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(b) Here we have to fnd a linear system whose solution is
⎛
⎝
x
1
x
2
x
3
⎞
⎠
=
t
⎛
⎝
5
2
3
⎞
⎠
=
⎛
⎝
5
t
2
t
3
t
⎞
⎠
.
From the last row we read o±
t
=
1
3
x
3
so that
x
1
=5
t
=
5
3
x
3
, and
x
2
=2
t
=
2
3
x
3
.
This system can be written more nicely as
½
3
x
1
−
5
x
3
=0
3
x
2
−
2
x
3
and corresponds to the matrix (i.e. linear trans²ormation)
T
=
µ
30
−
5
03
−
2
¶
.
8. We frst write the augmented coeﬃcient matrix and then per²orm GaussJordan elimi
nations (row operations):
⎛
⎝
11 1

1
12
k

2
14
k
2

3
⎞
⎠
subtract row I ²rom row II
subtract row I ²rom row III
⎛
⎝
11
1

1
01
k
−
1

1
k
2
−
1

2
⎞
⎠
subtract row II ²rom row I
subtract 3 times row II ²rom row III
⎛
⎝
10
−
k
+2

0
k
−
1

1
00
k
2
−
3
k
−
1
⎞
⎠
Let us observe that
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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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