This was a homework problem from Elementary Linear Algebra, Ninth Edition,
Howard Anton and Chris Rorres. Can someone please check it for me. I have
worked through it several times and gotten different answers.
Use Gauss-Jordan elimination to solve for x' and y' in terms of x and y.
x = x'cos(a) - y'sin(a)
y = x'sin(a) + y'cos(a)
First I established an augmented matrix: [ cos(a) -sin(a) | x ]
[ sin(a)
cos(a) | y]
Then I said R1--> cos(a) * R1; R2 --> sin(a) * R2:
s^2(a) -sin(a)cos(a) | x * cos(a)]
[sin^2(a) sin(a)cos(a) | y * sin(a) ]
So R1 --> R2 + R1: [ 1 0 | x cos(a) + y
sin(a)]
[ sin^2(a) sin(a)cos(a) | y
sin(a) ]
Next, R2 --> -sin^2(a) R1 + R2: [ 1 0 | x cos(a)
+ y sin(a) ]
[ 0 sin(a) cos(a)
| -xsin^2(a) cos(a) - y sin^3(a) + y sin(a)]
So R2 --> 1/(sin(a)*cos(a)) * R2 [ 1 0 | x cos(a) + y
]
[ 0 1 | -x
sin(a) - y sin(a) tan (a) + y / cos (a) ]
I think this is right, but would like someone to double check it for me.
Thanks for your time.
Respectfully,
Dustin
Paul Sperry - 22 Jan 2006 02:18 GMT
> This was a homework problem from Elementary Linear Algebra, Ninth Edition,
> Howard Anton and Chris Rorres. Can someone please check it for me. I have
[quoted text clipped - 33 lines]
> Respectfully,
> Dustin
Yes, that seems to be correct (but virtually impossible to read).
Note a little algebra gets y' = ycos(a) - xsin(a).
If you start over but multiply row 1 by sin(a) and row 2 by cos(a), you
can get y' easily.

Signature
Paul Sperry
Columbia, SC (USA)
standelds - 22 Jan 2006 02:34 GMT
Thank you for your time. Sorry it got so scrambled - it looked fine when I
was typing the message in.
Dustin
>> This was a homework problem from Elementary Linear Algebra, Ninth
>> Edition,
[quoted text clipped - 45 lines]
> If you start over but multiply row 1 by sin(a) and row 2 by cos(a), you
> can get y' easily.
G.E. Ivey - 25 Jan 2006 18:08 GMT
Are you aware that that transformation is "rotation about the origin by angle a"? What is the inverse of that transformation?