Let f: R^3 rightarrow R^3 with f(x, y, z) = (x - 2y + z, -2x - 2y + 2z, -3x - 3y
ID: 3030342 • Letter: L
Question
Let f: R^3 rightarrow R^3 with f(x, y, z) = (x - 2y + z, -2x - 2y + 2z, -3x - 3y + 3z) Does (1, 2, 3) belong to ker (f)? Does (1, 2, 3) belong to im(f)? Let f be as in question 4. Find a basis for im(f). Find a basis for ker(f). Check that your answers to (a) and (b) agree with the dimension theorem. You have received the message 1100011 coded by (7, 4)-Hamming. What 4-bit word did the sender want to convey? Explain your answer. Describe the following maps as composites of rotations, reflections, stretch maps, or projections. f: R^2 rightarrow R^2 with f(x, y) = (2x - 4y, x - 2y). f: R^2 rightarrow R^2 with f(x, y) = (2x - 4y, -x/4 + 2y).Explanation / Answer
. (a) The kernel of f is the set of all vectors v such that f (v)=0 . Now f (1,2,3) =
(1-2*2+3, -2*1 -2*2 +2*3 . -3*1-3*2+3*3)=(0,0,0).Therefore,(1,2,3) Ker(f).
(b) We know that im(f) = {f(x): x R3}. If (1,2,3) Im(f), then for some x,y,z R, we must
have x-2y +z = 1, -2x-2y +2z =2, and -3x -3y +3z = 3. The RREF of the augmented matrix
of this linear system has ( 1,0,-1/3, -1/3) , (0,1,-2/3, -2/3) and (0,0,0,0) as its 1st , 2nd and
3rd rows. Therefore, the given linear system is equivalent to x –z/3 = -1/3, y-2z/3 = -2/3.
Thus, if z = p, then x = p/3 -1/3 and y = 2p/3 – 2/3 so that (x,y,z) = (p/3-1/3, 2p/3 2/3, p),
where p is an arbitrary real number. Thus f(p/3-1/3, 2p/3 2/3, p) = (1,2,3). Therefore
(1,2,3) Im(f).
5. (a) We know that f(x,y,z) = (x-2y +z, -2x-2y +2z, -3x -3y +3z). Now, is A is the matrix with (1,-2,1),
(-2,-2,2) and (-3,-3,3) as its 1st , 2nd and 3rd rows, then the RREF of A has the columns (1,0,0)T,
(0,1,0)T and ( -1/3,-2/3,0)T . The 3rd column is apparently a linear combination of the first two
columns. Therefore, a basis for Im (f) is {(1,0,0)T, (0,1,0)T}.
(b)The kernel of f or Ker(f) is the solution to the equation AX= 0. The RREF of A has the columns
(1,0,0)T, (0,1,0)T and ( -1/3,-2/3,0)T . Thus, the equation AX = 0 is equivalent to x-z/3 = 0 and
y-2z/3 = 0 or x = z/3 and y = 2z/3. Thus (x,y,z) = p( 1/3,2/3,1) where p is an arbitrary real
number. Thus a basis for Ker(f) is { ( 1/3,2/3,1)T}.
( c) . The dimension of Im(f) is 2 and the dimension of Ker(f) is 1. The rank of A is the number of
its columns which is 3. Thus the dimension theorem stands verified.
Please post the remaining questions again separately.
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