Question #1452

A = 4 2
-5 -2
Find an invertible 2x2 matrix Q such that Q^-1AQ is equal to a rotation followed by scaling by a positive real number, and write down both the rotation matrix and the scaling matrix. What is the angle of rotation of the matrix?

Expert's answer

Rotation M = Scaling M

cos(theta) -sin(theta) = s_{x} 0

sin(theta) cos(theta) = 0 s_{y}

Thus

Q= 1 0

0 1

Scalling matrix is the same as well as rotation matrix. As cos(theta) = 1, rotation angle is equal to zero.

cos(theta) -sin(theta) = s

sin(theta) cos(theta) = 0 s

Thus

Q= 1 0

0 1

Scalling matrix is the same as well as rotation matrix. As cos(theta) = 1, rotation angle is equal to zero.

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