LINEARNA ALGEBRA 1-pismeni dio ispita-

-grupa A-19. 6. 2014.

1. (5 poena) Odrediti rang normalnu formu i odgovarajue matrice S i P matrice

A =

10 13 3 22 2 0 82 3 1 63 4 1 2

.2. (6 poena) Izraqunati determinantu

Dn =

n 1 0 0 . . . 0 0n 1 x 1 0 . . . 0 0n 2 0 x 1 . . . 0 0n 3 0 0 x . . . 0 0...

......

... . . ....

...2 0 0 0 . . . x 11 0 0 0 . . . 0 x

.

3. (6 poena) Nai bazu od L1 + L2 i L1 L2 ako je

L1 = Lin{(1, 0, 2, 1), (3, 1,2, 4), (2, 1, 0, 5)},

L2 = Lin{(0,1, 1,1), (2, 1, 3, 0), (2, 0, 4,1)}.4. (6 poena) Linearan operator A : R3 R3 u standardnoj bazi {e1, e2, e3} ima matricu

A =

15 11 512 13 11 7 9

.Odrediti mu matricu A1 u bazi

f1 = 2e1 + e2 + 2e3,

f2 = 3e1 e2 + e3,f3 = e1 + 5e2 3e3.

5. (6 poena) Dokazati da je operator A : R3 R3 definisan sa

A(x1, x2, x3) = (x1 + x2 x3, 2x1 + x3, x1 x2 + x3)

linearan, a zatim mu odrediti bazu i dimenziju jezgra. Izraqunati rang(A).

LINEARNA ALGEBRA 1-pismeni dio ispita-

-grupa B-19. 6. 2014.

1. (5 poena) Odrediti rang normalnu formu i odgovarajue matrice S i P matrice

A =

10 2 3 213 2 4 32 8 2 63 0 1 1

.2. (6 poena) Izraqunati determinantu

Dn =

1 0 0 0 . . . 0 x2 0 0 0 . . . x 1...

......

... . . ....

...n 3 0 0 x . . . 0 0n 2 0 x 1 . . . 0 0n 1 x 1 0 . . . 0 0n 1 0 0 . . . 0 0

.

3. (6 poena) Nai bazu od L1 + L2 i L1 L2 ako je

L1 = Lin{(1, 3, 0, 2), (0, 1,1, 1), (1, 4,1, 3)},

L2 = Lin{(2,2, 1, 3), (1, 4,1, 0), (3, 2, 0, 3)}.4. (6 poena) Linearan operator A : R3 R3 u standardnoj bazi {e1, e2, e3} ima matricu

A =

15 11 512 13 11 7 9

.Odrediti mu matricu A1 u bazi

f1 = 2e1 + 3e2 + e3,

f2 = e1 e2 + 5e3,f3 = 2e1 + e2 3e3.

5. (6 poena) Dokazati da je operator A : R3 R3 definisan sa

A(x1, x2, x3) = (x1 x3,x1 + x2 + 2x3, x1 + 2x2 + x3)

linearan, a zatim mu odrediti bazu i dimenziju jezgra. Izraqunati rang(A).