. Is it possible to have AB=In without B being the inverse of A?

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Homework Statement



30.
Is it possible to have AB= In without B being the inverse of A? Explain your reasoning.

Homework Equations


The Attempt at a Solution


yes if A=
|1 1 0|
|0 1 0|
B=

|1 -1|
|0 1|
|0 0|
AB=I(2) , BA=! I3 right?
 
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That would be true once A and B do not have the same dimensions (as you showed using an example)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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