Clarification on what is considered reduced row-echelon form

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Homework Statement


In a 2x2 matrix are these considered RREF?
(0 0, 0 0) and (0 1, 0 0)
 
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Well, is the definition of RREF?
 
According to wikipedia, it is:
In linear algebra a matrix is in row echelon form if

* All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes, and
* The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it.
 
Are you unsure whether these two matrices have those properties then?

And what is the third condition, for reduced row echelon form?
 
Every leading coefficient is 1 and is the only nonzero entry in its column.

From the looks of it, it looks like it follows those properties, but I'm just confused if it is still rref if everything is 0, since it doesn't have a leading 1.
 
The definition doesn't say that there must be nonzero rows (for your first matrix). In your second matrix
[0 1]
[0 0]
The leading entry in the first row is a 1.
 
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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