# simple matrix questionWatch

Thread starter 10 years ago
#1
the following matrix represents a rotation about the origin:

a -a
a a

where a = -1/sqrt(2).

find the angle, x, and the direction of the rotation.

right. i've only briefly looked at matrices today, so this might be blindingly obvious. if i'm correct, then cos x = a, sin x = a, sin x = -a and cos x = a. do i just have to find a value of x that appears in the above 4 equations or is it more complicated/simple than that?
0
10 years ago
#2
You're almost, but not quite correct. I assume you're equating a standard rotation matrix with the matrix you've been given. In which case, look closely at the sign of the elements of your standard matrix.
0
Thread starter 10 years ago
#3
(Original post by DFranklin)
You're almost, but not quite correct. I assume you're equating a standard rotation matrix with the matrix you've been given. In which case, look closely at the sign of the elements of your standard matrix.
ah... i didn't think the signs mattered, but i can see why they do now. so should it be "sin x = a" instead of "sin x = -a" in my 3rd equation?
0
10 years ago
#4
Yes. So you just need to solve for sin x = cos x = a.
0
10 years ago
#5
If the matrix represents a rotation of x degrees anticlockwise, you know that:
sin(x) = cos(x) = -1/sqrt(2)
==> x = arcsin(-1/sqrt(2)) = arccos(-1/sqrt(2))

Now you just need to find which values of arcsin(-1/sqrt(2)) and arccos(-1/sqrt(2)) are equal, since both expressions have two solutions.
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