Anyone know a good site that explains matrices, no solutions, infite, etc?

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Hello everyone, Anyone know any good tutorials that can explain to me how i can perform the following:
http://img499.imageshack.us/img499/9744/lastscan1jb.jpg
I know the basic jist of it, like all 000's in a row means infite, and that's about it :eek:
 
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Start with the Ax part. What does Ax mean, can you describe?
 
mr_coffee said:
I know the basic jist of it, like all 000's in a row means infite, and that's about it :eek:
Can you find a link with the determinant of the coefficient matrix?
 
Well A is a matrix and x = <x,y,z> if its a 3x3; when b is the value of the matrix. TD, i found lots of homework problems but no solutions :(
 
Have you seen the concept 'rank'? If the rank of the coefficient matrix is equal to the rank of the augemented matrix, then the system Ax = B has solutions.

Link with the determinant: if det(A) = 0 and A is an n x n matrix, then rank(A) < n, or: A is a singular matrix. If det(A) =! 0, then A is a regular matrix and its rank is n.
 
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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