 ropowiec2gkc

2022-04-03

Find a linear mapping G that maps $\left[0,1\right]×\left[0,1\right]$ to the parallelogram in the xy-plane spanned by the vectors $\left(2,3\right)$ and $\left(4,1\right)$. Lana Hamilton

Step 1
Given:
The maps $\left[0,1\right]×\left[0,1\right]$ to the parallelogram in the XY-plane spanned by the vectors $\left(2,3\right)$ and $\left(4,1\right)$
Step 2
Here, basis vectors are $\left(1,0\right)$$\left(0,1\right)$ of UV-plane mapping linearly to S and T.
The mapping $G\left(u,v\right)=\left(Au+Bv,Cu+Dv\right)$ whereA,B, Cand D are constaint.
$S=G\left(1,0\right)$
$=\left(Au+Bv,Cu+Dv\right)$$=\left(A,C\right)$
and
$T=G\left(0,1\right)$
$=\left(Au+Bu,Cu+Du\right)$
$=\left(B,D\right)$
We have given the vectors  in xy-plane.
$G\left(0,1\right)=\left(B,D\right)N=\left(2.3\right)$
$G\left(1,0\right)=\left(A,C\right)=\left(4,1\right)$
That is
So, the liner mapping G that maps $\left[0,1\right]×\left[0,1\right]$ to the parallelogram in the XY-plane spanned by the vectors
$\left(2,3\right)$ and $\left(4,1\right)$ is $G\left(u,v\right)=\left(4u+2v,u+3v\right)$ mhapo933its

Step 1
A simpler region in the uv-plane can be linearly mapped to a region described by two vectors r and s.
Step 2
r and s describe a parallelogram region. Step 3
Recall how the basis vectors $i=⟨1,0⟩,j=⟨0,1⟩i$ of the uv-plane map linearly tor and s. Write the vectors like this to see what is A, B, C, D:
$r=G\left(1,0\right)=⟨A,B⟩=⟨4,1⟩$
$s=G\left(0,1\right)=⟨C,D⟩=⟨2,3⟩$
Then write the mapping:

Step 4
It can be confirmed by checking the corners of the uv-rectangle $\left[0,1\right]×\left[0,1\right]$:
$G\left(0,0\right)=\left(0,0\right)$
$G\left(1,0\right)=\left(4,1\right)$
$G\left(0,1\right)=\left(2,3\right)$
$G\left(1,1\right)=\left(6,4\right)$
$G\left(u,v\right)=\left(4u+2v,u+3v\right)$