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 1 : ted shifri 1458 ## DESCRIPTION 2 : ## Linear Algebra 3 : ## ENDDESCRIPTION 4 : 5 : ## KEYWORDS ('linear algebra','vector space','linear transformation') 6 : 7 : ## DBsubject('Multivariable Mathematics') 8 : ## DBchapter('Linear Transformations') 9 : ## DBsection('Change of Basis') 10 : ## Date('') 11 : ## Author('Shifrin') 12 : ## Institution('UGA') 13 : ## TitleText1('') 14 : ## EditionText1('') 15 : ## AuthorText1('') 16 : ## Section1('') 17 : ## Problem1('') 18 : 19 : DOCUMENT(); # This should be the first executable line in the problem. 20 : 21 : loadMacros( 22 : "PG.pl", 23 : "PGbasicmacros.pl", 24 : "PGchoicemacros.pl", 25 : "PGanswermacros.pl", 26 : "PGmatrixmacros.pl", 27 : "PGmorematrixmacros.pl", 28 : "weightedGrader.pl", 29 : "MathObjects.pl", 30 : "PGauxiliaryFunctions.pl", 31 : ); 32 : 33 : 34 : TEXT(beginproblem()); 35 : $showPartialCorrectAnswers = 1; 36 : 37 : install_weighted_grader(); 38 : 39 : Context("Numeric")->variables->are(x=>'Real', y=>'Real', z=>'Real'); 40 : 41 : @normal=("1","2","2"); 42 : @choice=NchooseK(3,3); 43 :$sgn[1] = 1; 44 : $sgn[2] = random(-1,1,2); 45 :$sgn[3] = random(-1,1,2); 46 : foreach $i(1..3) {$a[$i]=$sgn[$i]*@normal[@choice[$i-1]]}; 47 : 48 : $g=Formula("$a[1]x+$a[2]y+$a[3]z")->reduce; 49 : 50 : $M = new Matrix(3,3); 51 :$R = new Matrix(3,3); 52 : $M -> assign(1,1,1);$R -> assign(1,1,1); 53 : $M -> assign(2,2,1);$R -> assign(2,2,1); 54 : $M -> assign(3,3,0);$R -> assign(3,3,-1); 55 : foreach $i(1..3) {foreach$j(1..3) {if($i!=$j) {$M -> assign($i,$j,0); 56 :$R -> assign($i,$j,0);}}} 57 : # $M -> assign(1,2,0);$M -> assign(2,1,0); $M -> assign(1,3,0);$M -> 58 : # assign(3,1,0); $M -> assign(2,3,0);$M -> assign(3,2,0); 59 : 60 : $T = new Matrix(3,3); 61 :$S = new Matrix(3,3); 62 : $denom =$a[1]**2+$a[2]**2+$a[3]**2; 63 : # $T = Matrix("[[($a[2]**2+$a[3]**2)/$denom,-$a[1]*$a[2]/$denom,-$a[1]*$a[3] # /$denom],[-$a[1]*$a[2]/$denom,($a[1]**2+$a[3]**2)/$denom,-$a[2]*$a[3]/$denom], 64 : # [-$a[1]*$a[3]/$denom,-$a[2]*$a[3]/$denom,($a[1]**2+$a[2]**2)/$denom]]"); 65 : $T -> assign(1,1,($a[2]**2+$a[3]**2)/$denom); 66 : $T -> assign(1,2,-$a[1]*$a[2]/$denom); $T -> assign(2,1,$T->element(1,2)); 67 : $T -> assign(1,3,-$a[1]*$a[3]/$denom); $T -> assign(3,1,$T->element(1,3)); 68 : $T -> assign(2,2,($a[1]**2+$a[3]**2)/$denom); 69 : $T -> assign(2,3,-$a[2]*$a[3]/$denom); $T -> assign(3,2,$T->element(2,3)); 70 : $T -> assign(3,3,($a[1]**2+$a[2]**2)/$denom); 71 : 72 : $S -> assign(1,1,(-$a[1]**2+$a[2]**2+$a[3]**2)/$denom); 73 :$S -> assign(1,2,-2*$a[1]*$a[2]/$denom);$S -> assign(2,1,-2*$a[1]*$a[2]/$denom); 74 :$S -> assign(1,3,-2*$a[1]*$a[3]/$denom);$S -> assign(3,1,-2*$a[1]*$a[3]/$denom); 75 :$S -> assign(2,2,($a[1]**2-$a[2]**2+$a[3]**2)/$denom); 76 : $S -> assign(2,3,-2*$a[2]*$a[3]/$denom); $S -> assign(3,2,-2*$a[2]*$a[3]/$denom); 77 : $S -> assign(3,3,($a[1]**2+$a[2]**2-$a[3]**2)/$denom); 78 : 79 : Context()->texStrings; 80 : BEGIN_TEXT 81 : 82 : Consider the $$2$$-dimensional subspace $$\mathcal P\subset\mathbb R^3$$ defined by 83 : $g = 0 .$ 84 :$PAR 85 : (a) Give a basis $$\mathcal B = \lbrace \mathbf v_1,\mathbf v_2,\mathbf v_3 \rbrace$$ so that $$\mathbf v_1,\mathbf v_2$$ span the plane $$\mathcal P$$ and $$\mathbf v_3$$ is orthogonal to it. 86 : $BR 87 : \{mbox('$$\mathbf v_1 =$$', ans_array(3,1,8), ', $$\quad \mathbf v_2 =$$', ans_array_extension(3,1,8), ', $$\quad \mathbf v_3 =$$', ans_array(3,1,8), '.') \} 88 :$PAR 89 : (b) Let $$T\colon \mathbb R^3\to\mathbb R^3$$ be the linear transformation defined by projecting onto the plane. Give the matrix representing $$T$$ with respect to the basis $$\mathcal B$$. 90 : $BR 91 : \{mbox('$$[T]_{\mathcal B} =$$', answer_matrix(3,3,3), '.')\} 92 :$PAR 93 : (c) Use the change-of-basis formula to calculate the standard matrix of $$T$$: 94 : $BR 95 : \{mbox ( '$$[T] =$$', answer_matrix(3,3,8), '.') \} 96 : (d) Now let $$S\colon \mathbb R^3\to\mathbb R^3$$ be the linear transformation defined by reflecting across the plane $$\mathcal P$$. Give the matrix representing $$S$$ with respect to the basis $$\mathcal B$$. 97 :$BR 98 : \{mbox('$$[S]_{\mathcal B} =$$', answer_matrix(3,3,3), '.')\} 99 : $PAR 100 : (e) Use the change-of-basis formula to calculate the standard matrix of $$S$$: 101 :$BR 102 : \{mbox ( '$$[S] =$$', answer_matrix(3,3,8), '.') \} 103 : 104 : END_TEXT 105 : 106 : Context("Vector"); 107 : WEIGHTED_ANS(basis_cmp([[$a[3],0,-$a[1]],[0,$a[3],-$a[2]]]),7); 108 : WEIGHTED_ANS(basis_cmp([[$a[1],$a[2],$a[3]]]),3); 109 : WEIGHTED_ANS(num_cmp($M->element(1,1)),1,num_cmp($M->element(1,2)),1,num_cmp($M->element(1,3)),1,num_cmp($M->element(2,1)),1,num_cmp($M->element(2,2)),1,num_cmp($M->element(2,3)),1,num_cmp($M->element(3,1)),1,num_cmp($M->element(3,2)),1,num_cmp($M->element(3,3)),1); 110 : WEIGHTED_ANS(num_cmp($T->element(1,1)),4,num_cmp($T->element(1,2)),4,num_cmp($T->element(1,3)),4,num_cmp($T->element(2,1)),4,num_cmp($T->element(2,2)),4,num_cmp($T->element(2,3)),4,num_cmp($T->element(3,1)),4,num_cmp($T->element(3,2)),4,num_cmp($T->element(3,3)),4); 111 : WEIGHTED_ANS(num_cmp($R->element(1,1)),1,num_cmp($R->element(1,2)),1,num_cmp($R->element(1,3)),1,num_cmp($R->element(2,1)),1,num_cmp($R->element(2,2)),1,num_cmp($R->element(2,3)),1,num_cmp($R->element(3,1)),1,num_cmp($R->element(3,2)),1,num_cmp($R->element(3,3)),1,); 112 : WEIGHTED_ANS(num_cmp($S->element(1,1)),4,num_cmp($S->element(1,2)),4,num_cmp($S->element(1,3)),4,num_cmp($S->element(2,1)),4,num_cmp($S->element(2,2)),4,num_cmp($S->element(2,3)),4,num_cmp($S->element(3,1)),4,num_cmp($S->element(3,2)),4,num_cmp(\$S->element(3,3)),4); 113 : 114 : ENDDOCUMENT(); # This should be the last executable line in the problem. 115 :