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and also
⎛
⎝
1
1
1
⎞
⎠
=
1
2
⎛
⎝
3
1
2
⎞
⎠
−
1
2
⎛
⎝
1
−
1
0
⎞
⎠
,
⎛
⎝
2
0
1
⎞
⎠
=
1
2
⎛
⎝
3
1
2
⎞
⎠
+
1
2
⎛
⎝
1
−
1
0
⎞
⎠
so that
⎛
⎝
1
1
1
⎞
⎠
,
⎛
⎝
2
0
1
⎞
⎠
and
⎛
⎝
3
1
2
⎞
⎠
,
⎛
⎝
1
−
1
0
⎞
⎠
span the same subspace.
Therefore we conclude that
⎛
⎝
0
−
4
k
⎞
⎠
must be a linear combination of
⎛
⎝
1
1
1
⎞
⎠
and
⎛
⎝
2
0
1
⎞
⎠
.
We solve the linear system
⎛
⎝
12

0
10
−
4
11

k
⎞
⎠
using GaussJordan elimination to obtain the RREF
⎛
⎝
2
01

2
00
k
−
2
⎞
⎠
Now we see that the system has a unique solution precisely when
−
k
−
2 = 0, i.e.
k
=
−
2.
3. The vectors
⎛
⎜
⎜
⎜
⎝
1
2
1
2
0
⎞
⎟
⎟
⎟
⎠
,
⎛
⎜
⎜
⎜
⎝
0
1
2
1
2
⎞
⎟
⎟
⎟
⎠
are linearly independent because they are not scalar multiples
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This note was uploaded on 04/12/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee

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