-1,
1. Obliczyć wskaźnik uwarunkowania macierzy A cond(A) = ||A|| ||A
-1 0 4
A= 0 -2 0
2 0 1
det(Ar A - A/) = det
0
-2
4-/1
0
-2
0
17-A
= 0 => (5 - A.)(4 - A.)(l7 -
(4-A)[(5-A)(17-A)-4] = 0 => A, =4
(5 - A)(17 - A) - 4 = 81 - 22A + A2 = 0 Norma ||A|| = Vmax A = 4,162
A2 = 4,675,
_ 4(4 _ X) = 0 A3= 17,325
A :
Macierz dopełnień macierzy A:
-2 0 |
0 0 |
0 -2 | ||
0 1 |
2 1 |
2 0 |
0 4
-1 0
2 0
1 0
[A] =
1 4
0 |
l |
2 |
1 | ||
0 |
4 |
-1 4 | |||
-2 |
0 |
0 |
0 |
Macierz dołączona: adj(A) = [A\ =
-2
0
8
_2 | |||
"_2 |
0 |
8 | |
T _ |
0 |
-9 |
0 |
4 |
0 |
2 |
'-1 |
0 |
2' |
"-1 |
0 |
2 |
'-1 |
0 |
4 | |||
II |
0 |
-2 |
0 |
, ata = |
0 |
-2 |
0 |
0 |
_2 |
0 |
= |
4 |
0 |
1 |
4 |
0 |
1 |
2 |
0 |
1 |
5 0-2
0 4 0
-2 0 17
Wartości własne macierzy AT A: '5-/1 0
Wyznacznik macierzy A: det(A) = (-l)(-2) 1 - 2 (-2) 4=18
adj(A)
det(M)
'-2 |
0 |
8 ' |
1 |
0 |
4" | |
78” |
18 |
9 |
9 | |||
0 |
-9 18 |
0 |
= |
0 |
1 2 |
0 |
4 .18 |
0 |
SS | ^ 1_ |
2 . 9 |
0 |
1 9 |
Macierz odwrotna: A =
{a~')ta=
1 |
0 |
2' |
1 |
0 |
4' |
5 |
0 |
2 ' | |
9 |
1 2 |
9 |
9 |
1 2 |
9 |
81 |
1 4 |
sl | |
0 |
0 |
0 |
0 |
= |
0 |
0 | |||
4 |
0 |
1 |
2 |
0 |
1 |
2 |
0 |
17 | |
9 |
9 |
. 9 |
9 |
~81 |
8T _ |
Wartości własne macierzy (A ')r A ':