MATH 2050
Assignment 9 Solutions
1 1 1
a b c
a2 b 2 c 2
(b a)(c a)(c b).
1. (a) detA =
=
1
1
1
0 ba
ca
0 b 2 a2 c 2 a2
=
Fall 2013
1
1
1
0 ba
ca
0
0
(c a)(c b)
=
b) We know that if and only if detA =
MATHEMATICS 2050:
Fall 2013
Solution of Midtest 2
2 3
1. A = 5 6 .
10 11
2. A + B, BA and B 1 dont exist since A, B have dierent size (dimension) in A + B; the
number of columns is 3 in B and the numb
DEPARTMENT OF MATHEMATICS AND STATISTICS
Assignment 8 Solutions
Mathematics 2050
Fall 2013
1. (a)
ABA1 B 1 = I ABA1 B 1 BA = BA
ABA1 (B 1 B)A = BA
ABA1 A = BA
AB = BA
(b)
(AB)2 = A2 B 2 ABAB = AABB
MATH 2050
Assignment 1
Fall 2013
Solutions
1. u v =
3 + 2
1
=
.
25
3
2. Let A = (x, y ). Then AB =
1
1x
=
, so 1 x = 1, 3 y = 2, and
3 y
2
A = (2, 1).
3
3. Since u1 = 2 u3 , vectors u1 and u3 are para
MATH 2050
Assignment 3
Fall 2013
Solutions
1. (a) w = u v is a vector perpendicular to both u and v . We compute:
2
ijk
w = u v = 1 1 0 = 2i 2j + 4k = 2 ;
4
312
the length of w is w = (2)2 + (2)2 + 42
HQ: 303:9 Assijmment =1 SGWHGMS :103
1; min = Jg-qfuahw «z Jm
a, 0L} 4 r , «3:
u 11- ....l..... a "u a» Emmi Veal-or m 'He. ciXtttHdh of {A $
s
u it u 53
A ,iicgrov 22$ \{ujirk 33w Hse. Asnchow
DEPARTMENT OF MATHEMATICS AND STATISTICS
Assignment 5 Solutions
2 cos
3
Mathematics 2050
4 cos 2
3
2
1
2
4+
= 4 1
2
1
6 2
1
2. We seek a, b and c such that 14 = a
2
the following system of