MHB Solve System of Equations: Unique Solution & Geometric Explanation

Click For Summary
The discussion focuses on solving a system of equations represented by a matrix and determining the conditions for unique solutions based on the parameter λ. It establishes that for λ ≠ 3, the system has a unique solution, with the intersection of the three planes represented by the equations being a single point. Conversely, when λ = 3, the system has no solution, as the third plane becomes parallel to the intersection line of the first two planes, resulting in an empty intersection. The importance of explicitly stating the geometric implications of the solutions is emphasized, particularly regarding the nature of the intersections. Overall, the analysis clarifies the relationship between the parameter λ and the system's solvability.
Guest2
Messages
192
Reaction score
0
I'm trying to answer the question below in the attachment. Could someone please check my answer to part (c), as I'm not sure. Is it correct? Is that that the right geometric explanation for the planes?

(a) The equation is $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} $

(b) $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} \to \begin{pmatrix}1 & -1& 2& 1 \\ 0 & 3 & -3 & -3\\ 0 & -1 & \lambda -2 & 2 \end{pmatrix} \to \begin{pmatrix}1 & -1& 2& 1 \\ 0 & 1 & -1 & -1\\ 0 & 0 & \lambda -3 & 1 \end{pmatrix}$.

Thus the condition for the system to have a unique solution is $\lambda - 3 \ne 0 \implies \lambda \ne 3$.

And if $\lambda \ne 3$ then $ z = \frac{1}{\lambda-3}$ then $y = \frac{1}{\lambda-3}-1$ and $x = -\frac{1}{\lambda-3}$ is the unique solution to the system.

(c) If $\lambda = 3$ then the system has no solution. For all other values of $\lambda$ there's exactly one unique solution (given above). Geometrically it means their intersection is a point.
 

Attachments

  • Screen Shot 2016-03-09 at 17.58.57.png
    Screen Shot 2016-03-09 at 17.58.57.png
    30.6 KB · Views: 137
Physics news on Phys.org
Guest said:
(a) The equation is $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} $
An equation must have the $=$ sign. In this case it is
\[
\begin{pmatrix}1 & -1& 2\\ 2 & 1 & 1 \\ 1 & -2 & \lambda \end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix}=
\begin{pmatrix}1\\-1\\3\end{pmatrix}.
\]
The matrix you wrote is the augmented matrix of the system of equations.

Guest said:
(c) If $\lambda = 3$ then the system has no solution. For all other values of $\lambda$ there's exactly one unique solution (given above). Geometrically it means their intersection is a point.
Whose intersection? It is clear from the question that you mean the planes defined by the equations, but I would not use only a pronoun in the answer.

You should also probably say explicitly that having infinitely many solutions is not possible. (It would be if the right-hand side of the last equation were 0.) If $\lambda\ne3$, then the tree planes intersect in a single point, and if $\lambda=3$, then the third plane is parallel to the intersection line of the first two planes, so the intersection of all three planes is empty.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K