Solve Simultaneous Equations: No/1/Inf Solns?

  • Context: MHB 
  • Thread starter Thread starter Guest2
  • Start date Start date
  • Tags Tags
    Simultaneous equations
Click For Summary

Discussion Overview

The discussion revolves around solving a system of simultaneous equations involving a parameter $\mu$. Participants explore conditions under which the system has no solutions, one solution, or infinitely many solutions, utilizing methods such as Gaussian elimination and matrix representation.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the system of equations and their solution using Gaussian elimination, expressing concern about the conditions for different types of solutions based on the parameter $\mu$.
  • Another participant questions the ability to derive a condition on $\mu$ from the determinants of the matrices involved.
  • A later reply suggests that there is no condition on $\mu$ as stated in the question, indicating a potential misunderstanding or lack of clarity in the problem setup.
  • One participant claims to have found that $\mu = 9$ is the only value for which the system has a solution, acknowledging a previous error in their reasoning.
  • Another participant confirms that the determinant of the matrix is zero, implying that the system may not have a unique solution.

Areas of Agreement / Disagreement

Participants express differing views on the conditions for $\mu$ and whether the problem as stated provides sufficient information to determine those conditions. There is no consensus on the implications of the determinant or the correct interpretation of the problem.

Contextual Notes

Participants note limitations in deriving conditions from the determinants and the potential ambiguity in the problem statement regarding the parameter $\mu$.

Guest2
Messages
192
Reaction score
0
$x+2y-z=2 \\
x-y+z =5 \\
3x+3y-z=\mu$

The question is for what value of the parameter has the equation (a) no solutions, (b) one solution, and (c) infinitely many solutions.

The trouble when I solve this system using guassian elimination I get $

(x, y, z) = ( -19-\frac{1}{2}\mu, 7, -7-\frac{1}{2}\mu)$
 
Physics news on Phys.org
Guest said:
$x+2y-z=2 \\
x-y+z =5 \\
3x+3y-z=\mu$

The question is for what value of the parameter has the equation (a) no solutions, (b) one solution, and (c) infinitely many solutions.

The trouble when I solve this system using guassian elimination I get $

(x, y, z) = ( -19-\frac{1}{2}\mu, 7, -7-\frac{1}{2}\mu)$

Write the system in matrix form as

$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 5 \\ \mu \end{matrix} \right] \end{align*}$

The system will have unique solution if $\displaystyle \begin{align*} \left| \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix} \right| \neq 0 \end{align*}$ and either no solution or infinite solutions if $\displaystyle \begin{align*} \left| \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix}\right| = 0 \end{align*}$.
 
But those determinants won't give me a condition on $\mu$.

Is there a way to read this off from the reduced row echelon form?
 
Guest said:
But those determinants won't give me a condition on $\mu$.

Is there a way to read this off from the reduced row echelon form?
As the question is stated there is no condition on [math]\mu[/math].

-Dan
 
topsquark said:
As the question is stated there is no condition on [math]\mu[/math].

-Dan
I found that $\mu = 9$ - and wolfram says that's the only value for which the system has solution.
 
Guest said:
I found that $\mu = 9$ - and wolfram says that's the only value for which the system has solution.
(Doh) That just shows that thinking the problem backward (as I did) is no substitute for thinking of it by simply solving the problem.

My apologies and thanks for the catch.

-Dan
 
I reduced the matrix to row reduced echelon form

$A =\begin{pmatrix} 1& 0& 1/3& 0\\ 0 &1 &-2/3 &0\\ 0& 0& 0& 1\end{pmatrix}$
 
Last edited:
If I compute the determinant I get det(A) = 0.

Could someone please explain what the answer to the problem is.
 

Similar threads

Replies
1
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
1K