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 x1=u and x3=v as the parameters, so the second equation gives x2=u+v.
The first equation then give u+(u+v)−v−x4=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,w3,w5).
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 x1=u and x3=v as the parameters, so the second equation gives x2=u+v.
The first equation then give u+(u+v)−v−x4=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,w3,w5).
No comments:
Post a Comment