Math 290: Answers to Review Problems I




1 −2
1
1 0 0 29
0



8 
2 −8
1. (a) A =  0
(b)  0 1 0 16 
(c) 3

−4
5
9 −9
0 0 1 3






x
29
1 −2
1





2 −8 
(d) x = 29, y = 16, z = 3 or  y  =  16 
(e) C =  0

z
3
−4
5
9



2
1 1
2 −1
1


0 
(b)  0
2. (a)  5 −2 1 −3

3
2 3 −5 12
0





7 
7 
11
x1
−3
− 3 − 3t
 x 
 −4 

 −4 − 4t 







(d)  2  =  31 20  =  20  t + 
 x3 
 3 

 3 + 3t 
t
1
x4
0 0
7/3
1 0
4
0 1 −20/3

− 11
3
−4 

(e)
31 
3

0

−11/3
−4 

31/3

 






=







is a particular solution and x0 = 
0
0
1
0

(b) REF but not RREF, x = 


2
−4
 

4. 
 1 , 0 ;
0
1


5. 


−9
4
1
0
 
 
 
,
 
8
−5
0
1
−7s + 2
3s + 1
−2s + 1
s




2
1 1
2


 5 −2 1 −3 
3
2 3 −5

3. (a) RREF, No solution
(c) 3




−7
3
−2
1
2
1
1
0



t

is





s+







2
−4



x=
 1 s +  0 t
0
1



;





x=
−9
4
1
0


5
1
0
0



6. x = x0 + x1 , where x1 = 





s + 


8
−5
0
1



t







s


7. (a) p = ±1 and q 6= p
(b) p 6= ±1 and q arbitrary
(c) p = ±1 and q = p
8. (a) k = −2/3
(b) k can be any number
10. b = (2 − 5s)a1 + (3 − 4s)a2 + sa3 for any s
11. b = 7a1 − 2a2


1
1 2


12. (a)  4 −2 3 
−7 −6 11


−5 + 2x1

5 + 2x2 
(b) 

−13 + 2x3
1
"
(c)
−9
2
#
(d) Undefined



+
the general solution to Ax = 0.
9. (a) k = −6

−4
0
0
1





−15/2

(e) x =  43/2 

−6
−12

8 
(f) x = 

3
13. x = 1,
z = −1/4,
y = 3/2,
w = −1
14. 4
15. a = 1 and b can be any number
"
16. X = 4B =
0 −4
8
4

#


12/5 −1/5

1/5 
17. X = 51 (A + 2B) =  6/5
,
−1/5 −1/5
"
18. X =
−6 −1
−2
3

6/5 −3/5

3/5 
Y = 15 (B − 2A) =  3/5

−3/5 −3/5
#
19. −20A − 26AT + 2B
2