Determinants

Solucionari

Unitat 2

Determinants Comencem

1 0 0 f) 0 1 0 = 1 0 0 1

s = –21 + 12 – 100 + 36 + 10 – 70 = –133.

1 2 1 g) 1 3 0 = 3; 0 4 -1

Exercicis 1. a) –1 b) 0.

2. |A| = 19, |B| = 26, |C| = 2, |A + B + C| = 17.

3. a)

2t - 10 t - 5 = t2 – 6t + 5 = (t – 1)(t – 5).

1+ t t

|

x 1

|

a 1 1 1 a 1 1 1 a

1 b) x

c)

1 2 1 1 1 1 1 1 1 1 3 0 = 1 2 0 + 1 1 0 = 2 + 1 = 3.

0 4 -1 0 3 -1 0 1 -1

|

12. a) 2a2 + 17a + 30 = 0 ® a = –5/2, a = –6 b) 7a + 35 = 0 ® a = –5.

= 1 – x 2 = (x + 1)(1 – x)

|

13. Resposta oberta. Per exemple, deixant la 1a fila igual i canviant les altres dues per f2' = f2 + f1 i f3' = = f3 – f1 i desenvolupant-lo pels elements de la segona columna nova queda:

= a 3 – 3a + 2 = (a – 1)2 (a + 2)

4. a) x3 + 3x2 = 0 ® x = 0, x = –3 b) –6x – 4 = –10 ® x = 1

–

3 6 = –12.

-5 -6

5. P(x) = –x3 + 2x – 1 ® P(–2) = 3 6. |A| = –14, |A*| = 196, |A*| = |A|2.

F GG H

I JJ K

14. 4 – 6 + 4 – 12 + 2 – 4 = –12

F GG H

I JJ K

1 -2 1 1 -2 1 7. A* = -2 4 -2 t(A*) = (tA)* = -2 4 -2 1 -2 1 .

1 -2 1

8. a) B* =

FG1 -2IJ H1 1K

b) No, perquè BB* =

FG 0 -3IJ i B*B = FG -3 -3IJ H 3 -3K H 3 0K

15. Rang A = 2, rang B = rang C = rang D = 3.

16. |A| = 0 ® –a2 + 3a – 2 = 0 ® a = 1, a = 2.

17. a) AX = B ®A-1(AX) = A-1B ® (A-1A)X = A-1B ® IX = = A-1B ® X = A-1B b) XA= C ® (XA)A-1 = CA-1 ® X(AA-1) = CA-1 ® XI = = CA-1 ® X = CA-1

c) |B| = |B*| = 3, |BB*| = |B*B| = 9.

9. a) 10 + 2 + 12 – 15 = 9 b) –(–2) + 3 · 4 + 5(–1) = 2 + 12 – 5 = 9 c) –2(–7) + 5(–1) = 14 – 5 = 9.

10. a) 3a columna ® –135 b) 3a fila ® –20 c) 2a fila ® –16.

11. Resposta oberta. Per exemple: 1 1 1 1 1 0 a) 1 -1 0 = 1 -1 1 = 3 0 1 -1 1 0 -1 1 1 1 1 1 1 b) 1 -1 0 = 3; -1 1 0 = –3 0 1 -1 1 0 -1 1 1 1 1 1 1 c) 2 -2 0 = 6; 2· 1 -1 0 = 2·3 = 6 0 1 -1 0 1 -1

FG -5 / 13 14 / 13 11 / 13IJ ; H -1 / 13 8 / 13 -1 / 13K F 5 / 13 -3 / 13I = G -11 / 13 4 / 13J .

GH 16 / 13 -7 / 13JK

c) X = A-1B =

X = CA-1

18. a) La matriu A b) A-1 =

FG 1 H -1 / 2

19. X = A-1(B + C) =

F GG H

IJ K

0 ; 1/ 4

AA-1 = A-1A =

FG 4 H -7 / 4

0 4 .

1 -7 / 4

FG 1 0IJ .

H 0 1K

IJ K

I JJ K

6 / 11 -8 / 11 -1 / 11 20. A-1 = -8 / 11 7 / 11 5 / 11 ; 7 / 11 -2 / 11 -3 / 11

F GG H

I JJ K

1 0 0 AA-1 = A-1A = 0 1 0 ; |A-1| = –1/11 = 1/|A|.

0 0 1

21. B = (–10/11

Matemàtiques Aplicades a les Ciències Socials. Batxillerat

6/11

9/11).

McGraw-Hill/Interamericana S.A.U.

F -1 / 2 GG 1 / 2 H 1/ 2 F -1 / 6 (AB) = (B )(A ) = G -1 / 6 GH 1 / 2 F1 0 23. La matriu A, A = G 0 GH -2 -41 F 7 -2 X = (2C – B)A = G 11 13 GH -2 -9 22. (BA)-1 = (A-1)(B-1) =

-1

-1

-1

-1

-1

I JJ K 1 / 2I 1 / 2J .

J -1 / 2K

0 1/ 2 2 / 3 -1 / 6 ; -1 / 3 -1 / 6 -1 / 3 2/3 0

I JJ K

0 0 ; 1

7. |A-1| = .

I JJ K

-1 -5 .

2

I JJ K

1

bad - bcg

8. B = 3I =

2

×

da - bc 1 d -b = = 2 -c a ad - bc ad - bc

C-1

I JJ K

3 1 1 1. A = 1 3 1 ® |A| = 20.

1 1 3

b

g

FG 3 0IJ .

H 0 3K

F GG H F -1 / 12 = G 1/ 6 GH 1 / 6

I JJ K

1 -2 7 9. A-1 = 0 1 -2 0 0 1

Acabem

F GG H

F GG H

-1 1 5 b) X = A-1B = -3 / 2 0 8 .

3 / 2 0 -1

F1 GG 1 H -1 7 / 12I -1 / 6J J -1 / 6K

B-1 =

-5 / 6 2/3 -1 / 3

I JJ K

-4 -3 -5 -3 6 4

F GG H

I JJ K

2. –6x2 + 3x = 0 ® x = 0, x = 1/2.

1 0 0 AA-1 = A-1A = BB-1 = B-1B = CC-1 = C-1C = 0 1 0 .

0 0 1 1 1 = .

10. |A-1| = A 7

3. p(x) = x3 – 112x + 384 = (x – 4)(x – 8)(x + 12).

11. No té matriu inversa perquè |B| = 0.

4. Rang A = 3, rang B = 2, rang C = 2 i rang D = 3.

5. k – 5k + 6 = 0 ® k = 2, k = 3.

Si k ¹ 2 i k ¹ 3, el rang és 3; i si k = 2 o k = 3, aleshores el rang és 2.

12. Les matrius A, B i D

2

6. a) X = CB-1 =

F -1 / 7 GG -3 / 7 H 25 / 21

I JJ K

3/ 7 5/7 2 / 7 -6 / 7 -47 / 21 57 / 21

McGraw-Hill/Interamericana S.A.U.

C-1 =

F 2 GG -1 H -4 / 3

13. C = A-1B =

I JJ K

F GG H

I JJ K

-1 0 1 0 0 1 0 CC-1 = C-1C = 0 1 0 .

2 / 3 1/ 3 0 0 1

FG 2 / 5 0IJ .

H -6 / 5 1K

Matemàtiques Aplicades a les Ciències Socials. Batxillerat