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which gives
⎡
⎣
⎤
⎦
D
1
D
1
L
21
D
1
L
31
···
D
1
L
n
1
0
D
2
D
2
L
32
···
D
2
L
n
2
0
0
D
3
···
D
3
L
3
n
U
=
(2.24)
.
.
.
.
.
.
.
0
0
0
···
D
n
We see that during decomposition of a symmetricmatrix only
U
hastobe stored,since
D
and
L
can beeasilyrecovered from
U
. ThusGauss elimination, which results in an
upper triangular matrix of the form shown in Eq. (2.24), issufficienttodecompose a
symmetric matrix.
There is an alternative storage schemethatcan beemployed during
LU
decom-
position. The ideaistoarrive at the matrix
⎡
⎣
⎤
⎦
D
1
L
21
L
31
···
L
n
1
0
D
2
L
32
···
L
n
2
00
D
3
···
L
n
3
U
∗
=
(2.25)
.
.
.
.
.
.
.
00 0
···
D
n
Here
U
can be recovered from
U
i j
=
D
i
L
ji
. It turnsoutthatthis schemeleadstoa
computationallymoreefficient solutionphase; therefore, we adopt itfor symmetric,
bandedmatrices.
Symmetric, Pentadiagonal Coefficient Matrix
Weencounterpentadiagonal(bandwidth
5) coefficient matrices in the solution of
fourth-order,ordinarydifferentialequationsbyfinitedifferences. Oftenthesematrices
aresymmetric, in which case an
n
=
×
n
matrix has the form
⎡
⎣
⎤
⎦
d
1
e
1
f
1
0
0
0
···
0
e
1
d
2
e
2
f
2
0
0
···
0
f
1
e
2
d
3
e
3
f
3
0
···
0
0
f
2
e
3
d
4
e
4
f
4
···
0
=
A
(2.26)
.
.
.
.
.
.
.
.
.
.
0
···
0
f
n
−
4
e
n
−
3
d
n
−
2
e
n
−
2
f
n
−
2
0
···
00
f
n
−
3
e
n
−
2
d
n
−
1
e
n
−
1
0
···
00 0
f
n
−
2
e
n
−
1
d
n
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