A qustion that my solution differs from the answer of the book

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You are misreading the question. They are not asking you to find vector v such that T-1= b. The "-1" is a sub script, not superscript. They have defined Ta and asked you to solve T-1v= b. In other words, Ta with a= -1. Just row reduce the augmented matrix with A itself,, not A inverse.

By the way, if they were asking you to solve A-1v= b when they had given you A, you certainly should not find A-1 and then solve the equation! If you know A, the you know that AA-1v= v= Ab. Just multiply vector b by A.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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