(b) Here we have to find 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 off
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
3
x
1
−
5
x
3
= 0
3
x
2
−
2
x
3
= 0
and corresponds to the matrix (i.e. linear transformation)
T
=
3
0
−
5
0
3
−
2
.
8. We first write the augmented coeﬃcient matrix and then perform GaussJordan elimi
nations (row operations):
⎛
⎝
1
1
1

1
1
2
k

2
1
4
k
2

3
⎞
⎠
subtract row I from row II
subtract row I from row III
⎛
⎝
1
1
1

1
0
1
k
−
1

1
0
3
k
2
−
1

2
⎞
⎠
subtract row II from row I
subtract 3 times row II from row III
⎛
⎝
1
0
−
k
+ 2

0
0
1
k
−
1

1
0
0
k
2
−
3
k
+ 2

−
1
⎞
⎠
Let us observe that
k
2
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 Fall '08
 lee
 3k, Row, 1 2 k, 0 1 k, 2 0 0 k

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