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(ii) Let
B

A

αI
2
. Then
B
2
= (
A

αI
2
)
2
=
A
2

2
αA
+
α
2
I
2
=
A
2

(
a
+
d
)
A
+ (
ad

bc
)
I
2
,
But by problem 3, chapter 2.4,
A
2

(
a
+
d
)
A
+ (
ad

bc
)
I
2
= 0, so
B
2
= 0.
(iii) Now suppose that
B
6
= 0. Then
BE
1
6
= 0 or
BE
2
6
= 0, as
BE
i
is the
i
–th column of
B
. Hence
BX
2
6
= 0, where
X
2
=
E
1
or
X
2
=
E
2
.
(iv) Let
X
1
=
BX
2
and
P
= [
X
1

X
2
]. We prove
P
is non–singular by
demonstrating that
X
1
and
X
2
are linearly independent.
Assume
xX
1
+
yX
2
= 0. Then
xBX
2
+
yX
2
=
0
B
(
xBX
2
+
yX
2
)
=
B
0 = 0
xB
2
X
2
+
yBX
2
=
0
x
0
X
2
+
yBX
2
=
0
yBX
2
=
0
.
Hence
y
= 0 as
BX
2
6
= 0. Hence
xBX
2
= 0 and so
x
= 0.
Finally,
BX
1
=
B
(
BX
2
) =
B
2
X
2
= 0, so (
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
 Fall '10
 Dreibelbis

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