Problem:
Solution:
For part (a), we can use MATLAB to compute the reduced row echelon form quickly.
rref([[2 3 -1;1 -1 0;3 7 -2] [9;1;17]])
That gives
1.00000 0.00000 -0.20000 2.40000
0.00000 1.00000 -0.20000 1.40000
0.00000 0.00000 0.00000 0.00000
which means
$ x - 0.2z = 2.4 $
$ y - 0.2z = 1.4 $
That give the parametrization $ (2.4 + 0.2t, 1.4 + 0.2t, t) $
For part (b), we could use MATLAB again, but it is simple so I will skip and do that manually.
It is obvious that we need two parameters, and it is easy to have $ x_1 = u $ and $ x_3 = v $ as the parameters, so the second equation gives $ x_2 = u + v $.
The first equation then give $ u + (u + v) - v - x_4 = 0 $, so the parametrization is $ (u, u + v, v, 2u) $.
For part (c), the answer is right in front of us if we simply let $ x = w $ be the parameter, so the parametrization is simply $ (w, w^3, w^5) $.
Solution:
For part (a), we can use MATLAB to compute the reduced row echelon form quickly.
rref([[2 3 -1;1 -1 0;3 7 -2] [9;1;17]])
That gives
1.00000 0.00000 -0.20000 2.40000
0.00000 1.00000 -0.20000 1.40000
0.00000 0.00000 0.00000 0.00000
which means
$ x - 0.2z = 2.4 $
$ y - 0.2z = 1.4 $
That give the parametrization $ (2.4 + 0.2t, 1.4 + 0.2t, t) $
For part (b), we could use MATLAB again, but it is simple so I will skip and do that manually.
It is obvious that we need two parameters, and it is easy to have $ x_1 = u $ and $ x_3 = v $ as the parameters, so the second equation gives $ x_2 = u + v $.
The first equation then give $ u + (u + v) - v - x_4 = 0 $, so the parametrization is $ (u, u + v, v, 2u) $.
For part (c), the answer is right in front of us if we simply let $ x = w $ be the parameter, so the parametrization is simply $ (w, w^3, w^5) $.
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