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Merge pull request #1167 from gajennings/main
Fix bug 4835, update to PGML, add solution.
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  • OpenProblemLibrary/Mizzou/Matrix_Theory/HW_1

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‎OpenProblemLibrary/Mizzou/Matrix_Theory/HW_1/prob_1.pg‎

Lines changed: 109 additions & 134 deletions
Original file line numberDiff line numberDiff line change
@@ -12,177 +12,152 @@
1212
## Level(2)
1313
## Tagged by cmd6a 5/3/06
1414

15-
1615
###########################################################################
1716
# initialization
1817
###########################################################################
1918
DOCUMENT();
2019
loadMacros(
2120
"PGstandard.pl",
2221
"MathObjects.pl",
22+
"PGML.pl",
2323
"parserMultiAnswer.pl",
2424
"PGcourse.pl"
2525
);
2626

27-
TEXT(beginproblem());
28-
$showPartialCorrectAnswers = 1;
29-
30-
3127
###########################################################################
3228
# setup contexts and variables
3329
###########################################################################
34-
Context("Matrix");
3530

36-
$sign1 = random(-1,1,2);
37-
$a = random(2,5,1) * $sign1;
38-
$b = random(3,9,1) * $sign1;
31+
Context("Vector");
3932

33+
$sign1 = random(-1,1,2);
4034
$sign2 = random(-1,1,2);
41-
$c = random(1,6,1) * $sign2;
35+
$sign3 = random(-1,1,2);
36+
$sign4 = random(-1,1,2);
4237
$d = random(3,6,1);
43-
$epd = random(-1,1,2);
44-
$e = $epd - $d;
45-
$f = random(3,9,1) * $sign2;
46-
$m = random(2,3,1) * random(-1,1,2);
47-
$g = $m * ($e + $d);
48-
$h = $m * ($e * $a + $b);
49-
$i = $m * ($e * $c + $f);
38+
$m = random(2,3,1)*$sign4;
39+
40+
# R1,R2,R3 are the rows of a 3x4 matrix with rank 2
41+
42+
$R1 = Vector(1,-1,random(2,5,1)*$sign1, random(1,6,1)*$sign2);
43+
$R2 = Vector($d, $sign3 - $d, random(3,9,1)*$sign1, random(3,9,1)*$sign2);
44+
$R3 = $m*($sign3-$d)*$R1 + $m*$R2;
45+
46+
# the 3x4 matrix
47+
48+
$M = Matrix([[$R1->value],[$R2->value],[$R3->value]]);
5049

51-
# matrix representing left hand side of equation
50+
# for latex display
5251

53-
$A = Matrix([
54-
[1 ,-1,$a],
55-
[$d,$e,$b],
56-
[$g, 0,$h]
57-
]);
52+
($a11,$a12,$a13,$a14) = ($R1->value);
53+
($a21,$a22,$a23,$a24) = ($R2->value);
54+
($a31,$a32,$a33,$a34) = ($R3->value);
5855

59-
# for answer
56+
$pm13 = ($a13 < 0) ? "-" : "+";
57+
$pm22 = ($a22 < 0) ? "-" : "+";
58+
$pm23 = ($a23 < 0) ? "-" : "+";
59+
$pm33 = ($a33 < 0) ? "-" : "+";
6060

61-
$k = $a + $epd * ($b - $a*$d);
62-
$l = $epd * ($b - $a*$d);
63-
$mm = $c + $epd * ($f - $c*$d);
64-
$n = $epd * ($f - $c*$d);
61+
# particular solution xp with xp3=0
6562

66-
#expected answers
67-
$V1=ColumnVector($mm,$n,0);
68-
$V2=ColumnVector(-$k,-$l,1);
63+
$xp2 = $sign3*($a24-$d*$a14);
64+
$xp1 = $xp2 + $a14;
6965

70-
#for tex display
66+
$xp = ColumnVector($xp1,$xp2,0);
7167

72-
$pm1 = ($a < 0) ? "-" : "+";
73-
$pm2 = ($e < 0) ? "-" : "+";
74-
$pm3 = ($b < 0) ? "-" : "+";
75-
$pm4 = ($h < 0) ? "-" : "+";
68+
# homogeneous solution xh with xh3 = 1
7669

77-
$c1 = abs($a);
78-
$c2 = abs($e);
79-
$c3 = abs($b);
80-
$c4 = abs($h);
70+
$xh2 = -$sign3*($a23-$d*$a13);
71+
$xh1 = $xh2 - $a13;
8172

73+
$xh = ColumnVector($xh1, $xh2, 1);
8274

83-
$multians = MultiAnswer($V1,$V2)->with(
75+
$ans = MultiAnswer($xp, $xh)->with(
8476
singleResult => 0,
8577
checker => sub {
8678
my ($correct, $student, $self)=@_;
87-
my ($stu1,$stu2)= @{$student};
88-
$check1 = ( $A*$stu1 == Matrix([[$c],[$f],[$i]]) );
89-
$check2a = ( $A*$stu2 == Matrix([[ 0],[ 0],[ 0]]) );
90-
$check2b = ($stu2->isZero);
91-
if ( ! $check1 ){
92-
$self->setMessage(1,"Doesn't satisfy the given equation.");
93-
}
94-
if ( ! $check2a){
95-
$self->setMessage(2,"This vector points in the wrong direction.");
96-
}
97-
elsif ( $check2b ){
98-
$self->setMessage(2,"Can't be zero; equation has infinitely many solutions.");
99-
}
100-
return [ $check1, $check2a && ! $check2b ];
101-
}
102-
);
79+
my ($stu_p,$stu_h)= @{$student};
80+
if ($stu_h->isZero){ $self->setMessage(2,"Can't be zero; equation has infinitely many solutions.");}
81+
# augment student vectors with a fourth component
82+
my @stu_p = ($stu_p->extract(1),$stu_p->extract(2),$stu_p->extract(3));
83+
my @stu_h = ($stu_h->extract(1),$stu_h->extract(2),$stu_h->extract(3));
84+
push @stu_p, -1;
85+
push @stu_h, 0;
86+
$stu_p = Vector(@stu_p);
87+
$stu_h = Vector(@stu_h);
88+
my @check = (0,0);
89+
if ( ($M*$stu_p)->isZero ){$check[0]=1;}
90+
else {$self->setMessage(1,"Doesn't satisfy the given equation.");}
91+
if ( ($M*$stu_h)->isZero ){$check[1]=1;}
92+
else {$self->setMessage(2,"Must satisfy the corresponding homogeneous equation.");}
93+
return @check;
94+
});
10395

96+
BEGIN_PGML
97+
Use Gauss-Jordan reduction to solve the following linear system:
10498

105-
###########################################################################
106-
# state the problem
107-
###########################################################################
108-
Context()->texStrings;
109-
BEGIN_TEXT
110-
Use the Gauss-Jordan reduction to solve the following linear system:
111-
$PAR
112-
\[ \left\{'\{'\} \begin{array}{rcrcrcr}
113-
x_1 & - & x_2 & $pm1 & $c1 x_3 & = & $c \cr
114-
$d x_1 & $pm2 & $c2 x_2 & $pm3 & $c3 x_3 & = & $f \cr
115-
$g x_1 & & & $pm4 & $c4 x_3 & = & $i
116-
\end{array} \right. \]
117-
$PAR
118-
$BCENTER
119-
\(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right]\)
120-
\(=\) \{$multians->ans_array\} \(\ +\ s\ \)\{$multians->ans_array\}
121-
$ECENTER
122-
END_TEXT
123-
Context()->normalStrings;
124-
125-
#\{ mbox('\(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right]\)',
126-
#'\(\quad =\)', ans_array(3,1,5), '\(+\)', ans_array_extension(3,1,5), '\(s\).' ) \}
99+
[`` \left\{\ \ \begin{array}{rcrcrcr} x_1 & - & x_2 & [$pm13] & [@ abs($a13) @] x_3 & = & [$a14] \\ [$a21] x_1 & [$pm22] & [@ abs($a22) @] x_2 & [$pm23] & [@ abs($a23) @] x_3 &=& [$a24] \\ [$a31] x_1 & & & [$pm33] & [@ abs($a33) @] x_3 &=& [$a34] \end{array} \right.``]
127100

101+
[`` \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} ``] = [_]*{$ans} + s [_]*{$ans}
102+
END_PGML
128103

129-
###########################################################################
130-
# check the answer
131-
###########################################################################
132-
#ANS(vec_solution_cmp([[$m, $n, 0], [- $k , -$l, 1]]));
104+
# used in solution
133105

134-
ANS($multians->cmp());
106+
$M1 = Matrix([[$R1->value],[($R2 - $d*$R1)->value], [($R3-$m*$sign3*$R1)->value]]);
107+
108+
$M2 = Matrix([[$R1->value],[($R2 - $d*$R1)->value], [0,0,0,0] ]);
109+
110+
BEGIN_PGML_SOLUTION
111+
The augmented matrix for this system is
112+
113+
[`` [$M] ``].
114+
115+
Add [@ -$d @] times row 1 to row 2, and [@ -$m*$sign3 @] times row 1 to row 3, to obtain
116+
117+
[`` [$M1] ``].
118+
119+
Add [@ -$m @] times row 2 to row 3
120+
121+
[`` [$M2] ``].
122+
123+
Thus the original system is equivalent to
124+
125+
[`` \begin{gather}
126+
x_1-x_2+[$a13]x_3 = [$a14] \\
127+
([$sign3])x_2 + ([@ $a23-$d*$a13 @])x_3= [@ $a24-$d*$a14 @]
128+
\end{gather} ``]
129+
130+
or equivalently, solving for [`x_2`] first then [`x_1`],
131+
132+
[`` \begin{align}
133+
x_2 &= [@ $sign3*($a24-$d*$a14) @] + ([@ ($a23-$d*$a13) @])x_3 \\
134+
x_1 &= [$a14] +x_2 + ([@ -$a13 @])x_3 = [@ $a14 + $sign3*($a24-$d*$a14) @] + ([@ -$a13 + ($a23-$d*$a13) @])x_3
135+
\end{align} ``]
136+
137+
Therefore [`x_3`] is a free variable so there are infinitely many solutions. One may parametrize them with the parameter [`s`] that is given in the problem by setting [`x_3 = s`], so the parametrized solution becomes
138+
139+
[``\begin{align} x_1 & = [@ $a14 + $sign3*($a24-$d*$a14) @] + ([@ -$a13 + ($a23-$d*$a13) @])s \\
140+
x_2 &= [@ $sign3*($a24-$d*$a14) @] + ([@ ($a23-$d*$a13) @])s \\
141+
x_3 &= s
142+
\end{align} ``],
143+
144+
or, equivalently,
145+
146+
[``\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
147+
= [$xp] + s [$xh] . ``]
148+
149+
In fact there are infinitely many different solutions. For given any two real numbers [`a`] and [`b`] with [`a\neq 0`] one could create another parameter [`t = as + b`] and parametrize the solutions by [`t`]:
150+
151+
[``\begin{align}
152+
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
153+
&= [$xp] + t [$xh] \\
154+
&= [$xp] + (as + b) [$xh] \\
155+
&= \left([$xp]+b[$xh]\right) + s \left(a[$xh]\right). \end{align} ``]
156+
157+
This means that, in the "official" solution given above, one could replace [`[$xp]`] with [`\left([$xp]+b[$xh]\right)`] and replace [`[$xh]`] with [`\left(a[$xh]\right)`], and get another solution that would work equally well. (The condition [`a\neq 0`] ensures that as [`s`] ranges over all the real numbers [`t`] does also.)
158+
159+
END_PGML_SOLUTION
135160

136-
###########################################################################
137-
# solution
138-
###########################################################################
139-
$e1 = -$k;
140-
$e2 = -$l;
141-
142-
Context()->texStrings;
143-
BEGIN_SOLUTION
144-
$PAR $BBOLD SOLUTION: $EBOLD
145-
By using the Gauss-Jordan reduction, we can find the reduced row echelon form of the augmented matrix of our system which turns out to be:
146-
$PAR
147-
\[ \left[
148-
\begin{array}{rrr|r}
149-
1 & 0 & $k & $mm \cr
150-
0 & 1 & $l & $n \cr
151-
0 & 0 & 0 & 0
152-
\end{array}
153-
\right] \]
154-
$PAR
155-
Hence, \(x_1,x_2\) are lead variables and \(x_3\) is the only free variable. By writing down the system corresponding to the RRE form, we get that
156-
$PAR
157-
\[
158-
x_1 = $mm + $e1 x_3 \qquad \mbox{and} \qquad x_2 = $n + $e2 x_3,
159-
\]
160-
$PAR
161-
where \(x_3\) is free (can take any value). Therefore every vector of the form
162-
$PAR
163-
\[
164-
\left[
165-
\begin{array}{r}
166-
x_1 \\ x_2 \\ x_3
167-
\end{array}
168-
\right]
169-
=
170-
\left[
171-
\begin{array}{r}
172-
$mm \\ $n \\ 0
173-
\end{array}
174-
\right]
175-
+
176-
\left[
177-
\begin{array}{r}
178-
$e1 \\ $e2 \\ 1
179-
\end{array}
180-
\right] s
181-
\]
182-
$PAR
183-
is a solution, and every solution has that form for some value of \(s\).
184-
END_SOLUTION
185-
Context()->normalStrings;
186161

187162
COMMENT('Edited and updated in 2012/2013 by Rick Lynch @ Mizzou. Originally taken from NPL.
188163
$BR

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