58 CHAPTER 3. VECTOR SPACES

Example 3.2.5 Show that  111

 ,

 133

 ,

 014

is a basis for R3.

There are two things to show, that the set of vectors is independent and that it spans R3.Thus we need to verify that there is exactly one solution to the system of equations

x

 111

+ y

 133

+ z

 014

=

 abc

for any choice of the right side. Recall how to do this. You set up the augmented matrixand then row reduce it.  1 1 0 a

1 3 1 b1 3 4 c

After some row operations, this yields 1 0 0 3

2 a− 23 b+ 1

6 c0 1 0 2

3 b− 12 a− 1

6 c0 0 1 1

3 c− 13 b

Thus there is a unique solution to the system of equations. This shows that the set of vectorsis a basis because one solution when the right side of the system equals the zero vector isx = y = z = 0. Therefore, from what was just done, it is the only solution and so the vectorsare linearly independent. As to the span of the vectors equalling R3, this was just shownalso.

Example 3.2.6 Show that  111

 ,

 113

 ,

 11−4

is not a basis for R3.

You can do it the same way. It is really a question about whether there exists a uniquesolution to the system

x

 111

+ y

 133

+ z

 11−4

=

 abc

