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I. 2. Write the matrix for reflection in line OA

Thus line of reflection pass through the origin the transformation is linear and could be expressed by 2 x 2 matrix. 1. Reflect base vectors e1, e2. 2. Images e1' and e2' are columns of the matrix of linear transformation. 3. Check the representation by means of the picture "lion".

Fixed points (construction step 6)

Method 1 (Experimental): Move with point B to get location, where B = B'. Discover all such position. Method 2 (Algebraic): Matrix equation of transformation X' = MX together with condition for fixed points X' = X give equation X = MX for unknown fixed point X. Rearrangement in homogenous form: could be solved by Gauss elimination. GeoGebra command ReducedRowEchelonForm(matrix) of the matrix (M - E) returns the equivalent echelon form. Matrix for reflection in line OA, O=(0,0), A =(1, 2) is M = {{-0.6, 0.8},{0.8, 0.6}}, thus

Using back-substitution, unknown coordinates x, y can be solved for. x - 0.5 y = 0, hence y = 2x.

Eigenvectors = Fixed directions (construction step 7)

Method 1 (experimental): Move with dynamic points B, C and investigate the relationship of line BC and its image B'C'. We look for position where line BC is parallel with B'C'. Method 2 (algebraic): Fixed direction (=eigenvectors) X fulfils equation X' = λX for some λ, i.e. we should find coefficient λ for which λX = MX has nonzero solution X. Rearrangement in homogenous form (M-λE)X = o gives the necessary condition for nonzero solution:

det(M-λE) = 0

Change the value of slider λ. Only two values induce singular matrix (M-λE).
  • λ=1 gives direction vector of mirror line (M-E)X = o; x + 2y = 0, X1=(-2t, t)
  • λ=-1 gives f normal vector of mirror line (M+E)X = o; x - 0.5y = 0, X2 = (t, 2t)