GAAL

1. (a) M11 = 7 • 4–(–1) • 1 = 29, M12 = 21, M13 = 27, M21 = –11, M22 = 13, M23 = –5, M31 = –19, M32 = –19, M33 = 19

(b) C11 = 29, C12 = –21, C13 = 27, C21 = 11, C22 = 13, C23 = 5, C31 = –19, C32 = 19, C33 = 19

2. (a)

(b)

(c)

(d)

MC 21 21 4 1 6 4 1 14 4 1 2 96; 72= − = − = −

MC 22 22 4 1 6 4 1 14 4 1 2 0;0 = = =

M

4 1 –6 4 1 14 4 1 2

C 2323 96; 96 = = − =

M

0 0 3 4 1 14 4 1 2

C 13 13 3 41 41 0;0 = = + ⋅ = =

103

3. (a)

(b) |A| = 1 • M11 – 6 • M21 – 3 • M23 = 152

(c) |A = 6 • M21 + 7 • M22 + 1 • M23 = 152

(d) |A| = 2 • M12 + 7 • M22 + 1 • M32 = 152

(e) |A| = –3 • M31 – 1 • M32 + 4 • M33 = 152

(f) |A| = 3 • M13 + 1 • M23 + 4 • M33 = 152

4. (a)

(b) From Exercise 3.(a), |A| = 152. Hence

5. Second column:

6. Second row:

A = − ⋅ −

+⋅ − = − + − =1 31 35 4 33 13 18 48 −66

A =⋅ − − = ⋅ − = −5 37 15 5 8 40

A− = − −1 29 152 11 152 19 152 21 152 13 152 1 / / / //99 152 27 152 5 152 19 152 / / / /          

=

/// /// / 29 152 11 152 1 8 21 152 13 152 1 8 27 15 − − 225/1521/8



    

   

adj A

CC

() =



    



11 21 31

12 22 32

13 23 33

C CCC CCC   

= − −



    



29 11 19 21 13 19 27 5 19

  

(see Exercise 1.(b.))

A =⋅ −

+⋅ − −

+⋅ −

1

71 14

2

61 34

3

67 331

29 42 81 152 =++=

104 Exercise Set 2.1

7. First column:

8. Second column: = –(k – 1)(2k – 20) + (k – 3)(k + 1) – 35_ – (k + 1)(4k – 10) = k3 – 8k2 – 10k + 95

9. Third column:

10. First row:

11.

12.

adj(A)= −

−−−



    

   

12 0 9 4 2 4 6 0 6 ;;;

/ ///AA = − = −−       −6 2 0 3 2 2 3 1 3 2 3 1 0 1 1    

adj(A)= −

−− −



    

    355 345 223 ;; AA = − =

−− −

−−



    

−1

355 3 4 5 223

1

   

A =⋅ − −⋅4

3 3 1 0 2 4 2 3 4 6 2 3 2 4 2 3

1

3330 1 2243 9463 2243

000 =+=

A = − ⋅ − − ⋅ − 3 3 3 5 2 2 2 2 10 2 3 3 3 5 2 2 2 4 110 240=−

Ak

k

k

k

k

k ( ) ( ) ( ) = − − ⋅ + − ⋅ + −[phone] ⋅⋅ +k 17 24

A

kk kk

kk kk

kk kk = ⋅ − ⋅ + ⋅ = 1 1 1 0 2 2 2 2 2 2

Exercise Set 2.1 105

13.

14.

15. (a)

(b) Same as (a).

(c) Gaussian elimination is significantly more efficient for finding inverses.

16.

x1 = 13/13 = 1, x2 = 26/13 = 2

AA 22 73 35 ;26=       =

AA 11 32 51 ;13= −      =

AA ,;= −            = 72 31 b13 b= 3 5

A− =

−− −

−−