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=
Finally, the equations
Ux
y
,or
⎡
⎣
⎤
⎦
1 4
1
7
U
y
=
0 2
−
2
6
00
−
9
18
are solvedbyback substitution. This yields
18
−
x
3
=
9
=−
2
6
+
2
x
3
2
6
+
2(
−
2)
x
2
=
=
=
1
2
x
1
=
7
−
4
x
2
−
x
3
=
7
−
4(1)
−
(
−
2)
=
5
EXAMPLE 2.6
Compute Choleski's decomposition of the matrix
⎡
⎣
⎤
⎦
4
−
22
A
=
−
22
−
4
2
−
4
11
Solution
First we note that
A
issymmetric. Therefore, Choleski's decompositionis
applicable, provided that the matrix is also positive definite.An
a priori
test forposi-
tive definiteness is not needed,since the decompositionalgorithm contains its own
test:if the square root of anegative numberisencountered, the matrix is not positive
definite and the decomposition fails.
Substituting the givenmatrix for
A
in Eq. (2.16), weobtain
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
4
−
22
L
11
L
11
L
21
L
11
L
31
L
21
+
L
22
−
22
−
4
L
11
L
21
L
21
L
31
+
L
22
L
32
L
31
+
L
32
+
L
33
2
−
4
11
L
11
L
31
L
21
L
31
+
L
22
L
32
Equating the elements in the lower (or upper)triangular portions yields
√
4
L
11
=
=
2
L
21
=−
/
L
11
=−
/
=−
2
2
2
1
L
31
=
2
/
L
11
=
2
/
2
=
1
2
2
L
21
=
L
22
=
−
−
1
2
=
1
−
4
−
L
21
L
31
L
22
−
4
−
(
−
1)(1)
L
32
=
=
=−
3
1
11
11
L
31
−
L
32
=
L
33
=
−
−
(1)
2
−
(
−
3)
2
=
1
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