Merge pull request #1274 from dlglin/updateTCNJLinInd

Convert problem to PGML and add solution

Author: dlglin

OpenProblemLibrary/TCNJ/TCNJ_LinearIndependence/problem3.pg

@@ -14,64 +14,53 @@
 ## EditionText1('1')
 ## Section1('2.3')
 ## Problem1('')
-DOCUMENT();        # This should be the first executable line in the problem.
+DOCUMENT();    # This should be the first executable line in the problem.
 
-loadMacros(
-  "PGstandard.pl",
-  "PGchoicemacros.pl",
-  "PGgraphmacros.pl",
-  "PGmatrixmacros.pl",
-  "PGcourse.pl"
-);
+loadMacros("PGstandard.pl", "PGML.pl", "PGcourse.pl");
 
-TEXT(beginproblem());
 $showPartialCorrectAnswers = 1;
 
-$a = non_zero_random(1,8,1);
-$b = non_zero_random(1,8,1);
-if($a<$b){
-  $c=$b;
-  $b=$a;
-  $a=$c;
+$a = non_zero_random(1, 8, 1);
+$b = non_zero_random(1, 8, 1);
+if ($a < $b) {
+    $c = $b;
+    $b = $a;
+    $a = $c;
 } else {
-if ($a == $b) { $a=$b+1; }}
-
+    if ($a == $b) { $a = $b + 1; }
+}
 
 ## $a is now the larger value
 ## $a and -$b will be the roots of the equation.
 
-$c= $a -$b;
-$d= $a * $b;
+$c = $a - $b;
+$d = $a * $b;
 
+Context("Numeric")->variables->are('k' => 'Real');
+Context()->noreduce('(-x)-y');
+$f    = Formula("$c*k+$d")->reduce();
+$g    = Formula("k^2-$c*k-$d")->reduce();
+$ans1 = List(-$b, $a);
 
-if( $a == $b+1){
-BEGIN_TEXT
-$BR
+BEGIN_PGML
+If [`k`] is a real number, then the vectors [`\begin{bmatrix}1\\ k\end{bmatrix},\begin{bmatrix}k\\[$f]\end{bmatrix}`]
+are linearly independent precisely when [`k \ne`][_]{$ans1}{5}
 
-If \(k\) is a real number, then 
-the
-vectors  \( (1, k),  (k,  k+ $d) \) are linearly independent precisely when
+If there is more than one value, enter a list separated by commas.
+END_PGML
 
-\( k \ne a, b \), $BR
-where  \(a= \)\{ans_rule(10)\},  \(b = \)  \{ans_rule(10)\}, and
-\( a<b \).
-END_TEXT
-}else{
-BEGIN_TEXT
-$BR
+BEGIN_PGML_SOLUTION
+The vectors will be linearly independent provided that if [`a\begin{bmatrix}1\\ k\end{bmatrix}+b\begin{bmatrix}k\\[$f]\end{bmatrix}=\begin{bmatrix} 0\\0\end{bmatrix}`], then [`a=0`] and [`b=0`].
 
-If \(k\) is a real number, then the
-vectors  \( (1, k),  (k, $c k+ $d) \) are linearly independent precisely when
-
-\( k \ne a, b \), $BR
-where  \(a= \)\{ans_rule(5)\},  \(b = \)  \{ans_rule(5)\}, and
-\( a<b \).
-END_TEXT
-}
+This is equivalent to checking that the system of equations given by [`\begin{bmatrix}1 & k\\k & [$f]\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}`] has a unique solution.
 
+One way to do this is to check that the determinant of the coefficient matrix is nonzero:
 
+[`\begin{array}{rcl}\begin{vmatrix}1 & k\\k & [$f]\end{vmatrix}\ne 0 & \Rightarrow & ([$f])-k^2\ne 0\\
+& \Rightarrow & [$g]\ne 0\\
+& \Rightarrow & (k+[$b])(k-[$a])\ne 0\\
+& \Rightarrow & k\ne-[$b],[$a]
+\end{array}`]
+END_PGML_SOLUTION
 
-$k=2;
-ANS(num_cmp(-$b));
-ANS(num_cmp($a));
-ENDDOCUMENT();       # This should be the last executable line in the problem.
+ENDDOCUMENT();    # This should be the last executable line in the problem.