Solutions 6 (1) - MAS212 Linear Algebra I Exercises 6 (with...

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Unformatted text preview: MAS212 Linear Algebra I Exercises 6 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday 17th November 2006 in the blue box on the ground floor. Your submission must have your FULL NAME and STUDENT NUMBER PRINTED CLEARLY at the top and separate sheets must be stapled together. Marks out of 100 are shown in square brackets after each question number. 1. [20] For each of the following linear maps, write down the matrixA of the map with respect to the standard bases for the domain and codomain: (a) out}?2 —> R, oc(x1,x2) =2x1—3 x2; (b) 0c : F; —> FE, oc(x1,x2,x3) =(x1,x1+x2,x1+x2 +x3); (c) 0c:lR3 —> R3, 0c x1,x2,x3) = (x2 +x3,x3 +x1,x1+x2); ((1) 0c : R2 —> R3, 0c(x1,x2) =(x1+ x2,x1—x2, 0). Solution Mapping the standard basis means setting each coordinate to 1 in turn, with all other coordinates set to 0, and the result is immediately a column of the matrix of the map, e. g. for (a) first column is OL( 1, 0) =2 and second column is 0c(0, 1) =—3. 100 011 11 (a)A=[2-3],(b)A=110,(c)A=101,(d)A=1—1 111 110 00 Maple Solution 2. [30] State, with brief justification, which of the maps in Question 1 is an isomorphism and for each of these maps construct its inverse map. Hence write down the matrix B of the inverse map with respect to the standard bases and calculate the matrix products A B and B A. Solution For a map to be an isomorphism, the dimensions of its domain and codomain must be the same and hence the matrix must be square, which rules out (a) and ((1). Claim that (b) and (c) are isomorphisms. (b) Letoc(x1, x2, x3) = (y1,y2, y3), so that (x1, x2, x3) = oc_1(y1, y2,y3). Solving 0L(x1, x2, x3) = (x1, x1 +x2, x1 +x2 +x3) = (y1,y2,y3) forxl, x2, x3 gives —1 x1=y1, x2 =y2 +y1, x3 =y3 +y2 over [132. Hence 0c (yl, y2,y3) = (y1,y2 +y1, y3 +y2). Therefore 100 100 100 100 B:= 1 10 .A= 1 1 0 fiolesoA.B= 01 0 andBA= 0 1 0 overIFz. 011 111 001 001 _1 I (e) Let 0L(x1, x2, x3) = (y1,y2, y3), so that (x1, x2, x3) = on (yl, y2,y3). Solvmg oc(x1, x2, x3) = (x2 +x3, x3 +x1, x1 +x2) = (y1,y2,y3) forxl, x2, x3 gives 1 1 1 x1= 3(71 +312 +y3),x2= 3(y1—y2+y3)sx3= 3(y1+y2‘y3)' Hence _1 1 06 (y1,y2,y3) =3(-yl+y2+y3,y1-y2+y3,y1+y2-y3)- Therefore _i i i 2 2 2 0 1 1 1 0 0 1 0 0 B:= % -% % .A= 101 fromQ1soA.B= 010 andBA= 010. i i _i 1 1 0 0 0 1 0 0 1 2 2 2 Maple Solution 3. [20] For each of the isomorphisms (a) 0: : R2 —> R2, 0c(x1,x2) = (x2,x1), (b) (X : R3 —> R3, (x(x1, x2, x3) =(x1+2x2 +x3,x1—x3,x2), . -1 . . . construct the inverse map on , then wrlte down the matrixA of the map 0t and the matrlx B of the . -1 . . . . inverse map 0: w1th respect to the standard bases for the domain and codoma1n and ver1fy that A B = B A = I, where I is the identity matrix. Solution 1 (a) The effect of 06 is to reverse the coordinates so obviously oc— must also reverse the . . -1 . coordmates 1n order to undo 0L. Therefore 0L (x1, x2) = (x2, x1 For formally, solv1ng . . —1 0L(xl,x2) = (x2, x1) = (yl,y2) forx1,x2 g1ves x1=y2, x2 =y1, 1.e. 0t (yl,y2) = (x1,x2) = (y2,y1). 0 1 0 1 1 0 l 0 1 0 1 0 0 1 0 1 HenceA= andB= ,soAB= andBA= (b) Solving oc(xl, x2, x3) = (x1 + 2 x2 +x3, x1 —x3, x2) = (y1,y2, ys) forxl, x2, x3 gives 1 1 1 1 . x1=-y3+ 3y2+ Eyl,x2=y3,x3=—y3—3y2+3yl, 1.e. —1 1 l 1 1 a (y1,y2,y3)= [-y3+3y2+3y1,y3,-y3-3y2+3y1]- 1 2 1 1 1 —2 1 0 0 l 0 0 HenceA= 1 0 —1 andB=% 0 0 2 ,soAB= 0 l 0 andBA= 0 l 0 . 0 l 0 l —1 —2 0 0 l 0 0 1 Maple Solution 4. [30] You may assume the rules of matrix transposition as given in the lecture notes. (a) Write down a simple 2 x 3 matrix A with integer elements. Compute its transpose AT. Choose a small integer k and compute B = kAAT and C = kATA. Show explicitly that BT = B and CT = C. Prove that if A is any matrix over a field K, k E [K and B = kAAT then BT = B. a b c d BT = B. Prove that if A is any square matrix over a field K, k E [K and B = k(A + AT) then BT = B. (b) Leta, b, c, d, k E [K and letA = . ComputeB = k(A + AT) and show explicitly that “1 (c) LetA= ,B=[ b1 b2 ]be matrices overafieldK. Show explicitly that (AB)T=BTAT and “2 (BA)T=ATBT. Solution 1 2 1 2 3 T (a) For example, letA= and letk=3. ThenA = 2 0 , 3 —1 l 2 1 2 3 14 —1 42 —3 B= 2 0 =3 = and 2 0 —l —1 5 -3 15 3 —1 1 2 5 2 1 15 6 3 l 2 3 T 42 —3 C=3 20 =3 246=61218.SoB= =Band 2 0 —1 —3 15 3 —1 1 6 7 3 18 21 15 6 3 CT: 6 12 18 =C. 3 18 21 T T In general, ifB = kAAT then BT = (kAAT) = k (AT) AT= kAAT = B. a b T a b a c 2kd k(b+c) T (b)A= =>B=k(A+A )=k + = =B. c d c d b d k(b+c) 2kd T In general, ifB=k(A +AT) thenBT= (k(A +AT)) =k(AT+A) =B. T “1 “1,91 “ibz “1’91 aZbl (c)IfA= ,12=[b1 b2 ]then(AB)T= = and “2 “2,91 azbz “1’92 azbz b1 “1’91 “zbl BTAT= [a1 a2 ]= so (AB)T =BTAT; b2 “1’92 “zbz (BA)T = (alb1 + azbz) T= alb1 + azb2 since a scalar of 1 x 1 matrix transposes to itself and b1 ATBT=[a1 a2] b =a1b1+a2b2so(BA)T=ATBT. 2 Maple Solution ...
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This note was uploaded on 02/11/2012 for the course MTH 141, 142, taught by Professor Mcallister during the Spring '08 term at SUNY Empire State.

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Solutions 6 (1) - MAS212 Linear Algebra I Exercises 6 (with...

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