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Unformatted text preview: Math 310: PRACTICE Hour Exam 1 (Solutions)
Prof. S. Smith: (Fall 94: orig. given 27 Sept 1993)
1 3 3 3:1 1
Problem 1: Find the general solution of 2 6 9 $2 : 5 .
—1 —3 3 x3 5
Az—2X11A;X1 1 3 3 1 M%X2 1 3 3 1 AI3X2,A;6X2 1 3 0 _2
—> 0 0 3 3 —> 0 0 1 1 —> 0 0 1 1
0 0 6 6 0 0 6 6 0 0 0 0
Thus :32 is free variable; With :02 = 1 in Ar = 0, special solution is (—3 1 0).
With 332 = 0 in A3: = b, particular solution is (—2 0 1). So general: (—2 0 1) + a(—3 1 0). Problem 2: 1 0 0
(a) Is A = ( 0 1 0 ) invertible? If so, ﬁnd A’l.
2 3 1 2 1 0
(b) Give the LDU decomposition of ( 0 4 2 )
6 3 5 (a) This is product of elementary matrices A§X1A§X2, so inverse is 1 O 0
(Agx2)—1(A§x1)—1 = A;3X2A3:2X1 : ( 0 1 0 )
—2 —3 1 _3X1 210 200 1% 0
A
(was 042,soD=040andU=01§.
005 005 0 0 1
1 0 0
AndL=(A;3><1)1=A§><1= 0 1 0
3 0 1 Problem 3: Let V be the space of realvalued functions of 30. Show the solution set S of f’(x) = xf(;v) is a subspace of V.
Iff,g E S then f’ =xf,g’=mg. So(f+g)’=f’+g’=:z:f+:cg=$(f+g);andalsof—l—gES.
Similarly iff E S and c is scalar, (cf)’ 2 C(f’) = C(mf) = {13(cf) so that also cf 6 S. Problem 4: 1 1 0 2
(a) For A = ( 1 3 1 2 ) ﬁnd a basis for the row space and the column space.
3 1 —1 6 Is Ax = b solvable for all I)? (b) Let V be the space of polynomials in a: of degree at most 3.
Give a basis for the subspace W of such polynomials with f (2) : 0. A—lxl A—3x1 1 1 0 2
(a) (rows) 2 —’>3 0 2 1 0
—1 0 0 —2
Then A§><2 converts third row to 0, so (1 1 O 2) and (0 2 1 0) give basis.
1 1 1 1 3
1 3 1 A—lxl A—2><1 0 2 _2
T _ 2 7 4
(columns) Work on A _ 0 1 _1 _> 0 1 _1
2 2 6 0 0 0 This time Azﬁﬂ kills third row; so basis is (1 1 3)T and (0 2 — 2)T. And Aw = b solvable only for b in this column space—here, NOT all h. (b) Easiest to note that V is 4di1nensi0nal With basis {1, m, 332, $3}, and W is 3di1nensional subspace With a: — 2 dividing f (m) ; so a nice basis is {x — 2, x($ — 2), w2($ — 2)}, that is, {m — 2, $2 — 23:, x3 — 23:2} ...
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 Spring '08
 Smith

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