Finding the Adjoint of a Matrix

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hey guys, just need a little help. say your got a matrix
1 2
3 4

how do you find the adjoint, i can get the right numbers but i can't work out y the signs change. any help would be helpful. thanks guys.
 
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The adjoint is just the transpose, isn't it? Or are you talking about the adjugate? They are different.
 
i think its the transpose of the cofactor matrix
but i don't know how to get the cofactor matrix?? thanks
 
That's actually the adjugate not the adjoint. Try the wikipedia entry http://en.wikipedia.org/wiki/Adjugate It's pretty good and not that hard. You can also look up cofactor matrix from the same source. Also nothing wrong with the entry. You can save me a lot of writing that way.
 
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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