Determinants as Area or Volume

Lonely Lemon
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Homework Statement



If S is a parallelepiped determined by v1=(1, 1, 0) and v2= (3, 2, 1) and v3=(6, 1, 2) and T: R3--> R3 by T(x)=Ax, find the volume of T(S)

Homework Equations



{volume of T(S)}=|det A|.{volume of S}

The Attempt at a Solution



A is [v1 v2 v3] and the |A| = 9 by my calculations. I thought this was the volume, but the answer to the questions is given as 24. Please help!
 
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So the vectors define both the solid S and the mapping A? Just want to make sure you've described the problem as assigned.
 
I assume so, that's the problem posed word for word. T: R3-->R3 is the linear transformation determined by a 3x3 matrix A, and S is the a parallelepiped in R3, so the vectors define both?
 
If that's true, then shouldn't the volume of T(S) be a perfect square? By the way, I get det A=3, so you might want to recheck your calculations.
 
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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