Find Inverse of Matrix Mod 2: Solving the Puzzle

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1 1 1
1 1 0
1 0 0

Find the inverse of this matrix Mod 2

For the standard inverse I got

0 0 1
0 1 -1
1 -1 0

Now do I change all those -1's to 1? Or should I have done that at each step and not leave it to the last?

Is finding the inverse mod n, the same as saying find the inverse over the field k, where k = Fn, where all values correspond to an integer in {0,1,2,..,n-1}? i.e in F3, {0,1,2} the number -1 would correspond to 2.
 
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If you change the -1's in the second matrix to 1's, and then multiply by the first matrix, you get I3. For your first question, I don't believe it makes a difference when you change, at intermediate steps or the end.
 
What is -1? That is, what is the additive inverse of 1 (mod 2)? That is Mark44's point.
 
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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