1. Obliczyć wskaźnik uwarunkow ania macierzy A cond(A) = ||A|| ||A_l||
-1 0 4
A =
0-2 0 2 0 1
'-1 |
0 2 |
'-i |
1 <N O |
f-1 0 |
4 |
1 Ul O 1 ro _i | |||
Ar = |
0 |
-2 0 |
, ATA = |
0 |
-2 0 |
<N 1 O |
0 |
— |
0 4 0 |
4 |
0 1 |
4 |
0 1 |
L 2 0 |
1 |
-2 0 17 | |||
T Wartości własne macierzy A |
A: | ||||||||
'5-/1 |
0 |
-2 | |||||||
det(/ł |
rA- |
XI) = det |
0 4 |
- X |
0 |
= 0 => (5 |
-X)(4 | ||
-2 |
0 |
17-A | |||||||
(4 - X)[(5 |
-X)(17- |
r>-4] = |
0 |
=> |
II << |
(5 - X)(17-X.) - 4 - 81 - 22X + X2 = 0 Norma ||A|| = Jma\X = 4,162
X2 = 4.675, h= 17,325
A"1:
Macierz dopełnień macierzy A:
O 1 |
0 0 |
(N 1 O | ||
0 1 |
2 1 |
2 0 |
0 4 0 1
-1
2
-1 0
2 0
-1 0
[A\ =
4
1
0 4 |
-1 4 | ||
-2 0 |
0 0 |
Macierz dołączona: adj(A) = [A] =
-2 0 4
0-9 0 8 0 2
-2 | |||
"_2 |
0 |
8 | |
T . |
0 |
-9 |
0 |
4 |
0 |
2 |
Wyznacznik macierzy A: det(A) = (-1 )(-2) 1 - 2 (-2) 4=18
Macierz odwrotna: A 1 =
adj(A)
det(/ł)
0
-9
18
0
9
— 0
0
i
4
9
0
l
9
_2
HT
0
4
18
18
0
2_
18
9
- 0
(A-')rA-' =
-I o ii |
[-i 0 i" |
1- <N 1 e 'O i_ | |
9 9 0 -- 0 ? i 0 i |
9 9 0 -- 0 2 2 ~ 1 0 |
81 81 0-0 4 -2 0 17 | |
— | |||
9 9j |
9 9 J |
—1 00 0© _1 |
-KT *-l
Wartości własne macierzy (A ) A
1