=========================preview======================
(MATH113)[2010](f)final~2769^_10026.pdf
Back to MATH113 Login to download
======================================================
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