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will contain a column with two zero elements, which could conflict with any pairing
we make at this stage.
Let's begin by choosing (
9.21
) and (
9.22
). The solution employs the following
strategy: Given the following matrix equation
a
1
x
y
c
1
c
2
b
1
=
a
2
b
2
then
x
y
1
=
=
c
1
b
1
a
1
c
1
a
1
b
1
.
c
2
b
2
a
2
c
2
a
2
b
2
Therefore, using the 1st and 2nd (
9.21
) and (
9.22
)wehave
v
2
v
3
1
=
=
−
(a
11
−
1
)a
13
a
12
−
(a
11
−
1
)
a
12
a
13
−
a
21
a
23
(a
22
−
1
)
−
a
21
(a
22
−
1
)a
23
v
1
=
a
12
a
23
−
a
13
(a
22
−
1
)
v
2
=
a
13
a
21
−
a
23
(a
11
−
1
)
a
12
a
21
.
Similarly, using the 1st and 3rd (
9.21
) and (
9.23
)wehave
v
3
=
(a
11
−
1
)(a
22
−
1
)
−
v
1
=
a
12
(a
33
−
1
)
−
a
13
a
32
v
2
=
a
13
a
31
−
(a
11
−
1
)(a
33
−
1
)
1
)
−
a
12
a
31
and using the 2nd and 3rd (
9.22
) and (
9.23
)wehave
v
3
=
a
32
(a
11
−
v
1
=
(a
22
−
1
)(a
33
−
1
)
−
a
23
a
32
v
2
=
a
23
a
31
−
a
21
(a
33
−
1
)
v
3
=
a
21
a
32
−
a
31
(a
22
−
1
).
Now we have nine equations to cope with any eventuality. In fact, there is nothing
to stop us from choosing any three that take our fancy, for example these three
equations look interesting and sound:
v
1
=
(a
22
−
1
)(a
33
−
1
)
−
a
23
a
32
(9.24)
v
2
=
(a
33
−
1
)(a
11
−
1
)
−
a
31
a
13
(9.25)
v
3
=
(a
11
−
1
)(a
22
−
1
)
−
a
12
a
21
.
(9.26)
T
. Note that the sign of
Therefore, the solution for the eigenvector is
[
v
1
v
2
v
3
]
v
2
has been reversed to maintain symmetry.
Let's test (
9.24
), (
9.25
) and (
9.26
) with the transforms used above:
