How can you find the inverse of a matrix A when given A and AB?

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General question:
How do you determine a matrix B when given A and AB?
 
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If you know the inverse of A, you can multiply on the left side by A^-1 to get B, i.e. (A^-1)AB = B.
 
Oh I see. I understand the computation, but could you explain the theory behind this?
 
This isn't exactly a thorough explanation, but basically if you multiply a matrix by its inverse, you get the identity matrix I. If you multiply any matrix by the identity matrix, you get the original matrix back again, e.g. IA = AI = A.
 
The theory? A guess it's things like "associative law" and "existence of the multiplicative identity"!
If you were given two numbers a, c, a not equal to 0, and told that ab= c, how would you solve for b?

If A has an inverse, then A-1(AB)= (A-1A)B= IB= B.

Notice the condition "If A has an inverse". There are plenty of different matrices, A, B, not having inverses, such that AB= 0. If A does not have an inverse, there may be many different matrices B such that AB= C.
 
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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