How to find the determinant of this matrix?

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    Determinant Matrix
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Homework Statement
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Relevant Equations
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I think you all can see that ##a_{(i+1,j+1)} = a_{i,j} + a_{i+1,j} + a_{i,j+1}##

Now the determinant always give me problem. I have and idea to reduce this matrix by Chio to a 2x2 matrix and find the determinant of this 2x2.

Put i was not able to see any pattern to find what how the 2x2 matrix would be (beside symmetric)

Any tips?
 
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I would first try to write the matrix as a product of two simpler matrices, because the construction rule is similar to matrix multiplication. If this would be too complicated, I'd try the polynomial method: ##\det = \sum_{\sigma\in S_n}(-1)^n \ldots##
 
I want to suggest using row operations to reduce the matrix to something more manageable.

What could be helpful is the following.

Adding or subtracting any two rows of a matrix does not change the determinant.
Exchanging two rows of a matrix changes the sign of the determinant.
 
Another idea is to prove by induction that the determinant equals ##2^{(n^2-n)/2}=2^{\binom n 2}##.
 
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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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