Rozdział 2
10. Rozwiązać równania macierzowe z niewiadomą macierząX:
1 0 -2
A = 12 3
a )AX-B= C
A-11 = C + 5
X= A~X(C + B) Wyznaczmy macierz ,4_1
10-2 10 12 3 1 2 ;
-1 -1 -1 -1 -1 -2 + 2-4 + 3 = -1*0
2 3
= (-D
Au = (-D
^13 = (-1)
A21 = (-1)
^22 = (-1)
^23 = (-1)
Asi = (-D
As 2 = (-1)
-1 -1
1 3
-1 -1
1 2 -1 -1
0 -2 -1 -1
1 -2 -1 -1
1 0 -1 -1
0 -2
2 3
1 -2 1 3
1 0 1 2
= 1
= -2
= 1
= 2
= -3
= 1
= 4
= -5
B =
C =
-1 - |
1 - |
1 |
0 2 |
-2 | |
5 1 |
0 | |
0 -3 |
1 | |
-12 |
0 |
6 |
7 |
-2 |
0 |
0 |
0 |
-1 |
^33 = (-1)
= 2
1 -2 1 |
T |
1 2 4 | |
Ad = |
2 -3 1 |
= |
-2 -3 -5 |
4-5 2 |
1 1 2 |
A'1 = (-1)
-2 -3 -5 1 1 2
-1 -1 -2
-12 |
0 |
6 | ||
C + B = |
7 |
-2 |
0 |
+ |
0 |
0 |
-1 |
0 2-2 5 1 0
0 -3 1
12 -1 0 0-3 0
r i <N I T—H 1 1_ |
-12 2 4 | ||
X = |
2 3 5 |
• |
12 -1 0 |
-1 -1 -2 |
0-3 0 |
d) A-XC = BC -XC = 5C-.4 XC = -BC + A X= (~BC + A)C-1 Wyznaczmy macierz C_1
C =
-12 |
0 |
6 |
-12 |
0 |
7 |
-2 |
0 |
7 |
_2 = -24 * 0 |
0 |
0 |
-1 |
0 |
0 |
-2 0 0 -1
= 2
C12 = (-1)
Cl3 = (-1)
C21 = (-1)
C22 = (-1)
C23 = (-1)
C31 = (-1)
= 7
= 0
= 0
= 12
= 0
= 12