SOLUTION FOR HW 3
1.5.10 x = x2 (3, 1, 0, 0)T + x3 (0, 0, 1, 0)T + x4 (4, 0, 0, 1)T .
1.5.24
(a.) F lse. A nontrivi l sol tion of Ax = 0 is ny nonzero x th t s tisfies the
eq tion.
(b.) Tr e.
(c.) Tr e.
(d.) Tr e.
(e.) F lse. The st tement is tr e only wh

Solutions 4.1
2.
x
x
cx
a. If u =
is in W , then the vector cu = c
=
is in W because (cx)(cy) =
y
y
cy
c2 (xy) 0 since xy 0.
1
2
b. Example: If u =
and v =
, then u and v are in W but u + v is not in
7
3
W.
8. Yes. The zero vector is in the set H. I

Solution for HW9
April 4, 2016
Section 3.2
6. -24
16. 35
24. Linearly dependent
31. It suffices to show that (det A)(det A1 )=1. According to the fact that det (AB)=det
A det B, it is equivalent to prove that det (AA1 )=1. This is obvious because AA1 = I,