A shear matrix rotates an image like a dilation matrix enlarges/shrinks an image

Question

a) A triangle is created by these coordinates A(1,3) B(5,0) and C(1,-3). Find the measure of angle CBA.

b) Multiply the shear matrix

1	2
0	1

by triangle coordinates A(1,3) B(5,0) and C(1,-3) to create triangle A’B’C’. Find the measure of angle C’B’A’.

c) Multiply the shear matrix

1	3
0	1

by triangle coordinates A(1,3) B(5,0) and C(1,-3) to create triangle A’’B’’C’’. Find the measure of angle C’’B’’A’’.

d) Compare the measure of angles CBA, C’B’A’ and C’’B’’A’’. Does the shear matrix modify the angle of the triangle? Explain why or why not.  

Answer 1

a) There are two ways to determine the angle CBA.

Triangle ABC is isosceles, and the perpendicular from vertex B to AC bisects AC. Call this midpoint N, then tanABN=tanNBC=¾. Since ABN=NBC=½CBA. So CBA=2arctan(¾)=73.74º approx.

AB=BC=5 by Pythagoras Theorem on right triangles ABN and NBC, where angle N is the right angle. Using the cosine rule AC²=AB²+BC²-2AB.BCcosCBA; 36=25+25-50cosCBA; cosCBA=(50-36)/50=0.28. CBA=arccos(0.28)=73.74º.

b) The shear matrix transforms the coords (x,y) to (x+2y,y). So A'=(7,3), B'=(5,0)=B, C'=(-5,-3). From the geometry angle C'B'A'=arctan(3/(5-(-5)))+180-arctan(3/(7-5))=140.39º approx.

c) This time the transformation is (x,y) to (x+3y,y). So A"=(10,3), B"=(5,0), C"=(-8,-3). Angle C"B"A"=arctan(3/(5-(-8)))+180-arctan(3/(10-5))=162.03º approx.

d) The shear matrix considerably modifies the angle because B stays the same throughout (because it lies on the x-axis) while the other two vertices are further displaced from it due to the shear (A is displaced more to the right and C more to the left) The angle increases towards 180º as the shear factor increases. However, the area of each triangle is the same at 12 square units. AC, A'C', A"C" all pass through (1,0) because this point is unaffected by the shear matrix.