- #1
jinksys
- 123
- 0
Let S = {v1, v2, v3, v4, v5}
v1 = <1,1,2,1>
v2 = <1,0,-3,1>
v3 = <0,1,1,2>
v4 = <0,0,1,1>
v5 = <1,0,0,1>
Find a basis for the subspace V = span S of R^4.
----
My attempt:
I place the five vectors into a matrix, where each vector is a row of the matrix.
I solve for row-echelon (not RREF). I get:
Therefore the basis of V = span S of R4 contains the vectors:
<1,1,2,1>, <0,1,5,0>, <0,0,1,0>, <0,0,0,1>
---
Question, besides the question "Is this correct?", I'd like to know the difference between solving for REF and RREF. I have two textbooks that I'm using to get through LinAlg and they differ on this section. One tells me to solve for RREF and one REF. Using the former method I get the identity matrix for this problem, and using the latter I get the basis above.
v1 = <1,1,2,1>
v2 = <1,0,-3,1>
v3 = <0,1,1,2>
v4 = <0,0,1,1>
v5 = <1,0,0,1>
Find a basis for the subspace V = span S of R^4.
----
My attempt:
I place the five vectors into a matrix, where each vector is a row of the matrix.
I solve for row-echelon (not RREF). I get:
Code:
1 1 2 1
0 1 5 0
0 0 1 0
0 0 0 1
0 0 0 0
Therefore the basis of V = span S of R4 contains the vectors:
<1,1,2,1>, <0,1,5,0>, <0,0,1,0>, <0,0,0,1>
---
Question, besides the question "Is this correct?", I'd like to know the difference between solving for REF and RREF. I have two textbooks that I'm using to get through LinAlg and they differ on this section. One tells me to solve for RREF and one REF. Using the former method I get the identity matrix for this problem, and using the latter I get the basis above.
Last edited: