- #1
robertjford80
- 388
- 0
Homework Statement
The Attempt at a Solution
I don't understand where 2v1 + 3v2
and
4v1 - 3v2
came from.
Coordinates and change of base
- Thread starter robertjford80
- Start date
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- Tags
- Base Change Coordinates
Click For Summary
Homework Help Overview
The discussion revolves around understanding vector equations and the change of basis in a coordinate system. Participants are attempting to clarify the origins of specific vector combinations and their implications in solving linear equations.
Discussion Character
- Exploratory, Conceptual clarification, Problem interpretation
Approaches and Questions Raised
- Participants are questioning the derivation of specific vector expressions and their transposed forms. There is also a request for additional context, such as a figure, to aid understanding. Some are interpreting the equations related to vector combinations and seeking solutions for the coefficients involved.
Discussion Status
The discussion includes various interpretations of the vector equations, with some participants providing insights into the relationships between the coefficients and the vectors. There is a recognition of progress, as one participant indicates they have achieved understanding.
Contextual Notes
One participant notes the absence of a referenced figure, which may be critical for visualizing the problem. The discussion appears to be constrained by the need for clearer examples and definitions related to the vectors involved.
I don't understand where 2v1 + 3v2
and
4v1 - 3v2
came from.
- #2
robertjford80
- 388
- 0
I see that the vectors are raised to T which I'm pretty sure means they're transposed but that still doesn't get me anywhere.
- #3
DryRun
Gold Member
- 837
- 4
Where's Figure 3.29a? You should post the whole example to help us get a clearer picture of the problem.
- #4
HallsofIvy
Science Advisor
Homework Helper
- 42,895
- 983
You are solving the equations
[tex]x= \begin{bmatrix}11 \\ 7\end{bmatrix}= a\begin{bmatrix}1 \\ 2\end{bmatrix}+ b\begin{bmatrix}3 \\ 1\end{bmatrix}[/tex]
which is the same as the two equations a+ 3b= 11 and 2a+ b= 7. Can you solve for a and b?
And
[tex]y= \begin{bmatrix}-5 \\ 5\end{bmatrix}= c\begin{bmatrix}1 \\ 2\end{bmatrix}+ d\begin{bmatrix}3 \\ 1\end{bmatrix}[/tex]
which is the same as the two equations c+ 3d= -5 and 2c+ d= 5. Can you solve for a and b?
[tex]x= \begin{bmatrix}11 \\ 7\end{bmatrix}= a\begin{bmatrix}1 \\ 2\end{bmatrix}+ b\begin{bmatrix}3 \\ 1\end{bmatrix}[/tex]
which is the same as the two equations a+ 3b= 11 and 2a+ b= 7. Can you solve for a and b?
And
[tex]y= \begin{bmatrix}-5 \\ 5\end{bmatrix}= c\begin{bmatrix}1 \\ 2\end{bmatrix}+ d\begin{bmatrix}3 \\ 1\end{bmatrix}[/tex]
which is the same as the two equations c+ 3d= -5 and 2c+ d= 5. Can you solve for a and b?
Last edited by a moderator:
- #5
robertjford80
- 388
- 0
ok, thanks, I got it now.
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