How do I find the values of b and y?[A][b = [3 y] 4]

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How do I find the values of b and y?
[A][b = [3
y] 4]
 
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You are going to format that better, aren't you?
 
[A][b y]=[3 4]
y is below b
4 is below 3
 
And what is A?
 
That's a lot like asking: "What is Ax= 4?" without telling us what A is. The only possible answer is "x= 4/A" or "X= A-14".

The only possible answer to your question is
\left[\begin{array}{c} b \\ y \end{array}\right]= A^{-1}\left[\begin{array}{c} 3 \\ 4\end{array}\right]
until we know what "A" is.

Of course, if the problem really was what you said, that may be the correct answer!
 
thats what i thought. it was a question in an exam i recently had.
 
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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