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(MATH113)[2010](f)final~2407^_10025.pdf
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Math113 Introduction to Linear Algebra
Fall 2010
C Final Examination (All Sections) C

Name:

Student ID:

Lecture Section:

.
There are FIVE questions in this .nal 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.


1. Let

.
.
.....

02 2 42
44 8 40

.....

A =

8 2 10 0 2
63 9 21

(a)
Find a subset of the set of columns of A, which forms a basis for Col A.

(b)
Find a subset of the set of rows of A, which forms a basis for Row A.


(20 points) Solution:
(a)
.
..
..
.
02242 44840 11210

.....

44 8 40

8 2 10 0 2

.....

.

.....

02 2 42

8 2 10 0 2

.....

.

.....

02 2 42

8 2 10 0 2

.....

63921 63921 63921

.
..
..
.
11210 11210 11210

.

.....

01 1 21

0 .6 .6 .82

.....

.

.....

01 1 21

0 .6 .6 .82

.....

.

.....

01121

00048

.....

0 .3 .3 .41 00 0 00 00000
The 1st, 2nd, and 4th columns are pivot columns. It follows that
{(0, 4, 8, 6), (2, 4, 2, 3), (4, 4, 0, 2)} is a basis for Col A. (10 points)
.
(b) Row A = Col AT 0486 2423 2423
.
.
.
..
AT
=

.......

24 2 3

2 8 10 9

44 0 2

.......



.......

04 8 6

2 8 10 9

44 0 2

.......



.......

0486

0042

0000

.......

2021 2021 0000
The last matrix is in echelon form which indicates that the 1st, 2nd, and 3rd columns are
pivot columns.
It follows that {(0, 2, 2, 4, 2), (4, 4, 8, 4, 0), (8, 2, 10, 0, 2)}
is a basis for Col AT and Row A. (10 points)
2. Let P3 be the vector space of polynomials of degree at most 3. Consider the following .ve polynomials in P3:
33 3
p1(t)=1+3t + t2 +2t, p2(t)=1+2t + t2 + t, p3(t)=3+7t +3t2 +4t, 33
p4(t) = 2+10t +3t2 +8t, p5(t) = 4+17t +7t2 + 13t
(a)
Let W = Span {p1,p2,p3,p4,p5} be a subspace of P3. Find a subset of {p1,p2,p3,p4,p5}which forms a basis for W . What is dim W ?

(b)
Set q(t)=1+4t + t2 +2t3 . Is q in W ? If yes, .nd the coordinate vector of q relative to the basis obtained in (a).


(20 points)
Solution:
(a) Take the standard basis B = {1, t, t2,t3} for P3. Under the B-coordinate mapping, W will be mapped to the column space of the following matrix A:
.
..
..
.
1132 4 11 324 11 324

A =

.....

3 2 7 10 17

1133 7

.....

.

.....

0 .1 .245

00 013

.....

.

.....

0 .1 .245

00 013

.....

214813 0 .1 .245 00 000
So the 1st, 2nd, 4th columns of A form a basis for Col A. Correspondingly, we obtain {p1,p2,p4} as a basis for W . Since W has a basis with 3 elements, dim W = 3. (12 points)
(b) We check if one can .nd suitable scalars c1, c2, c4 such that c1p1 + c2p2 + c4p4 = q. Under the B-coordinate mapping, it is equivalent to check the consistency of the linear