Eliminacja Gaussa i macierze odwrotne

Eliminacja Gaussa i macierze odwrotne
1.
(a)
(c)
2.
Stosuj¡c eliminacj¦ Gaussa, rozwi¡» nast¦puj¡ce ukªady równa«


 x1 +x2 +x3
2x1 +x2 −4x3

 x +2x +7x
1
2
3


 2x1 +x2 −x3
x1
+2x3


3x1 +x2 +x3
[
2 −1
4 −3
(b)

4x1 +5x2 +6x3
x1 +2x2 +3x3
0 −2
−1
5
]

1
 2
(h) 
 −1
3
0
1
2
1
[
(c)

2 −1
0
1 −1 
(e)  0
−1 −1
2

= 6
= 4

x1 +x2 +x3



 −x
−x2 +2x3
1
 x1 −2x2 −x3



3x1 −x2 +2x3

3 −2 −1

−4
3
2 
(g)
8 −6 −3

2 −1
1
3 

1
0 
2 −1
2 1
5 3
=
0
=
3
= −2
= −1
]

1
0 0
(f )  2 −1 0 
−2
1 1


6
3 2
 4
3 1
(i) 
 1 −1 2
3
2 1

1
3 
 .
5 
3
Korzystaj¡c z podanej macierzy odwrotnej do macierzy ukªadu, rozwi¡» poni»sze
równania.
(c)
(d)
[

1 1 1
(d)  1 2 3 
1 3 4
(b)
= 1
= 2
= 3
]

(a)
(b)
Znajd¹ macierze odwrotne do poni»szych (o ile istniej¡).
(a)
3.
{
=
0
=
1
= −1
[
]
2 1
5 3
{


1 2 1
1 4 2
1 3 2

= −3

 2x1 −x2
+x2 −x3 =
1


−x1 −x2 +2x3 =
1

x1 −24x2 −5x3 +32x4



 −x +25x +5x −33x
1
2
3
4

−x1 +37x2 +8x3 −50x4



−5x2 −x3 +7x4

6 3 2
4 3 1

1 −1 2
3 2 1

1
3

5
3
3x1 −x2
−5x1 +2x2
= 0
= 1
= −9
=
15
= −2
=
0
4.
Rozwi¡» równania macierzowe:
[
(a)
[
(c)
2
1
−5 −2
−1 0
−1 2
]
[
X=
]
[
X
2 3
1 5
3
2
−2 −1
[
]
(b) X
]
[
=
0 2
1 4
]
[
(d)
2 3
1 5
3 −2
4
2
[
[
]
=
]
[
X
] [
]2 [
3
2
1 −2
5
(e)
X
=
−2 −1
1
0
−3
[
] [
]
[
1 −1
−1 −2
3
(f )
−
X=
2
0
0
1
0
] [
[
] [
6
2
1
2 −1
=
+
(g) X
−8
0 −2
0
3



−1
0
1
1 −2
1 −1  X =  0 −2 
(h)  0
1
1
1 −2
0




−1 2
0 1
2
(j)  1 1 −1  X =  6 7 
2 5
1 1
1

4 −4
2 −9
]
3
1
−2 −2
−26
14
]
1
3
]
2
8
]
[
=
2 0
−1 3
]

]
[
0 1
2
8 7 −6


(i) X 1 1 −1 =
12 14
2
1 1
1

]