MA122 Mock Midterm
Answers
(Full Solutions will NOT be posted;
use the MAC’s drop-in help centre if you have any questions.)
**** Please remember that the mock test was meant as a means of providing an extra set of practice
questions and basis for a review class. Do not study for the midterm based solely on the topics covered by
the mock test! Go back through notes/labs/homework to ensure you have reviewed all concepts discussed
in the course.
1. A is a subspace of R4
2 3
2 3
2
3
2. (a) !
x = 415 + t 4 4 5, t 2 R
0
1
3. (a)
p9
902
4. (a)
p4
21
p
(b)
(b)
1
1
; 1;
2
2
(b)
821
3x
3y + 6z =
27 (or any multiple, such as x + y
(c) the line is parallel to the plane
2z = 9)
0
5. — — – ! use properties of cross product
!
6. — — – ! use !
v1 (k1 !
v1 + k2 !
v2 + k3 !
v3 ) = !
v1 0 to show that k1 = 0
! similarily, show that k2 = 0 and k3 = 0
2
7. — — – ! use property that k!
xk =!
x !
x to show LS =
2 3
1=3
6 7
6 2 7
6 7
8. (a) !
x =6 7
6 1 7
4 5
0
9. (a) a = 2
2
2
6
16
6
!
(b) v = 6
36
4
(b) a =
1
3s
6 1+s
6
6
6
!
s
10. (a) x = 6
6
6
t
4
t
(b) for ex:
2t
3
2
2 3 2
0
7
6 7 6
627 6
07
7
6 7 6
7 + 26 7 + 6
37
637 6
5
4 5 4
6
1
3
3
(c) a 2 R, a 6=
2
6
t 7
7 6
7 6
7 6
7=6
7 6
7 6
5 4
3
2
1
6
17
7
6
7
6
7
0 7 + s6
6
7
6
7
6
05
4
0
3
3
2
2
0
3
7
27
7
7
27
5
0
(justify each answer!)
3
2
17
7
7
07
7, s; t 2 R
7
17
5
6
17
7
6
7
6
7
1 7 + t6
6
7
6
7
6
05
4
0
1
= RS
if s = 0 = t then a sol’n would be !
x = ( 1; 1; 0; 0; 0)
if s = 1 and t = 2 then a sol’n would be !
x = ( 8; 0; 1; 2; 2)