Extras din seminar
SEMINARUL NR.1 ALGEBRØA LINIARØA 1
an univ. 2006/2007
1. SØa se calculeze determinant¸ii:
a)
2 1 3
3 2 0
2 1 2
b)
2 2 1 1
1 3 3 2
1 0 9 1
3 4 2 0
c)
1 1 0 2
2 1 1 1
3 0 0 1
1 1 2 1
2. SØa se calculeze rangul matricelor:
a)
3 2 1
2 1 1
6 2 4
; b)
1 2 1 1
1 2 1 1
1 2 1 5
;
c)
1 1 1 1 1
3 2 1 1 3
0 1 2 2 6
5 4 3 3 1
; d)
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
3. Folosind metoda eliminØarii sØa se calculeze inversele urmØatoarelor matrice, dacØa
acestea existØa.
a)
2 2 3
1 1 0
1 2 1
; b)
1 -4 -3
1 -5 -3
-1 6 4
; c)
1 1 1 1
1 1 -1 -1
1 -1 -1 1
-1 -1 1 1
d)
1 1 1
0 2 1
1 0 1
; e)
1 0 1 1
0 0 1 0
1 1 1 0
1 0 0 2
4. SØa se rezolve, folosind metoda eliminØarii, urmØatoarele sisteme:
a)
x1 + 2x2 + 3x3 2x4 = 6
2x1 x2 2x3 3x4 = 8
3x1 + 2x2 x3 + 2x4 = 4
2x1 3x2 + 2x3 + x4 = 8
b)
x1 + 2x2 + 3x3 x4 = 1
3x1 + 2x2 + x3 x4 = 1
2x1 + 3x2 + x3 + x4 = 1
2x1 + 2x2 + 2x3 x4 = 1
5x1 + 5x2 + 2x3 = 2
c)
x1 + x2 + x3 + x4 = 1
2x1 x2 + x3 x4 = 2
x1 2x2 2x4 = 1
d)
x1 + x2 + x3 = 2
2x1 3x2 + x3 = 6
4x1 x2 + 3x3 = 10
5. SØa se rezolve sistemele:
SEMINARUL NR.1 ALGEBRØA LINIARØA 2
a)
(
x1 3x2 + x4 = 0
x2 + x3 x4 = 0
, b)
2x1 + 3x2 x3 = 0
x2 + 2x3 3x4 = 0
2x1 4x3 + x4 = 0
c)
2x1 + x2 x3 = 0
3x1 2x2 + x3 = 0
x1 x2 + x3 = 0
6. SØa se studieze compatibilitatea sistemului AX = B, Æ1n caz de compatibilitate
sØa se expliciteze solut¸iile, Æ1n urmØatoarele situat¸ii:
a) A =
2 3 1
3 1 1
1 2 2
, B =
1
3
2
;
b) A =
4 0 3 1
0 3 1 4
0 3 0 1
1 2 0 3
, B =
10
1
17
25
;
c) A =
5 7
3 2
5 5
2 9
11 3
, B =
4
6
10
2
16
;
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- algsem2_rezolvari.pdf
- algsem3_rezolvari.pdf
- algsem4_rezolvari.pdf
- algsem5_rezolvari.pdf
- algsem6_rezolvari.pdf
- als1.pdf
- als10.pdf
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- als3.pdf
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