HP 15c - Manual (Page 92)

92 Section 4: Using Matrix Operations
92
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1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
6 . 2014 and
6 . 2014
6 . 2014
6 . 2014
6 . 2014
6 . 2014
1 - E X .
Substituting, you find
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00147 . 0
00146 . 0
00146 . 0
00146 . 0
00146 . 1
EX .
Upon checking (using >7), you find that 1/||E
−1 || ≈ 9.9 × 10 −5
which is very small
compared with ||E|| ≈ 1.6 × 10
4
(or that the calculated condition number is large—
||E|| ||E
−1 || ≈ 1.6 × 10 8
).
Choose any row vector u
T
= (1, 1, 1, 1, 1) and calculate
u
T
E
−1
≈ 10,073 (1,1,1,1,1).
Using a = 10
−4
v
T
= a u
T
E
−1
≈ 1.0073 (1,1,1,1,1)
r
T
= (1,1,1,1,1)
|| r
T
E|| ≈ 5 × 10
-4
|| r
T
|| ||E|| ≈ 8 × 10
4
As expected, ||r
T
E|| is small compared to || r
T
|| ||E||.
Now replace the first row of E by
10
7
r
T
E = (1000, 1000, 1000, 1000, 1000)
and the first row of B by 10
7
r
T
B = 10
7
. This gives a new system equation AX=D, where
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0
0
0
0
10
and
1000 1000 1000 1000 1000 7
D A
x y y y y
y x y y y
y y x y y
y y y x y
.