Implementing change of basis outcome in MX

source/linear-algebra/exercises/bank.xml

@@ -196,16 +196,17 @@
             <path>outcomes/MX/MX2</path>
             <slug>MX2</slug>
             <description>
-                Determine if a matrix is invertible, and if so, compute its inverse.
+                Determine if a matrix is invertible, and if so, compute its inverse
+                and use it to solve an appropriate system of equations.
             </description>
         </outcome>
         <outcome>
-            <title>Solving systems with inverse matrices</title>
+            <title>Change of basis</title>
             <path>outcomes/MX/MX3</path>
             <slug>MX3</slug>
             <description>
-                Find the unique solution for a system of equations using an
-                inverse matrix.
+                Calculate the change-of-basis matrix for the standard basis to a
+                non-standard basis of <m>\mathbb R^n</m>.
             </description>
         </outcome>
         <outcome>

source/linear-algebra/exercises/outcomes/MX/MX2/generator.sage

@@ -7,20 +7,33 @@ class Generator(BaseGenerator):
         shuffle(labels)
         # invertible matrix
         A=CheckIt.simple_random_matrix_of_rank(4,rows=4,columns=4)
+        solution = column_matrix(
+            vector(QQ, [randrange(1,5)*choice([-1,1]) for _ in range(4)])
+        )
+        vars = column_matrix(var("x_1 x_2 x_3 x_4"))
+        constants = A*solution
+        m = A.augment(constants, subdivide=True)
         matrices = [{
             "matrix": A,
             "rref": A.rref(),
             "invertible": True,
             "inverse": A^(-1),
             "label": labels[0],
+            "vars": vars,
+            "solution": solution,
+            "constant_vector": constants,
+            "vector_eq": TBIL.VectorEquation(m),
         }]
         # non-invertible matrix
         A=CheckIt.simple_random_matrix_of_rank(choice([2,3]),rows=4,columns=4)
+        constants = A*solution
+        m = A.augment(constants, subdivide=True)
         matrices += [{
             "matrix": A,
             "rref": A.rref(),
             "invertible": False,
             "label": labels[1],
+            "vector_eq": TBIL.VectorEquation(m),
         }]
 
         shuffle(matrices)

source/linear-algebra/exercises/outcomes/MX/MX2/template.xml

@@ -68,6 +68,24 @@ of an appropriate augmented matrix).
 <!-- {{/invertible}} -->
 <!-- {{^invertible}} -->
 N/A
+<!-- {{/invertible}} -->
+                </p>
+            </outtro>
+        </knowl>
+        <knowl>
+            <content>
+                <p>
+If the matrix is invertible, explain how to use it with
+technology to solve the vector equation <me>{{vector_eq}}.</me>
+                </p>
+            </content>
+            <outtro>
+                <p>
+<!-- {{#invertible}} -->
+<me>{{inverse}}{{constant_vector}}={{solution}}</me>
+<!-- {{/invertible}} -->
+<!-- {{^invertible}} -->
+N/A
 <!-- {{/invertible}} -->
                 </p>
             </outtro>

source/linear-algebra/exercises/outcomes/MX/MX3/generator.sage

@@ -5,22 +5,25 @@ class Generator(BaseGenerator):
     def data(self):
         # create a 3x3 invertible matrix
         A = CheckIt.simple_random_matrix_of_rank(3,rows=3,columns=3)
-        solution = column_matrix(
+        new_vector = column_matrix(
             vector(QQ, [randrange(1,5)*choice([-1,1]) for _ in range(3)])
         )
-        vars = column_matrix(var("x_1 x_2 x_3"))
-        constants = A*solution
-        result = {
-            "vars": vars,
-            "solution": solution,
-            "matrix": A,
-            "inverse": A^(-1),
-            "constant_vector": constants,
+        old_vector = A*new_vector
+        lin_combo_exp = TBIL.LinearCombination(
+            [
+                new_vector[i]
+                for i in range(3)
+            ],
+            [
+                LatexExpr(f"\\vec b_{i+1}")
+                for i in range(3)
+            ],
+        )
+        return {
+            "M_B": A^(-1),
+            "v": old_vector,
+            "lin_combo_exp": lin_combo_exp,
+            "b_1": column_matrix(A.columns()[0]),
+            "b_2": column_matrix(A.columns()[1]),
+            "b_3": column_matrix(A.columns()[2]),
         }
-        m = A.augment(constants, subdivide=True)
-        if choice([True,False]):
-            result["system"] = CheckIt.latex_system_from_matrix(m)
-        else:
-            result["vector_eq"] = TBIL.VectorEquation(m)
-        return result
-

source/linear-algebra/exercises/outcomes/MX/MX3/template.xml

@@ -2,40 +2,35 @@
 <knowl mode="exercise" xmlns="https://spatext.clontz.org" version="0.2">
     <intro>
         <p>
-Consider the
-<!-- {{#system}} -->
-system of equations <me>{{system}}</me>
-<!-- {{/system}} -->
-<!-- {{#vector_eq}} -->
-vector equation <me>{{vector_eq}}</me>
-<!-- {{/vector_eq}} -->
-with a unique solution.
+Let <m>\vec{v}= {{v}}</m> and <m>\mathcal B=\{\vec b_1,\vec b_2,\vec b_3\}</m>
+where <m>\vec b_1 = {{b_1}}</m>, <m>\vec b_2= {{b_2}}</m>,
+and <m>\vec b_3 = {{b_3}}</m>.
         </p>
     </intro>
     <knowl>
         <content>
             <p>
-Explain and demonstrate how this problem can be restated
-using matrix multiplication.
+Explain and demonstrate how to verify that <m>\mathcal{B}</m> is a basis of <m>\mathbb R^3</m>
+and how to calculate <m>M_\mathcal{B}</m>, the change-of-basis matrix from the standard basis of
+<m>\mathbb R^3</m> to <m>\mathcal{B}</m>.
             </p>
         </content>
         <outtro>
-            <p>
-<me>{{matrix}}{{vars}}={{constant_vector}}</me>
-            </p>
+            <p><m>M_{\mathcal B} = {{M_B}}</m></p>
         </outtro>
     </knowl>
+
     <knowl>
         <content>
             <p>
-Use the properties of matrix multiplication to find the
-unique solution.
+Explain and demonstrate how to use <m>M_\mathcal{B}</m> to express <m>\vec{v}</m> in terms of
+<m>\mathcal{B}</m>-basis vectors. 
             </p>
         </content>
         <outtro>
             <p>
-<me>{{inverse}}{{constant_vector}}={{solution}}</me>
+                <m>\vec v = {{lin_combo_exp}}</m>
             </p>
         </outtro>
     </knowl>
-</knowl>
+</knowl>

source/linear-algebra/source/04-MX/02.ptx

@@ -234,6 +234,36 @@ Is the matrix <m>\left[\begin{array}{ccc} 2 &amp; 3 &amp; 1 \\ -1 &amp; -4 &amp;
     </statement>
 </observation>
 
+<observation>
+    <statement>
+        <p>
+            Let <m>T:\IR^n\to\IR^n</m> be a linear bijection with standard matrix <m>A</m> and suppose <m>\vec{b}\in\IR^n</m>.
+            By definition of the inverse map and inverse matrix, the vector <m>\vec{x}=A^{-1}\vec{b}</m> is the unique solution to the equation <m>A\vec{x}=\vec{b}</m>. 
+        </p>
+        <p>
+            In other words, when the matrix <m>A</m> is invertible, we have a new method for solving the equation <m>A\vec{x}=\vec{b}</m>: we can first calculate <m>A^{-1}</m> and then calculate the product <m>\vec{x}=A^{-1}\vec{b}</m>.
+        </p>
+    </statement>
+</observation>
+
+ <activity>
+        <introduction>
+          <p> Consider the vector equation <me>x_{1} \left[\begin{array}{c} 0 \\ 1 \\ 1\\1 \end{array}\right] + x_{2} \left[\begin{array}{c} 0 \\ 0 \\ 1\\-1 \end{array}\right] + x_{3} \left[\begin{array}{c} 0 \\ -1 \\ 0\\-1 \end{array}\right] +x_{4}\left[\begin{array}{c} -1 \\ -4 \\ -4\\2 \end{array}\right]= \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1\end{array}\right]</me>  with a unique solution. </p>
+        </introduction>
+          <task>
+            <statement>
+              <p> Use technology to both verify that the coefficient matrix is invertible and calculate its inverse.</p>
+            </statement>
+          </task>
+          <task>
+            <statement>
+              <p> Explain and demonstrate how to use the inverse to find the unique solution to the given vector equation. </p>
+            </statement>
+          </task>
+    </activity>
+<sage language="octave">
+</sage>
+
 <activity estimated-time='10'>
     <introduction>
         <p>
@@ -308,6 +338,77 @@ Is the matrix <m>\left[\begin{array}{ccc} 2 &amp; 3 &amp; 1 \\ -1 &amp; -4 &amp;
             </statement>
         </task>
     </activity>
+    <activity>
+        <introduction>
+            <p>
+                Solving linear systems using matrix multiplication is most useful when we are working with one common coefficient matrix, and varying the right-hand side. 
+                That is, when we have <m>A\vec{x}=\vec{b}</m> for several different values of <m>\vec{b}</m>. 
+            </p>
+            <p>
+                In the following, let <m>A=\left[\begin{matrix}2 &amp; -1 &amp; -6\\ 2 &amp; 1 &amp; 3\\ 1 &amp; 1 &amp; 4\end{matrix}\right]</m> and consider the following questions about various equations of the form <m>A\vec{x}=\vec{b}</m>?
+            </p>
+        </introduction>
+        <task>
+            <statement>
+                <p>
+                    Suppose that <m>\vec{b}=\left[\begin{matrix} 1\\1\\1\end{matrix}\right]</m>. 
+                    If asked to solve the equation <m>A\vec{x}=\vec{b}</m>, which of the following approaches do you prefer?
+                    <ol marker="A.">
+                        <li>
+                            <p>
+                                Calculate <m>\RREF[A|\vec{b}]</m>.
+                            </p>
+                        </li>
+                        <li>
+                            <p>
+                                Calculate <m>A^{-1}</m> and then compute <m>\vec{x}=A^{-1}\vec{b}</m>
+                            </p>
+                        </li>
+                    </ol>
+                </p>
+            </statement>
+        </task>
+        <task>
+            <statement>
+                <p>
+                    Suppose that <m>\vec{b}_1,\vec{b}_2,\vec{b}_3=\left[\begin{matrix} 1\\1\\1\end{matrix}\right],\left[\begin{matrix} 2\\1\\3\end{matrix}\right],\left[\begin{matrix} -1\\3\\5\end{matrix}\right]</m>. 
+                    If asked to solve each of the equations <m>A\vec{x}=\vec{b}_1, A\vec{x}=\vec{b}_2, A\vec{x}=\vec{b}_3</m>, which of the following approaches do you prefer?
+                    <ol marker="A.">
+                        <li>
+                            <p>
+                                Calculate <m>\RREF[A|\vec{b}_1]</m>, <m>\RREF[A|\vec{b}_2]</m>, and <m>\RREF[A|\vec{b}_3]</m>
+                            </p>
+                        </li>
+                        <li>
+                            <p>
+                                Calculate <m>A^{-1}</m> and then compute <m>\vec{x}=A^{-1}\vec{b}_1</m>, <m>\vec{x}=A^{-1}\vec{b}_2</m>, and <m>\vec{x}=A^{-1}\vec{b}_3</m>
+                            </p>
+                        </li>
+                    </ol>
+                </p>
+            </statement>
+        </task>
+        <task>
+            <statement>
+                <p>
+                    Suppose that <m>\vec{b}_1,\dots, \vec{b}_{10}</m> are 10 distinct vectors. 
+                    If asked to solve each of the equations <m>A\vec{x}=\vec{b}_1, \dots,  A\vec{x}=\vec{b}_{10}</m>, which of the following approaches do you prefer?
+                    <ol marker="A.">
+                        <li>
+                            <p>
+                                Calculate <m>\RREF[A|\vec{b}_1]</m>, ... <m>\RREF[A|\vec{b}_{10}]</m>.
+                            </p>
+                        </li>
+                        <li>
+                            <p>
+                                Calculate <m>A^{-1}</m> and then compute <m>\vec{x}=A^{-1}\vec{b}_1</m>, ... <m>\vec{x}=A^{-1}\vec{b}_{10}</m>.
+                            </p>
+                        </li>
+                    </ol>
+                </p>
+            </statement>
+        </task>
+    </activity>
 </subsection>
     <subsection>
         <title>Videos</title>
@@ -355,6 +456,33 @@ Is the matrix <m>\left[\begin{array}{ccc} 2 &amp; 3 &amp; 1 \\ -1 &amp; -4 &amp;
              </ul>
          </statement>
      </exploration>
+<exploration>
+            <statement>
+                Use row reduction to find the inverse of the following general matrix. Give conditions on which this inverse exists.
+                <me>\left[\begin{array}{ccc}1 &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; i \end{array}\right]</me>
+             </statement>
+         </exploration>
+
+<exploration>
+            <statement>Assume that <m>H</m> is invertible, and that <m>HG</m> is the zero matrix. Prove that <m>G</m> must be the zero matrix. Would this still be true if <m>H</m> were not invertible?
+                </statement>
+         </exploration>
+
+<exploration>
+            <statement>If <m>H</m> is invertible and <m>r \in \mathbb{R}</m>, what is the inverse of <m>rH</m>?
+                </statement>
+         </exploration>
+
+<exploration>
+            <statement>If <m>H</m> and <m>G</m> are invertible, is <m>H^{-1} + G^{-1} = (H+G)^{-1}</m>?
+                </statement>
+         </exploration>
+
+<exploration>
+            <statement>Prove that if <m>A</m>, <m>P</m>, and <m>Q</m> are invertible with <m>PAQ = I</m>, then <m>A^{-1} = QP</m>.
+             </statement>
+         </exploration>
+
     </subsection>
     <subsection>
         <title>Sample Problem and Solution</title>