Solutions to quiz
#
5 (October 7)
1.
Writing
X
=
bracketleftbigg
a b
c d
bracketrightbigg
, we obtain
bracketleftbigg
1
3
3
1
bracketrightbigg bracketleftbigg
a b
c d
bracketrightbigg
=
bracketleftbigg
a b
c d
bracketrightbigg bracketleftbigg
1
3
3
1
bracketrightbigg
⇐⇒
bracketleftbigg
a
+ 3
c b
+ 3
d
3
a
+
c
3
b
+
d
bracketrightbigg
=
bracketleftbigg
a
+ 3
b
3
a
+
b
c
+ 3
d
3
c
+
d
bracketrightbigg
⇐⇒
a
+ 3
c
=
a
+ 3
b
b
+ 3
d
=
3
a
+
b
3
a
+
c
=
c
+ 3
d
3
b
+
d
=
3
c
+
d
⇐⇒
b
=
c
a
=
d.
Answer:
X
=
bracketleftbigg
a b
b a
bracketrightbigg
,
where
a
and
b
can be any numbers.
2.
To find a basis of the kernel, we solve the system of linear equations:
x
1
x
2
x
3
x
4
1
2
0
3

0
1
2
1
2

0
1
2

1
4

0
→
1
2
0
3

0
0
0
1

1

0
0
0

1
1

0
→
1
2
0
3

0
0
0
1

1

0
0
0
0
0

0
,
from which the kernel consists of the vectors
x
1
=

2
t
2

3
t
1
x
2
=
t
2
x
3
=
t
1
x
4
=
t
1
=
t
1

3
0
1
1
+
t
2

2
1
0
0
,
where
t
1
and
t
2
can be any numbers. Hence vectors
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 Fall '08
 Conger
 Linear Algebra, Algebra, Elementary algebra, Hyperplane

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