Matrix Invertibility: RREF to Identity

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Prove that a matrix A is invertible if and only if its reduced row echelon row is the identity matrix.
 
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Even though I was never taught linear algebra fully, to do this problem I would consider what would make the matrix A invertible and what would it mean if the RRE form wasn't the identity matrix.

But I am not sure if that would be a valid proof.
 


This isn't too hard to prove. You can start by asking yourself what a row operation on a matrix translates to in matrix algebra. And what do the matrices corresponding to the row-operations amount to when they row-reduce A to I?

As for the "forward" conjecture, well I can think of something some might find objectionable. If it does not row-reduce A to I, it the RRE form has a row of zeros. That means that the determinant is 0 and hence it is not invertible. I'm sure there's a better way to do this.
 

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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