How can I describe geometrically the transformation ACA-1 ?

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If A represents the matrix of a rotation of 45 degrees anti-clockwise around (0,0), B a shear with x-axis invariant and shear factor of 2, and C a reflection in the x-axis.

How can I describe geometrically the transformation ACA-1 ?

My answer is a rotation of 90 degrees anti-clockwise around (0,0) but surely there is more than that? :frown:
 
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Natasha1 said:
If A represents the matrix of a rotation of 45 degrees anti-clockwise around (0,0), B a shear with x-axis invariant and shear factor of 2, and C a reflection in the x-axis.

How can I describe geometrically the transformation ACA-1 ?

My answer is a rotation of 90 degrees anti-clockwise around (0,0) but surely there is more than that? :frown:

There is no B in this problem? Think about what happens to the unit vectors in the x and y directions. Since A corresponds to a 45 degree rotation anti-clockwise, A-1 corresponds to its opposite, a 45 degree rotation clockwise. The vector i is changed by A-1 to the vector \frac{sqrt{2}}{2}(i- j), then by C to \frac{sqrt{2}}{2}(i+ j), then by A to j. Okay, that's been rotated 90 anti-clockwise. The vector j is changed by A<sup>-1</sup> to \frac{\sqrt{2}}{2}(i+ j)[\itex], then by C to \frac{\sqrt{2}}{2}(i- j) and finally by A to i. No, that's <b>not</b> a rotation by 90 degrees anti-clockwise! It is a reflection about what line?
 
HallsofIvy said:
There is no B in this problem? Think about what happens to the unit vectors in the x and y directions. Since A corresponds to a 45 degree rotation anti-clockwise, A-1 corresponds to its opposite, a 45 degree rotation clockwise. The vector i is changed by A-1 to the vector \frac{sqrt{2}}{2}(i- j), then by C to \frac{sqrt{2}}{2}(i+ j), then by A to j. Okay, that's been rotated 90 anti-clockwise. The vector j is changed by A<sup>-1</sup> to \frac{\sqrt{2}}{2}(i+ j)[\itex], then by C to \frac{\sqrt{2}}{2}(i- j) and finally by A to i. No, that's <b>not</b> a rotation by 90 degrees anti-clockwise! It is a reflection about what line?
<br /> <br /> line y = x <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/redface.gif" class="smilie" loading="lazy" alt=":redface:" title="Red Face :redface:" data-shortname=":redface:" />
 
Way to go!
 
That done I need to show that if P' is the image of P under D=BCB-1. If P is not on the x-axis, then PP' is bisected by the x-axis and is at a constant angle to the x-axis, for any choice of P?

I can visually see what's happening and can see that the angle is 90 degrees but how can I show it? :rolleyes:
 
Could anyone help me with the last question? Please:cry:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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