HP HP-15C - Page 111

Section 4: Using Matrix Operations 109
Substituting, you find
""1.00146
0.00146
EX- 0.00146
0.00146
0.00147
Upon checking (using I MATRIX] 7), you find that I/HE"1!! «
9.9 X 10~5, which is very small compared with ||E|| « 1.6 X 104 (or
that the calculated condition number is large — ||E|| HE"1!! *»
1.6 X108).
Choose any row vector UT = (1, 1, 1, 1, 1) and calculate
uTE'1~ 10,073(1, 1,1,1,1).
Using a = W
vT=auTE~l « 1.0073 (1, 1, 1, 1, 1)
rr= (1,1, 1,1,1)
||rTE||«5Xl(r4
||rl||E||«8X104.
As expected, ||rTE|| is small compared with ||rT||||E||.
Now replace the first row of E by
107rTE = (1000, 1000, 1000, 1000, 1000)
and the first row of B by 107rTB = 107. This gives a new system
equation AX — D, where
A =
1000 1000 1000 1000 1000
y x y y y
y y x y y
y y y x y
y y y y x
andD =
107
0
0
0
0