matematika - zadaci s matricama

Zadaci iz matematike - zadaci s matricama

1.) Izračunaj determinantu matrice reda 4

A =

⎣⎢⎢⎢⎢⎡ 0 0 -1 2

-1 1 -3 1

1 2 0 0

1 1 2 1

⎦⎥⎥⎥⎥⎤

Determinantu reda 4 možemo izračunati na 2 načina: 1.) Laplaceovim razvojem po npr. 1. stupcu 2.) svođenjem na trokutastu matricu LAPLACEOV RAZVOJ PO 1. stupcu

detA = (-1)1+1

∙ 0 �

1 −3 1

2 0 0

1 2 1

�

1 -3

2 0

1 2

+ (-1)2+1

∙ (-1) �

0 −1 2

2 0 0

1 2 1

�

0 -1

2 0

1 2

+ (-1)3+1

∙ 1

�

0 −1 2

1 −3 1

1 2 1

�

0 -1

1 -3

1 2

+ (-1)4+1

∙ 1 �

0 −1 2

1 −3 1

2 0 0

�

0 -1

1 -3

2 0

�

1 −3 1

2 0 0

1 2 1

�

1 -3

2 0

1 2

= 0 + 0 + 4 - ( 0 + 0 - 6) = 10

�

0 −1 2

2 0 0

1 2 1

�

0 -1

2 0

1 2

= 0 + 0 + 8 - ( 0 + 0 - 2 ) = 10

�

0 −1 2

1 −3 1

1 2 1

�

0 -1

1 -3

1 2

= 0 - 1 + 4 - ( - 6 + 0 -1 ) = 10

�

0 −1 2

1 −3 1

2 0 0

�

0 -1

1 -3

2 0

= 0 - 2 + 0 - ( - 12 + 0 + 0 ) = 10

detA = (1 ∙ 0 ∙ 10 ) + ( (-1) ∙ (-1) ∙ 10 ) + ( 1 ∙ 1 ∙ 10 ) + ((-1) ∙ 1 ∙ 10) = 0 + 10 + 10 - 10 = 10 detA = 10


2.) Riješite sustav jednadžbi AX = B pomoću inverzne matrice

A = �

1 3 1

2 2 1

3 1 0

� , X = �

�

�

�

� i B = �

1

0

2

�

A-1 =

�

�� ∙�∗, �∗ = ��

�

detA = �

1 3 1

2 2 1

3 1 0

�

1 3

2 2

3 1

= 0 + 9 + 2 - ( 6 + 1 + 0 ) = 4

detA = 4

a11 = (-1)1+1

∙ � 2 1

1 0 � = 1 ∙ (-1) = -1

a21 = (-1)2+1

∙ � 3 1

1 0 � = (-1) ∙ (-1) = 1

a31 = (-1)3+1

∙ � 3 1

2 1 � = 3 - 2 = 1 ∙ 1 = 1

a12 = (-1)1+2

∙ � 2 1

3 0 � = (-1) ∙ (-3) = 3

a22 = (-1)2+2

∙ � 1 1

3 0 � = 1 ∙ (-3) = -3

a32 = (-1)3+2

∙ � 1 1

2 1 � = (-1) ∙ (-1) = 1

a13 = (-1)1+3

∙ � 2 2

3 1 � = 1 ∙ ( - 4 ) = - 4

a23 = (-1)2+3

∙ � 1 3

3 1 � = (-1) ∙ (-8) = 8

a33 = (-1)3+3

∙ � 1 3

2 2 � = 1 ∙ (-4) = - 4

�∗ = �

−1 3 −4

1 −3 8

1 1 −4

�

�

= �

−1 1 1

3 −3 1

−4 8 −4

� A-1

= �

� ∙ �

−1 1 1

3 −3 1

−4 8 −4

�

AX = B X = A

-1 ∙ B


�

� ∙ �

−1 1 1

3 −3 1

−4 8 −4

� ∙ �

1

0

2

�

3.) Zadana je matrica

M =

⎣⎢⎢⎢⎡

0 2 1 0

−1 2 0 1

0 2 3 −2

5 0 0 1

⎦⎥⎥⎥⎤

Odredite detA

detA = (-1)1+1

∙ 0 �

2 0 1

2 3 −2

0 0 1

�

2 0

2 3

0 0

+ (-1)2+1

∙ (-1) �

2 1 0

2 3 −2

0 0 1

�

2 1

2 3

0 0

+ (-1)3+1

∙0

�

2 1 0

2 0 1

0 0 1

�

2 1

2 0

0 0

+ (-1)4+1

∙ 1 �

2 1 0

2 0 1

2 3 −2

�

2 1

2 0

2 3

�

2 0 1

2 3 −2

0 0 1

�

2 0

2 3

0 0

= 6 + 0 + 0 - ( 0 + 0 + 2 )= 4

�

2 1 0

2 0 1

2 3 −2

�

2 1

2 0

2 3

= 0 + 2 + 0 - ( 0 + 6 - 4 ) = 0

detM = 0 + ((-1) ∙ (-1) ∙ 4 ) + 0 + ((-5) ∙ 5 ∙ 0 ) = 4 detM = 4


4.) Odredite detH

H =

⎣⎢⎢⎢⎡

2 2 −1 1

4 3 −1 2

8 5 −3 4

3 3 −2 2

⎦⎥⎥⎥⎤

detH = (-1)1+1

∙ 2 �

3 −1 2

5 −3 4

3 −2 2

�

3 -1

5 -3

3 -2

+ (-1)2+1

∙ 4 �

2 −1 1

5 −3 4

3 −2 2

�

2 -1

5 -3

3 -2

+ (-1)3+1

∙8

�

2 −1 1

3 −1 2

3 −2 2

�

2 -1

3 -1

3 -2

+ (-1)4+1

∙ 3 �

2 −1 1

3 −1 2

5 −3 4

�

2 -1

3 -1

5 -3

= ( 1 ∙ 2 ∙ 2 ) + ((-1) ∙ 4 ∙ 1 ) + ( 1 ∙ 8 ∙ 1 ) + ( (-1) ∙ 3 ∙ 2) = 4 - 4 + 8 - 6 = 8 -6 = 2

�

3 −1 2

5 −3 4

3 −2 2

�

3 -1

5 -3

3 -2

= -18 - 12 - 20 - ( - 18 - 24 - 10 ) = 2

�

2 −1 1

5 −3 4

3 −2 2

�

2 -1

5 -3

3 -2

= -12 - 12 - 10 - ( -9 - 16 - 10 ) = 1

�

2 −1 1

3 −1 2

3 −2 2

�

2 -1

3 -1

3 -2

= - 4 - 6 - 6 - ( -3 - 8 - 6 ) = 1

�

2 −1 1

3 −1 2

5 −3 4

�

2 -1

3 -1

5 -3

= - 8 - 10 - 9 - ( -5 - 12 - 12 ) = 2

detH = 2


5.) Zadana je matrica A . Odredite detA

A =

⎣⎢⎢⎢⎡

2 −4 1 −2

−3 6 −1 10

5 −7 2 0

4 −5 1 −3

⎦⎥⎥⎥⎤

detA = (-1)1+1

∙ 2 �

6 −1 10

−7 2 0

−5 1 −3

�

6 -1

-7 2

-5 1

+ (-1)2+1

∙ (-3) �

−4 1 −2

−7 2 0

−5 1 −3

�

-4 1

-7 2

-5 1

+ (-1)3+1

∙5

�

−4 1 −2

6 −1 10

−5 1 −3

�

-4 1

6 -1

-5 1

+ (-1)4+1

∙ 4 �

−4 1 −2

6 −1 10

−7 2 0

�

-4 1

6 -1

-7 2

detA = ( 1 ∙ 2 ∙ 15 ) + ((-1) ∙ (-3) ∙ (-3)) + ( 1 ∙ 5 ∙ (-6)) + ((-1) ∙ 4 ∙ 0 ) = 30 -9 - 30 + 0 = -9

�

6 −1 10

−7 2 0

−5 1 −3

�

6 -1

-7 2

-5 1

= - 36 + 0 - 70 - ( - 100 + 0 - 21 ) = 15

�

−4 1 −2

−7 2 0

−5 1 −3

�

-4 1

-7 2

-5 1

= 24 + 0 + 14 - (20 - 0 + 21 )= 38 - 20 - 21 = 38 - 41 = - 3

�

−4 1 −2

6 −1 10

−5 1 −3

�

-4 1

6 -1

-5 1

= - 12 - 50 - 12 - ( - 10 - 40 - 18 ) = - 6

�

−4 1 −2

6 −1 10

−7 2 0

�

-4 1

6 -1

-7 2

= 0 - 70 - 24 - ( - 14 - 80 + 0 ) = 0

det A = - 9


6.) Zadana je matrica A. Odredite detA

A =

⎣⎢⎢⎢⎡−2 1 1 1

1 −2 1 1

1 1 −2 1

1 1 1 −2

⎦⎥⎥⎥⎤

(-1)1+1

∙ (-2) �

−2 1 1

1 −2 1

1 1 −2

�

-2 1

1 -2

1 1

+ (-1)2+1

∙ 1 �

1 1 1

1 −2 1

1 1 −2

�

1 1

1 -2

1 1

+ (-1)3+1

∙ 1

�

1 1 1

−2 1 1

1 1 −2

�

1 1

-2 1

1 1

+ (-1)4+1

∙ 1 �

1 1 1

−2 1 1

1 −2 1

�

1 1

-2 1

1 -2

=

�

−2 1 1

1 −2 1

1 1 −2

�

-2 1

1 -2

1 1

= - 8 +1 + 1 - ( -2 - 2 - 2 ) = 0

�

1 1 1

1 −2 1

1 1 −2

�

1 1

1 -2

1 1

= 4 + 1 + 1 - ( - 2 + 1 - 2 ) = 9

�

1 1 1

−2 1 1

1 1 −2

�

1 1

-2 1

1 1

= - 2 + 1 - 2 - ( 1 + 1 + 4 ) = - 9

�

1 1 1

−2 1 1

1 −2 1

�

1 1

-2 1

1 -2

= 1 + 1 + 4 - ( 1 - 2 - 2 ) = 9

detA = (1 ∙ (-2) ∙ 0 ) + ((-1) ∙ 1 ∙ 9) + (1 ∙ 1 ∙ (-9)) + ((-1) ∙ 1 ∙ 9 = 0 + ( - 9 ) + ( - 9 ) + (- 9 ) = - 27 detA = - 27


7.) Zadana je matrica A reda 3. Izračunajte. a)

aij = |� − �| �� > �

� + � �� ≤ � �11 = 1 ≤ 1 = � + � = 1 + 1 = 2 a12 = 1 ≤ 2 = 1 + 2 = 3 a13 = 1 ≤ 3 = 1 + 3 = 4 a21 = � > � = |� − �|= |2 − 1|= 1 a22 = � ≤ � = 2 ≤ 2 = 2 + 2 = 4 a23 = � ≤ � = 2 ≤ 3 = 2 + 3 = 5 a31 = � > � = |� − �|= |3 − 1|= 2 a32 = � > � = |� − �|= |3 − 2|= 1 a33 = � ≤ � = 3 ≤ 3 = 3 + 3 = 6

A = �

2 3 4

1 4 5

2 1 6

�

b) Odredite f (x) = 2x

2 - x + 3 f (A) = ?

f(A) = 2A

2 - A + 3I

A2 = A ∙ A = �

2 3 4

1 4 5

2 1 6

� ∙ �

2 3 4

1 4 5

2 1 6

� = �

15 22 47

16 24 54

17 16 49

�

f (A) = 2 ∙ �

15 22 47

16 24 54

17 16 49

� - �

2 3 4

1 4 5

2 1 6

� + �

3 0 0

0 3 0

0 0 3

�

= �

30 44 94

32 48 108

34 32 98

� - �

2 3 4

1 4 5

2 1 6

� + �

3 0 0

0 3 0

0 0 3

� = �

31 41 90

31 47 103

32 31 95

�


8.) Rješavanje sustava Cramerovim pravilom - x1 + 3x2 - 5x3 = 44 4x1 - 10x2 + 7x3 = - 75 -3x1 + 2x2 + 4x3 = - 36

D = ��

-1 3 -5

4 -10 7

-3 2 4

��

-1 3

4 -10

-3 2

= 40 - 63 - 40 - ( - 150 - 14 + 48 ) = 53

D1 = ��

44 3 -5

-75 -10 7

-36 2 4

��

44 3

-75 -10

-36 2

= - 1760 - 756 + 750 - ( - 1800 + 616 - 900 ) = 318

D2 = ��

-1 44 -5

4 -75 7

-3 -36 4

��

-1 44

4 -75

-3 -36

= 300 + 924 + 720 - ( - 1125 + 252 + 704 ) = 265

D3 = ��

-1 3 44

4 -10 -75

-3 2 -36

��

-1 3

4 -10

-3 2

= - 360 + 675 + 352 - ( 1320 + 150 - 432 ) = - 371

Rj: =( D1

D,

D2

D,

D3

D) , (

31853

, 26553

, -37153

), ( 6, 5, -7 )

9.) Riješite sustav Cramerovim pravilom 2x - 2y + 3z = - 1 x + y + z = 2 - 2x + 2y - 2z = 0

D = ��

2 -2 3

1 1 1

-2 2 -2

��

2 -2

1 1

-2 2

= - 4 + 4 + 6 - ( - 6 + 4 + 4 ) = 4 D3= ��

2 -2 -1

1 1 2

-2 2 0

��

2 -2

1 1

-2 2

= 0+8-2-(2+8+0)= - 4

D1 = ��

-1 -2 3

2 1 1

0 2 -2

��

-1 -2

2 1

0 2

= 2 + 0 + 12 - ( 0 - 2 + 8 ) = 8 Rj: =( D1

D,

D2

D,

D3

D) , (

84

, 44

, -44

), ( 2,1,-1 )

D2 = ��

2 -1 3

1 2 1

-2 0 -2

��

2 -1

1 2

-2 0

= - 8 + 2 + 0 - ( - 12 + 0 + 2 ) = 4


10.) Riješite sustav Cramerovim pravilom. 3x + y + z = 2 4x + 2y + z = 0 3x + y = 6

D = �

3 1 1

4 2 1

3 1 0

�

3 1

4 2

3 1

= 0 + 3 + 4 - ( 6 + 3 + 0 ) = 7 - 9 = - 2

D1 = �

2 1 1

0 2 1

6 1 0

�

2 1

0 2

6 1

= 0 + 6 + 0 - ( 12 + 2 + 0 ) = 6 - 14 = - 8

D2 = �

3 2 1

4 0 1

3 6 0

�

3 2

4 0

3 6

= 0 + 6 + 24 - ( 0 + 18 + 0 ) = 30 - 18 = 12

D3 = �

3 1 2

4 2 0

3 1 6

�

3 1

4 2

3 1

= 36 + 0 + 8 - ( 12 + 0 + 24 ) = 44 - 36 = 8

Rj: =( D1

D,

D2

D,

D3

D) , (

-8

-2,

12

-2,

8

-2 ), ( 4, -6, -4 )


11.) Ispišite donjetrokutastu matricu B = [��] reda 3 gdje je bij = 2i - j za � ≥ � B11 = 1 B12 = 0 B13 = 0 B21 = 3 B22 = 2 B23 = 0 B31 = 5 B32 = 4 B33 = 3 B11 = 1 ≥ 1 = 2 ∙1 − 1 = 2 − 1 = 1 B21 = 2 ≥ 1 = 2 ∙2 − 1 = 4 − 1 = 3 B22 = 2 ≥ 2 = 2 ∙2 − 2 = 4 − 2 = 2 itd.

⎣⎢⎢⎢⎡

1 0 0

3 2 0

5 4 3

⎦⎥⎥⎥⎤

12.) Odredite inverz matrice

A = �

−1 3 −5

4 −10 7

−3 2 4

� det A = �

−1 3 −5

4 −10 7

−3 2 4

�

−1 3

4 −10

−3 2

= 53

a11 = (-1)1 + 1

∙ �−10 7

2 4� = 1 ∙ ( - 54 ) = - 54

a21 = (-1)2 + 1

∙ �3 −5

2 4� = (-1) ∙ 22 = - 22

a31 = (-1)3 + 1

∙ �3 −5

−10 7� = 1 ∙ ( - 29 ) = - 29

a12 = (-1)1 + 2

∙ �4 7

−3 4� = (-1) ∙ 37 = - 37

a22 = (-1)2 + 2

∙ �−1 −5

−3 4� = 1 ∙ ( - 19 )= - 19

a32 = (-1)3 + 2

∙ �−1 −5

4 7� = (-1) ∙ 13 = - 13

a13 = (-1)1 + 3

∙ �4 −10

−3 2� = 1 ∙ ( - 22 ) = - 22

a23 = (-1)2 + 3

∙ �−1 3

−3 2� = (-1) ∙ 7 = - 7


a33 = (-1)3 + 3

∙ �−1 3

4 −10� = 1 ∙ ( - 2 ) = - 2

A*

= �

−54 −37 −22

−22 −19 −7

−29 −13 −2

�

�

= �

−54 −22 −29

−37 −19 −13

−22 −7 −2

�

A-1

= �

��

−54 −22 −29

−37 −19 −13

−22 −7 −2

�

13.) Izračunajte a)

det�1

2 ∙��

��

Naša matrica je reda 4 te je izračunata detA = -16

det��1

2�

�

∙ (−16)�

��

= -1 Broj potencije je 4 jer je matrica reda 4 b)

det �� ∙ 1

3∙ �� ∙ ��

izračunata determinanta iznosi - 9

�(−9)� ∙ �1

3�

�

∙ (−9)�� ∙ (−9)�

= - 9 c)

��

�� − 3��(��) reda 4, detH iz zadatka iznosi 2

��1

2�

�

∙2� − 3 ∙ 2� = − 95

8


d)

�� 1

2��

�

+ 2��(�� ∙ ��)

detA = 10 detA = detA

T

��1

2�

�

∙10� + 2 (10 ∙10��) = 21

8

e) ��(��)− ��(3��)+ ��(��) determinanta M iznosi - 9 (−9)� − ((3)� ∙ (−9)��)+ (−9) = 81


14.) Zadani su:

� = �2 1

4 3�, � = �

0 1

−1 4�

Odredite: f (x) = x

3 - x + 2

f (A) = A

3 - A + 2I

A

3 = A∙A∙A

A2 = �

2 1

4 3� ∙ �

2 1

4 3� = �

8 5

20 13�

A3 = A

2 ∙ A = �

8 5

20 13�∙ �

2 1

4 3� = �

36 23

92 59�

f ( A ) = �36 23

92 59� - �

2 1

4 3� + 2 �

1 0

0 1� = �

36 22

88 58�


15.) Zadan je polinom: a) f(x) = -3x

2 + x + 2

f(A) = ? f(A) = - 3A

2 + A + 2I

A = �

0 1 2

1 0 1

2 1 0

�

A2 = �

0 1 2

1 0 1

2 1 0

� ∙ �

0 1 2

1 0 1

2 1 0

� = �

5 2 1

2 2 2

1 2 5

�

- 3A2 = - 3 ∙ �

5 2 1

2 2 2

1 2 5

� = �

−15 −6 −3

−6 −6 −6

−6 −6 −15

�

f(A) = - 3A

2 + A + 2I

f(A) = �

−15 −6 −3

−6 −6 −6

−6 −6 −15

� + �

0 1 2

1 0 1

2 1 0

� + �

2 0 0

0 2 0

0 0 2

� = �

−13 −5 −1

−5 −4 −5

−1 −5 −13

�

b) odredite det��

��, gdje je B = f(A)

�

−13 −5 −1

−5 −4 −5

−1 −5 −13

�

−13 −5

−5 −4

−1 −5

= - 676 - 25 - 25 - ( - 4 - 325 - 325 ) = -72

detB = - 72

�� 1

2�� = �

1

2�

�

∙ (−72) = −9


16.) Zadana je simetrična matrica A reda 4 sa a) aij = |� − �|, za � ≥ � A11 = 0 A12 = 0 A13 = 0 A14 = 0 A21 = 1 A22 = 0 A23 = 0 A24 = 0 A31 = 3 A42 = 2 A43 = 1 A44 = 0 A41 = 3 A42 = 2 A43 = 1 A44 = 0

A = �

0 0 0 01 0 0 02 1 0 03 2 1 0

�

b) zadan je polinom f(x) = 4x

2 - x + 2 f(A) = ?

f(A) = 4A

2 - A + 2I

A2 = A ∙ A = �

0 0 0 01 0 0 02 1 0 03 2 1 0

� ∙ �

0 0 0 01 0 0 02 1 0 03 2 1 0

� = �

0 0 0 00 0 0 01 0 0 04 1 0 0

�

4A2 = 4 ∙ �

0 0 0 00 0 0 01 0 0 04 1 0 0

� = �

0 0 0 00 0 0 04 0 0 0

16 4 0 0

�

f(A) = �

0 0 0 00 0 0 04 0 0 0

16 4 0 0

� - �

0 0 0 01 0 0 02 1 0 03 2 1 0

� + 2 �

1 0 0 00 1 0 00 0 1 00 0 0 1

�


17.)

B = �

1 0 −2

−2 3 1

−2 2 0

� i C = �−1 3 5

0 1 −2�

Neka je A matrica tipa (2,3) čiji su elementi zadani formulom aij = 3j - ij a11 = 2 a12 = 4 a13 = 6 a21 = 1 a22 = 2 a23 = 3 a11 = 3 ∙ 1 - 1 ∙ 1 = 3 - 1 = 2 a12 = 3 ∙ 2 - 1 ∙ 2 = 6 - 2 = 4 a13 = 3 ∙ 3 - 1 ∙ 3 = 9 - 3 = 6 a21 = 3 ∙ 1 - 2 ∙ 1 = 3 - 2 = 1 a22 = 3 ∙ 2 - 2 ∙ 2 = 6 - 4 = 2 a23 = 3 ∙ 3 - 2 ∙ 3 = 9 - 6 = 3

A = �2 4 6

1 2 3�

b) Odredite ako postoji C ∙ B - A

�−1 3 5

0 1 −2� ∙ �

1 0 −2

−2 3 1

−2 2 0

� - �2 4 6

1 2 3� = �

−17 19 5

2 −1 1� - �

2 4 6

1 2 3� = �

−15 15 −1

1 0 −2�

c) Odredite B

-1, inverz matrice B

�� = 1

�� ∙ �∗, �∗ = [��]�

detB = �

1 0 −2

−2 3 1

−2 2 0

�

1 0

−2 3

−2 2

= 0 + 0 + 8 - ( 12 + 2 + 0 ) = - 6

b11 = (-1)1+1

∙ �3 12 0

� = 1 ∙ (-2) = - 2

b21 = (-1)2+1

∙ �0 −22 0

� = (-1) ∙ 4 = - 4

b31 = (-1)3+1

∙ �0 −23 1

� = 1 ∙ 6 = 6

b12 = (-1)1+2

∙ �−2 1−2 0

� = (-1) ∙ 2 = - 2


b22 = (-1)2+2

∙ �1 −2

−2 0� = 1 ∙ (- 4 ) = - 4

b32 = (-1)3+2

∙ �1 −2

−2 1� = (-1) ∙ (-3) = 3

b13 = (-1)1+3

∙ �−2 3−2 2

� = 1 ∙ 2 = 2

b23 = (-1)2+3

∙ �1 0

−2 2� = (-1) ∙ 2 = - 2

b33 = (-1)3+3

∙ �1 0

−2 3� = 1 ∙ 3 = 3

�∗ = �−2 −2 2−4 −4 −26 3 3

�

�

= �−2 −4 6−2 −4 32 −2 3

�

�� = 1

−6 �

−2 −4 6−2 −4 32 −2 3

�

18.) Izračunajte ��(3�)∙��(��)− 4��(��) Matrica reda 4 = 3� (−27) ∙ (−27)�� − 4 ∙ (−27) = 189


19.) Odredite inverz matrice

A = �

2 0 1

−2 3 1

4 −4 0

� det A = �

2 0 1

−2 3 1

4 −4 0

�

2 0

−2 3

4 −4

= 4

a11 = (-1)1 + 1

∙ �3 1

−4 0� = 1 ∙ 4 = 4

a21 = (-1)2 + 1

∙ �0 1

−4 0� = (-1) ∙ 4 = - 4

a31 = (-1)3 + 1

∙ �0 1

3 1� = 1 ∙ ( - 3 ) = - 3

a12 = (-1)1 + 2

∙ �−2 1

4 0� = (-1) ∙ (-4) = 4

a22 = (-1)2 + 2

∙ �2 1

4 0� = 1 ∙ ( - 4 )= - 4

a32 = (-1)3 + 2

∙ �2 1

−2 1� = (-1) ∙ 4= - 4

a13 = (-1)1 + 3

∙ �−2 3

4 −4� = 1 ∙ ( - 4 ) = - 4

a23 = (-1)2 + 3

∙ �2 0

4 −4� = (-1) ∙ (-8) = 8

a33 = (-1)3 + 3

∙ �2 0

−2 3� = 1 ∙ 6 = 6

�∗ = �

4 4 −4

−4 −4 8

−3 −4 6

�

�

= �

4 −4 −3

4 −4 −4

−4 8 6

�

�� = 1

4 �

4 −4 −3

4 −4 −4

−4 8 6

�

matematika - zadaci s matricama

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