, the that is easy, k = 4, but for a unique solution, could not that be anything but for in the field of k?
 
I'm sorry,i really didn't understand your question.Could you please rephrase it using other words...?

Daniel.
 
I really don't know what a unique solution is. One of the answers it gives in the answers section is k cannot equal 4, which is the restriction should there be a solution to the system, if k = 4 detA = 0. So if k=4 means the system has no unique solutions, what does it mean if it does have unique solutions? How could you figure that out? Would not any value of k other than 4 bring about a system with unique solutions?
 
  • #10
Yes,any value different than 4 would make unique solution.

Daniel.
 
  • #11
I am sorely tempted to ask "Do you know what a "dictionary" is"?
(but I won't).

"Solution" a value (in this case a vector or set of numbers) that satisfies the equation.

"unique" only one.

You don't actually need to know anything about matrices to answer these questions.

kx+ 2y= 1
3x+ (k-1)y= 1

Try to solve the pair of equations. For what values of k can you NOT get a single (unique) solution? If you can get a single solution what is it?

Yes, you could write this as a matrix equation. You can solve for a single (unique) result as long as you can take the inverse of the matrix- as long as the determinant is not 0. What value of k makes the determinant 0?
 
  • #12
Also keep in mind that for a system of Linear Equations, there are only 3 possibilities:
1) No Solutions
2) 1 Solution (The "Unique" Solution)
3) Infinite Number of Solutions

~~
 
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