Linear Transformations question

johnnyboy2005
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I'm just wondering if someone can let me know if I'm on the right path here...

this question asks to show that the Function T: R^3 ----> R^2 given by the formula T(X1, X2, X3) = (2X1 - X2 + X3, X2 - 4X3) is a linear transformation.

soln' the definition of a L.T. is that T(u + v) = T(u) + T(V)

and T(cU) = cT(U)

so i show that T[(X1, X2, X3)+(Y1, Y2, Y3)] = T(X1, X2, X3) + T(Y1, Y2, Y3)...but what do i use as Y1, Y2, Y3? thanks for the time and help
 
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Just leave them like that. Using the definition of T, write out explicitely what T[(X1, X2, X3)+(Y1, Y2, Y3)] is, and write explicitely what T(X1, X2, X3) + T(Y1, Y2, Y3) is, and then compare.
 
but what happens to (2X1 - X2 + X3, X2 - 4X3)?? do i just write the X's as Ys?
 
If f(x)= x2, what is f(y)?
 
y as in yes, i got it!
 
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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