Algebraic/Matrix Manipulation (linear algebra)

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Homework Statement



I have attached the relevant question as an image (for sake of ease)

Homework Equations





The Attempt at a Solution



Also attached, in blue.


Thanks a lot for any help at all!
 

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You could consider ##B^k Y = 0##. Which property does B need to get non-trivial solutions for Y? How is that related to a?
##Y=(B-I_2)X##. How do you get Y=0 (which is a solution to the equation above) with non-trivial X? How is that related to a?
 
Perhaps the very beginning is where I'm having trouble, I don't know what times a matrix would be equal to zero. Is it the transpose? Or perhaps a matrix whose determinant is zero?
 
If you have studied matrices at all then you should know this basic property: the equation Ax= y has a unique solution if and only if A is invertible: x= A^{-1}y. And that is only true if A has non-zero determinant.

The equation Ax= 0 always has the "trivial" solution, x= 0. It has other solutions if and only if its determinant is 0.
 
You're right, I should know that basic property :) I've been cramming too much this semester so I tend to forget.

Thanks a lot it makes good sense.
 
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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