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(MATH113)Final06Spring-Solution.pdf
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Math113: Final Exam, Spring 2006
Name:
ID No. Lecture Section:

Problem 1 (10 pts) 2 (13 pts) 3 (14 pts) 4 (14 pts) 5 (14 pts) 6 (14 pts) 7 (21 pts) Total (100 pts)
Score

1. (10 pts) Show that for all real numbers , , , we always have
.
sin cos cos( + ) .
sin cos cos( + ) =0. sin cos cos( + )
Solution: Since cos( + ) = cos cos . sin sin , we have
.
sin cos cos( + ) ..
sin cos cos cos . sin sin .
C3 + (sin )C1 sin cos cos( + ) = sin cos cos cos . sin sin sin cos cos( + ) sin cos cos cos . sin sin C3 . (cos )C2
.
sin cos 0 .
= sin cos 0 =0.
sin cos 0

2. (13 pts) Let P3 be the vector space of all real polynomials in one variable t with degree at most 3. Let
S = {f1,f2,f3,f4,f5}, where
f1(t)= .2+ t +3t2 + t3 ,
f2(t)= .5+3t + 11t2 +7t3 ,
f3(t)= 8 . 5t . 19t2 . 13t3 ,
f4(t)= t +7t2 +5t3 ,
f5(t)= .17 + 5t + t2 . 3t3 .

(a)
Find a subset B of S so that B is a basis of Span (S).

(b)
Express each polynomial in S as a linear combination of the polynomials in B.

(c)
Is Span (S)= P3? Justify your statement.


Solution. (a) Since {1, t, t2,t3} is basis of P3, the coordinate vectors of the polynomials f1(t),f2(t),... ,f5(t) are columns of the matrix
..
. ..
. ..
. ..
.
R2 +2R1


R3 . 3R1
R4 . R1 R1 . R2
.2 .5 8 0 .17
1 3 .5 1 5
3 11 .19 7 1
1 7 .13 5 .3

1 3 .5 1 5
.2 .5 8 0 .17
3 11 .19 7 1
1 7 .13 5 .3



..
. ..
. ..
. ..
.
1 0 1 .5 26
0 1 .2 2 .7
0 0 0 0 0
0 0 0 .4 20

1 3 .5 1 5
0 1 .2 2 .7
0 2 .4 4 .14
0 4 .8 4 .8

R1 . 3R2


R3 . 2R2
R4 . 4R2
(.1/4)R4


R3 . R4
..
. ..
.
1 0 1 .5 26
0 1 .2 2 .7
0 0 0 1 .5
0 0 0 0 0

(.1/4)R4


R3 . R4
..
. ..
.
1 0 1 0 1
0 1 .2 0 3
0 0 0 1 .5
0 0 0 0 0

.

It follows that the set .f1(t),f2(t),f4(t). form a basis of Span (S).
(b)
Thus f1(t)= f1(t), f2(t)= f2(t), f4(t)= f4(t), and f3(t)= f1(t) . 2f2(t),f5(t)= f1(t)+3f2(t) . 5f4(t).

(c)
No. The space P3 is of dimension 4 and Span (S) is of dimension 3.


3. (14 pts) Let A =

. ..
. ..
230

020

240

010

142

012

., B = ..
(a)
(9 pts) Determine which of the two matrices are diagonalizable. Justify your answer.

(b)
(5 pts) For the matrix which is diagonalizable, compute its 100th power.


Solution. (a) The matrix A has only one eigenvalue = 2, whose eigenspace has dimension one. Then A is not diagonalizable. The matrix B has eigenvalues =2, 1.
. =
..
..
..
For = 2, the system

. .. . .. . .. . .. . ..
0 40

0

x1
0 .10

0

x2
0 10

0

x3
has basic solutions v1 = [1, 0, 0]T , v2 = [0, 0, 1]T .
For = 1, the system

..
..