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Unformatted text preview: Kuwait University.
Department of Mathematics and Computer Science
Math. 111, Linear Algebra :‘ Date: 05/3/2007 Final Exam. ‘ .- Time Allowed: 2 Hrs. I
—'"“———_-—v—r-——— Answer all questions. Calculators and Mobile Phones are not allowed.
1. [(3+2) pts.] Let P1(2, —3,2), 132(3, —1,5) and P3(1, —5,’-—1) be three points in R3. (a) Show that the points P1, P2, P3 lie on the same line. .
(b) Show that the vector P2P3 is a linear combination of the two vectors P1 P2 and < 1, 1, l >. 2. [(2+2) pts.) Determine whether W is a subspace : (a) W: {(x,y) 6122:3291}.
(b) W = {X z X é R4, X.Y = 0}, where Y is a fixed vector in R4. 11 2 2 0 111 3.[(3+2+l)pts.]LetA: 12 3 3
- 1122 12;; 3 (a) Finda‘basis for the column space of A.
_ (b) Find a basis for the row space of A that contains only the rows of A.
(c) What is the rank of A? 4. [Spits] Find all values of a. and b for which the vector (0., 1, 2) does not belong to
5pan{(15413)1(0i 111)1(b2! _3bi 2)}' a '5. [(-1+1 + 3)_pts.] .Let {1,2, 3} be the set- of eigenvalues of the 3 x 3 matrix A and
‘ ' S 3' {.(1. 1, 1), (0, 1, I), (0, 1, 2)} be the set of. the corresponding eigenvsctors. (a) Show was is linearly independent.
(b) Give a reason that the matrix A is diagonalizable.
(e) Find the matrix A. 6. ..[('4+2) pm] (a) Let .S' = {X1,X2,X3} be a basis for a vector space V. Determine whether
T = {Y1,Y2,Y3} is a; basis £01“ V, where Y1 = 2X1 + X2 — 2X3, Y2 = 4X1+ 3X2 and
Y3 =3X1+2X2'—X3. .
- {b} Let A be a 3 x 3 nonsingular matrix. Show that -adj(ade) = iA[A.
T. [1.5 pt. each] Answer each of the following as True or False (justify your answer): a. If A is an n IX 11. nonsingular upper triangular matrix satisfying B{A”1)T = In -— AT,
then -B is lewer triangular. b. If A and B are 2 5k 2 matrices satisfying. AB = 0, then A a: D or B x O. c. If the system AX = B, (B 7E 0) has ‘two distinct solutions, then AX = O has only the
: trivial solution. - d. If A- and B are n x n nonsingular matrices, then the matrices AB‘1 and BA—1 have the same eigenvalues.
e. If A is a 3 x 4 matrix, then nullity of A = I.
f. If A and B are n i< n nonsingu-lar similar matrices, then A‘1 and B‘1 are similar.
'1’ ‘.—.——- j 00 09 1. (a) The equation of the line through P1, P2 is: :1: = 2+t, y = —3+2t, 5;: 2+3t.
1ft : —1 we get P3. ’
(b) P2P3 = (—2, —4, "6) 1-— —2P1P2. 2. (a) W is not a subSpace. (2,1) 6 W but (—2, —1) e W.
(b) W is asubspace. X1,X2 E W: (X1+X2)-Y=X1-Y+X2-Y=O=>
X1 + X2 E W. AISOV(CX1) - Y 2 C(Xl ' Y) = 0 2“} CX1 E W. 0
1 T_.
3.A — 1
1 :0me
town—114
bathtub—A l
1
2
2 0
(a) basis for Column(A)= basis for Row(AT)={(1,0,1,1,1),(0,1,1,0,1),(0,0,0,0,1)}.
. (1)) basis for R0w(A)=basis for Colum'n(AT)= {(1,1,2,2), (0,1,1,1),(1,2,4,3)}. (c) Rank/i=3 10 b2
4. 411—31; 3 1 2 0 1 l 1
1 ,1 O 1
0 O 0 1
0 0 0 0 a 1
1 m 0 “3b —— 4b2
2 0 0 b2 + 31; + 2 _
There is no'sqution if b2+3b+2 = 0 and 1+a 7E O, that is b = —1,~m2 and a: 7& —1.
1 01 0 1 0 0
5. Let P: l l 1 . Then P'"1= —-1 2 —1
1 1 2 O —1 1
(a) Det(P)=1, therefore the columns are independent. (b) Since A has distinct eigenvalues it is diagonalizable. 1 0 0 1 0 0
(e)P‘1AP=D=>A=PDP_1,whereD: 0 2 O ThenA: H1 1 1 .
' 0 0 3 —l —-2 4
6. (a) Not a basis. Check for independence. Consider cl Y1 + €2Y2 + (2ng 2 0. :
(2C1-i— 4C2 “i- 3C3)X1 + (C21 + 302 + 263)X2 + (—201 — C3)X3 = 0 22$
2 4 3 0 1 3 2 0
1 3 2 0 m 0 2 1 0 .
—2 0 _1 0 O 0 0 0
This system has a. nonutrivial solution, therefore {Y1,Y2,Y3} are dependent.
(b) adj(A) = [11114-1 :> [adj(A)| : [Afifi : 11412 and (adj(A))’1 : %- Then
adjmdjw) = 1adj<Ail(adj(A))-1: |A|2fifi = :AIA. O. 09 ...
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