Finding the Order of a 2x2 Matrix

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I have a specific 2 by 2 matrix, but obviously I don't want the answer to my actual problem, so can someone just explain to me the general process of finding the order of a 2x2 matrix??
 
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What is the definition of "order of a matrix"?
 
You know as well as I do. The problem I am given states:

"Given a linear transformation given by the matrix equation:

(x',y') = (x,y) (a specific 2x2 matrix with some number b included, which I will leave out for now),

show that this transformation has order 3, no matter what b is."
 
No, I don't know what definition you are using for "order of a matrix" which is what you asked. The only definition I know would say that the order of a 2 x 2 matrix is "2 x 2".

There is, however, an "order" of a LINEAR TRANSFORMATION which you now mention. It is the smallest power, n, such that Tn is the identity. If that is the definition you are using just calculate the square and third power of your matrix. If the square is not the identity but the third power is, then the linear transformation has "order" 3.
 
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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