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(MATH113)[2010](f)midterm~883^_10450.pdf
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Math113 Introduction to Linear Algebra
Fall 2010
C Midterm Examination (All Sections) C

Name:

Student ID:

Lecture Section:

.
There are FOUR questions in this midterm examination.

.
Answer all the questions.

.
You may write on both sides of the paper if necessary.

.
You may use a HKEA approved calculator. Calculators with symbolic calculus functions are not allowed.

.
The full mark is 100.


..
..
..

2

..

..

, a3
=

..

6

..

, b =

..

b2
..

1. Let a1
=

, and A =[ a1 a2 a3 ].
, a2
03 .3 b3

(a) Does the homogeneous equation Ax = 0 have a nontrivial solution? If yes, .nd the solution. Solution: Yes. Row reduction of [A 0] yields
..
.
..
..
..
.
12 10 1210 12 10 10 30

..

20 60

..



..

0 .4 40

..



..

01 .10

..



..

01 .10

..

03 .30 03 .30 00 00 00 00 from which the general solution is found to be
.
...
x1 .3x3
..

x2
..

=

..

x3
..

x3 x3
where x3 is a free variable. Any nonzero x3 leads to a nontrivial solution.
(b) If a vector u is in Span{a1, a2, a3}, then is it also in Span{a1, a2}?
Solution: Yes. [In Span{a1, a2, a3}, u = c1a1 + c2a2 + c3a3. Since .3a1 + a2 + a3 = 0, i.e., a3 =3a1 . a2,
u = c1a1 + c2a2 + c3(3a1 . a2)=(c1 +3c3)a1 +(c2 . c3)a2
which shows that u is in Span{a1, a2}.]
(c) Determine the condition for b to be in Span{a1, a2, a3}. Solution: Row reduction of [A b] yields
.
..
.
12 1 b1 121 b1
..

20 6 b2
..



..

0 .44 b2 . 2b1
..



03 .3 b3 0
3 .3

.
.

b3
.

.

12 1 b1 12 1 b1
..

01 .1 .b2/4+ b1/2
..



..

01 .1 .b2/4+ b1/2
..

01 .1 b3/3 00 0 b3/3+ b2/4 . b1/2
from which the condition is determined to be b3/3+ b2/4 . b1/2=0 or 6b1 . 3b2 . 4b3 = 0.
(d) Consider the matrix transformation T de.ned by x T (x)= Ax. Is T one-to-one and why?
Is T onto the codomain and why?
Solution: It is not one-to-one because of the existence of nontrivial solutions in (a). It is
not onto the codomain because {a1, a2, a3} dont span R3 according to (d).

[30 points]
2. (a) Find the inverse (if it exists) of the following matrix:
.
.
1 .10 0
.11 .10

A =

.

.....

.....

0 .11 .1
00 .11

Solution: Perform row operations to the combined matrix [ A | I4 ]: 1 .10 0
1000 1 .100 1000
.
.
.
.

.....

.11 .10
0100
0 .11 .1
0010
.....

.r4+r3
.

.r1+r2
.....

00 .10
1100
0 .100
0011
.....

00 .11
0001 00 .11
0001
.
.
.
.
10 00
10 .1 .1 1000
10 .1 .1
.r2+r4
.


.r3+r1
.....

00 .10
1100
0 .100
0011
.....

r2.r3
.


scalings

.....

0100
00 .1 .1
0010
.1 .10 0
.....

.

00 01
.1 .1 0 1 0001
.1 .10 1
Hence A is invertible, and A.1 is given by:
.
.
.....

10 .1 .1
00 .1 .1

.1 .10 0
.1 .10 1

.....

.

A.1
=
(b) Consider the following linear system:

. ..
. .
..

x1 . x3 . x4 = b1 x3 + x4 =