Map or recover points and finite coordinate sets

Coordinates and the origin

Write the point as a coordinate column. A linear map multiplies that column by a matrix and always fixes the origin. Its first column is the image of the first standard basis vector and its second is the image of the second standard basis vector. A translation adds a vector to the point. A nonzero translation cannot be represented by a 2×2 linear matrix alone

Translate P=(−2,3) by t=(5,−4). Find the image x-coordinate

Original point

\(\displaystyle P=\begin{bmatrix}-2\\3\end{bmatrix}\)

Translation

\(\displaystyle t=\begin{bmatrix}5\\-4\end{bmatrix}\)

Find its image y-coordinate

Read matrix columns, then multiply

The displayed matrix A sends (1,0) to (2,0) and (0,1) to (1,−1). Therefore (x,y) maps to (2x+y,−y). Reflection in y=x exchanges the coordinates; reflection in y=−x exchanges and negates them

Matrix

\(\displaystyle A=\begin{bmatrix}2&1\\0&-1\end{bmatrix}\)

For this A, find the x-coordinate of the image of (−1,3)

Which map can be represented solely by a 2×2 linear matrix?

Invert an affine map

For p′=Ap+t with det(A)≠0, subtract t first and then multiply by A⁻¹. If det(A)=0, different points can share an image and the map has no inverse on the plane

Under (x,y)↦(2x+1,−y+3), the image is (7,5). Find the original y.

Explain and sketch a map, ungraded

Sketch the triangle with vertices (0,0),(2,0),(0,1) and its image under (x,y)↦(2x+1,−y+3). Label corresponding vertices and recover each original point. Explain why this is affine but not linear. The sketch and reasoning are ungraded

Model: map all vertices

The original vertices are (0,0), (2,0) and (0,1). Their images are (1,3), (5,3) and (1,2). The origin maps to (1,3), so this is not linear. The displayed A is the matrix part and t is the translation column. Corresponding straight edges join the corresponding images

Transform a function graph into a relation, ungraded

Start with the full graph v=u². Translate every point by (2,−1), then reflect in the line y=x. Derive an equation for each image and decide whether each is a function of its horizontal coordinate. Sketch both images, labelling their vertices and the images of (−1,1), (0,0) and (1,1). Write the composite map and its inverse, then show that the inverse recovers the whole original graph. The sketch and reasoning are ungraded

Model: use the inverse map in the curve equation

Start with the displayed square function. Translate by (2,−1), then reflect in the line where the coordinates are equal. After translation the image is y=(x−2)²−1, a function with vertex (2,−1). Reflection exchanges coordinates, giving x=(y−2)²−1 with vertex (−1,2). This is y=2±√(x+1), x≥−1, so it is not a function of x: at x=0 there are outputs 1 and 3. The three translated points are (1,0), (2,−1), (3,0); their reflected images are (0,1), (−1,2), (0,3). With the displayed F and t, the composite is q=F(p+t), not Fp+t. Thus (u,v) maps to (X,Y)=(v−1,u+2). Its inverse is p=Fq−t=(Y−2,X+1). If X=(Y−2)²−1, the recovered point satisfies v=X+1=(Y−2)²=u². Conversely every original point maps to that relation, so no branch is lost. Draw two smooth sideways branches through the labelled points, not just one square-root branch