I believe @ML2000-LT's answer is correct. I'll try to provide some intuition for why its the case. First, we can notice that the height of each point is not changed, since the image is not vertically stetched. So, we make [0 1] the bottom row. Focusing on the left edge, we see points are shifted more right the higher they are (linearly with height), so we use x'=x+yt for some t>0. This makes the first row [1 t] in our matrix. Using this view, its also easy to see we make t<0 to shear to the left, and use the matrix [1 0], [t 1] to shear vertically.
Staffrishiu
Nice work @ML2000-LT and @CarneAsadaFry! You guys are totally right and great intuition
matrix = [1, t, 0, 1]with t > 0
I believe @ML2000-LT's answer is correct. I'll try to provide some intuition for why its the case. First, we can notice that the height of each point is not changed, since the image is not vertically stetched. So, we make [0 1] the bottom row. Focusing on the left edge, we see points are shifted more right the higher they are (linearly with height), so we use x'=x+yt for some t>0. This makes the first row [1 t] in our matrix. Using this view, its also easy to see we make t<0 to shear to the left, and use the matrix [1 0], [t 1] to shear vertically.
Nice work @ML2000-LT and @CarneAsadaFry! You guys are totally right and great intuition