I have completed the question with explanations, however i would like confirm that my thought process is correct.
Please supply any corrections that i could make.
Question
Determine the solution space, described as the span of a set of vectors, for the following linear system:
1x + 5y + 0z = 0
6x + 7y + 9z = 0
0x + 5y + 5z = 0
State, with justification, the dimension of the solution space.
My Attempt:
Row echelon form of the augmented matrix:
| 1 4 0 0 |
| 0 1 19/3 0 |
| 0 0 1 0 |
therefore :
x = -4y;
y = -19/3z;
z = 0;
Let A = the solution set of the augmented matrix.
A= {(x,y,z) | x,y,z ∈ R^3}
= {(-4y,-19/3z, 0) | x,y,z ∈ R^3}
= {(-4y, 0, 0) + (0, -19/3z, 0) | x,y,z ∈ R^3}
= {y(-4, 0, 0) + z(0, -19/3, 0) | x,y,z ∈ R^3}
= span{(-4, 0, 0), (0, -19/3, 0)}
the solution space is thus a subspace on R^3
I'm second guessing my self on the validity of this, any suggestions will be gladly accepted.
Please supply any corrections that i could make.
Question
Determine the solution space, described as the span of a set of vectors, for the following linear system:
1x + 5y + 0z = 0
6x + 7y + 9z = 0
0x + 5y + 5z = 0
State, with justification, the dimension of the solution space.
My Attempt:
Row echelon form of the augmented matrix:
| 1 4 0 0 |
| 0 1 19/3 0 |
| 0 0 1 0 |
therefore :
x = -4y;
y = -19/3z;
z = 0;
Let A = the solution set of the augmented matrix.
A= {(x,y,z) | x,y,z ∈ R^3}
= {(-4y,-19/3z, 0) | x,y,z ∈ R^3}
= {(-4y, 0, 0) + (0, -19/3z, 0) | x,y,z ∈ R^3}
= {y(-4, 0, 0) + z(0, -19/3, 0) | x,y,z ∈ R^3}
= span{(-4, 0, 0), (0, -19/3, 0)}
the solution space is thus a subspace on R^3
I'm second guessing my self on the validity of this, any suggestions will be gladly accepted.