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**Unformatted text preview: **Throughout this solution set, unless otherwise stated, large boldface uppercase letter
signify a matrix, small boldface letters signify a vector, small italics letters signify vector components and small Greek letters (aﬁy) signify scalar. 1) a)
b) c)
d) True. The kernel of A is the vector space spanned by the solutions to Ax=0, which is the same as the solution to Bx=0, B=rref(A). False. Image(A) is the column span of A, but we obtain rref(A) by elementary row operations that do not preserve the properties (e. g. linear independence) of the columns. True. AAAB = AABA = ABAA = BAAA. True. Image(A) = Kemel(A) implies A(Ay) = 0 for any y. This means A2 = 0. So
0 O 0 O 0 O 1 0
0 1 O 0 directly, by creating a transformation that sends everything to some (zw in this case) plane but wipes out everything in that plane. False. All spaces contain 0 by definition. So if Kemel(A)={0}, then it is contained in the Image(B) but both A and B can be invertible. False. If A is invertible then its dimension must be equal to its rank, so
rank(A)=2. 0 O
A could be e. g. 0 O . Note that we could also construct the matrix 1] iii Wt: want [i] ril‘li'i'L' _-‘L.\'={l. 2m in: first row reduce. A 10 gel: _| .i.‘ _ l —3 l} 3 [ll ll
'.'-.
l] i} J l ['l ' ll
.1': = _
i} {l [l [l l ' l]
.'.
1'} {J' ii in [J " i1
[-15 _, _
which git-Uri LIH lilL‘ L‘LlutiliiiiiH:
III-11'" I 114:1} :3 1,1: 3|} — 3U.
Dry.
.1: .- 1-2 i.;-=fl :31}: -_','u.
.1'.;=U.
.i'.:[l
}"i|._‘]t§:ll|l:1 mlill imix iii' Ilic l'm'm:
'x,’ '—_i 7:" 2 3'
.i'. U | l [l
.i'-, =H 3-“. :fi' [1 stiKc1'ricli.-U:ap;m[ l]. ---1 :u
.i'_. l U E'] ]
.1, . ii U u [l b) We know that the Image of A is the column span on A. From part a), we know that we
need three columns, and that they must be the ones with the leading 1, so the Image(A) = 1 1 o
1‘ —1 0 WM 3 ,_1, _1}
—1 0 0 c) Just row reduce the augmented matrix to get: 1 —2 O 3 04 0 0 1 2 0—1 0 0 0 O 112 0 0 0 0 0 0 representing
x1 1 —-2 0 3 0 4
x2 0 0 1 2 O —1
x3 =‘ 0 0 0 0 1 12
x4 0 0 0 O O 0
x5 yielding the equations: x5=12 X4=(X X3+2.X4=-1 =>X3=-1-20L X2=B x1—2x2+3x4=4 =>x1=4+ZB-30t 3) “1; 3] 81: 3] PS [3 3] 22 56
note that PS*PS=PS, so P550 = PS so (PS)5°[17] = [ 0 ] 4) a) 2 1 _ 2 2 1 V— 2 1 O 0
A: — 3 3 5 so consider the augmented matrix: — 3 3 5 0 1 0
-1 2 2 —l 2 2 0 . O 1 1 O 0 6 -11
and row reduce it to get: A'1 = 0 1 0 - - 2 4
0 0 l 5 — 9 b) Just multiply it out to get AA'1=I .4 6 —11 2 39
c)usingx=A'1b wegetx: —1 —2 4 7 = —12
3 5 -9 1 32 5) Spot the linear system: let so, mo, lobe the numbers of small, medium and large cars respectively in 1990. If S], m], 11 are the numbers of small, medium and large cars respectively in 2000 then:
31 .5 So + .3 mo .4So+.6m0+.4lo ll .180+.1mo+.610 but since the number of cars did not change, this is 5
II II II so =.5 so+.3 mo => -.5 so+y.3 mo =0
mo=.4so+.6m0+.4lo =>.4So-.4mo+.4lo =0
10 =.lso+.1m0+.6lo =>.lso+.lmo-.4lo =0
rowreduce: l 0 — 1.5 s0
0 1 — 2.5 m0 = 0
0 0 0 l0
yielding
10 = t
mo = 2.5t
So = 1.5t
the survey was conducted among 1000 drivers, so t+2.5t+1.5t = 1000 => St = 1000 =>
t=200
so 10 = 200, mo = 500, so =300. ' 6) a) To compute the matrix of the transformation, all we need to know is what the projection does to the basis vectors. Using the formula for orthogonal projection onto a
line: 2, ~21
e =l ‘ " W we get the matrix: ‘1 2 -2 A:1 2 4' -4
9 _—‘2 —4 4 b) We could compute the kernel by solving Ax=0, but there is an easier way. Since A is a
matrix associated with an orthogonal projection onto L, only lines perpendicular to L will
be reduced to 0. From dot product properties: 1 s1
2 s2 = 0
|_— 2' s3 _. we get 81 +282 —283 = 0 set s1=0 we get k1=[0,1,—1] set sz=0 we get k2=[2,0,1] set 53:0 we get k3=[-2,1,0] but note that k1 = k2 + k3 So Kernel(A) = span {[2,0,1],[-2,1,0]} c) First note that we have calculated p, or rather the matrix that transforms x into p in part
a) of this problem. So we can write p=Ax. Let B be the matrix of the orthogonal
projection onto the plane S. Then, from the hint:
Bx=x—p=x—Ax=Ix—Ax=(I-A)x
’ 8 — 2 2"
— 2 5 4
2 4 5 SoB=I-A=l
9 ...

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