Cho ¡nh x⁄ f : R2 ! R3 (x1; x2) 7! f(x1; x2) = (x1 + 2x2; −x1; 0).
a) Chøng minh f là ¡nh x⁄ tuy‚n t‰nh;
b) T…m ma tr“n cıa f đŁi với cặp cơ sở B = fu1; u2g trong R2 và B0 = fv1; v2; v3g
trong R3 với u1 = (1; 3); u2 = (−2; 4); v1 = (1; 1; 1); v2 = (2; 2; 0); v3 = (3; 0; 0);

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

step-by-step explanation:

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