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Line breaks, Chapter M

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src/section-CRS.xml

 \beta_1\vect{B}_1+
 <![CDATA[&]]>
 \beta_2\vect{B}_2+
-\beta_3\vect{B}_3+
 \cdots+
 \beta_s\vect{B}_s+
 \cdots+
 <![CDATA[&=]]>
 \beta_1\vect{A}_1+
 \beta_2\vect{A}_2+
-\beta_3\vect{A}_3+
 \cdots+
 \beta_s\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \beta_1\vect{A}_1+
 \beta_2\vect{A}_2+
-\beta_3\vect{A}_3+
 \cdots+
 \beta_s\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \beta_1\vect{A}_1+
 \beta_2\vect{A}_2+
-\beta_3\vect{A}_3+
 \cdots+
 \beta_s\vect{A}_s+\left(\beta_t\alpha\right)\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \beta_1\vect{A}_1+
 \beta_2\vect{A}_2+
-\beta_3\vect{A}_3+
 \cdots+
 \left(\beta_s+\beta_t\alpha\right)\vect{A}_s+
 \cdots+
 <alignmath>
 <![CDATA[\gamma_1&\vect{A}_1+]]>
 \gamma_2\vect{A}_2+
-\gamma_3\vect{A}_3+
 \cdots+
 \gamma_s\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \gamma_1\vect{A}_1+
 \gamma_2\vect{A}_2+
-\gamma_3\vect{A}_3+
 \cdots+
 \gamma_s\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \gamma_1\vect{A}_1+
 \gamma_2\vect{A}_2+
-\gamma_3\vect{A}_3+
-\cdots+
-\left(-\alpha\gamma_t\vect{A}_s\right)+\gamma_s\vect{A}_s+
-\cdots+
-\left(\alpha\gamma_t\vect{A}_s+\gamma_t\vect{A}_t\right)+
-\cdots+
-\gamma_m\vect{A}_m\\
-<![CDATA[&=]]>
-\gamma_1\vect{A}_1+
-\gamma_2\vect{A}_2+
-\gamma_3\vect{A}_3+
 \cdots+
 \left(-\alpha\gamma_t+\gamma_s\right)\vect{A}_s+
 \cdots+
 <![CDATA[&=]]>
 \gamma_1\vect{B}_1+
 \gamma_2\vect{B}_2+
-\gamma_3\vect{B}_3+
 \cdots+
 \left(-\alpha\gamma_t+\gamma_s\right)\vect{B}_s+
 \cdots+

src/section-MINM.xml

 <theorem acro="CUMOS" index="unitary matrices!columns">
 <title>Columns of Unitary Matrices are Orthonormal Sets</title>
 <statement>
-<p>Suppose that $A$ is a square matrix of size $n$ with columns $S=\set{\vectorlist{A}{n}}$.  Then $A$ is a unitary matrix if and only if $S$ is an orthonormal set.</p>
+<p>Suppose that $S=\set{\vectorlist{A}{n}}$ is the set of columns of a square matrix $A$ of size $n$.  Then $A$ is a unitary matrix if and only if $S$ is an orthonormal set.</p>
 
 </statement>
 

src/section-MISLE.xml

 <![CDATA[ -2 & -3 & -1 & -3 & -2 & 0\\]]>
 <![CDATA[ -1 & -3 & -1 & -3 & 1 & 0]]>
 \end{bmatrix}
-<![CDATA[&\rref&]]>
+<![CDATA[\rref]]>
 \begin{bmatrix}
 <![CDATA[\leading{1} & 0 & 0 & 0 & 0 & -3\\]]>
 <![CDATA[0 & \leading{1} & 0 & 0 & 0 & 0\\]]>
 <![CDATA[0 & 0 & 0 & \leading{1} & 0 & 1\\]]>
 <![CDATA[0 & 0 & 0 & 0 & \leading{1} & 1]]>
 \end{bmatrix}
-<![CDATA[&\text{ so}&]]>
+<![CDATA[;]]>
 \vect{B}_1=\colvector{-3\\0\\1\\1\\1}\\
 <intertext>Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_2}$,</intertext>
 \begin{bmatrix}
 <![CDATA[ -2 & -3 & -1 & -3 & -2 & 0\\]]>
 <![CDATA[ -1 & -3 & -1 & -3 & 1 & 0]]>
 \end{bmatrix}
-<![CDATA[&\rref&]]>
+<![CDATA[\rref]]>
 \begin{bmatrix}
 <![CDATA[\leading{1} & 0 & 0 & 0 & 0 & 3\\]]>
 <![CDATA[0 & \leading{1} & 0 & 0 & 0 & -2\\]]>
 <![CDATA[0 & 0 & 0 & \leading{1} & 0 & 0\\]]>
 <![CDATA[0 & 0 & 0 & 0 & \leading{1} & -1]]>
 \end{bmatrix}
-<![CDATA[&\text{ so}&]]>
+<![CDATA[;]]>
 \vect{B}_2=\colvector{3\\-2\\2\\0\\-1}\\
 <intertext>Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_3}$,</intertext>
 \begin{bmatrix}
 <![CDATA[ -2 & -3 & -1 & -3 & -2 & 0\\]]>
 <![CDATA[ -1 & -3 & -1 & -3 & 1 & 0]]>
 \end{bmatrix}
-<![CDATA[&\rref&]]>
+<![CDATA[\rref]]>
 \begin{bmatrix}
 <![CDATA[\leading{1} & 0 & 0 & 0 & 0 & 6\\]]>
 <![CDATA[0 & \leading{1} & 0 & 0 & 0 & -5\\]]>
 <![CDATA[0 & 0 & 0 & \leading{1} & 0 & 1\\]]>
 <![CDATA[0 & 0 & 0 & 0 & \leading{1} & -2]]>
 \end{bmatrix}
-<![CDATA[&\text{ so}&]]>
+<![CDATA[;]]>
 \vect{B}_3=\colvector{6\\-5\\4\\1\\-2}\\
 <intertext>Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_4}$,</intertext>
 \begin{bmatrix}
 <![CDATA[ -2 & -3 & -1 & -3 & -2 & 1\\]]>
 <![CDATA[ -1 & -3 & -1 & -3 & 1 & 0]]>
 \end{bmatrix}
-<![CDATA[&\rref&]]>
+<![CDATA[\rref]]>
 \begin{bmatrix}
 <![CDATA[\leading{1} & 0 & 0 & 0 & 0 & -1\\]]>
 <![CDATA[0 & \leading{1} & 0 & 0 & 0 & -1\\]]>
 <![CDATA[0 & 0 & 0 & \leading{1} & 0 & 1\\]]>
 <![CDATA[0 & 0 & 0 & 0 & \leading{1} & 0]]>
 \end{bmatrix}
-<![CDATA[&\text{ so}&]]>
+<![CDATA[;]]>
 \vect{B}_4=\colvector{-1\\-1\\1\\1\\0}\\
 <intertext>Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_5}$,</intertext>
 \begin{bmatrix}
 <![CDATA[ -2 & -3 & -1 & -3 & -2 & 0\\]]>
 <![CDATA[ -1 & -3 & -1 & -3 & 1 & 1]]>
 \end{bmatrix}
-<![CDATA[&\rref&]]>
+<![CDATA[\rref]]>
 \begin{bmatrix}
 <![CDATA[\leading{1} & 0 & 0 & 0 & 0 & -2\\]]>
 <![CDATA[0 & \leading{1} & 0 & 0 & 0 & 1\\]]>
 <![CDATA[0 & 0 & 0 & \leading{1} & 0 & 0\\]]>
 <![CDATA[0 & 0 & 0 & 0 & \leading{1} & 1]]>
 \end{bmatrix}
-<![CDATA[&\text{ so}&]]>
+<![CDATA[;]]>
 \vect{B}_5=\colvector{-2\\1\\-1\\0\\1}\\
 </alignmath>
 We can now collect our 5 solution vectors into the matrix $B$,

src/section-MM.xml

 <proof>
 <p>We are assuming $A\vect{x}=B\vect{x}$ for all $\vect{x}\in\complex{n}$, so we can employ this equality for <em>any</em> choice of the vector $\vect{x}$.  However, we'll limit our use of this equality to the standard unit vectors, $\vect{e}_j$, $1\leq j\leq n$ (<acroref type="definition" acro="SUV" />).  For all $1\leq j\leq n$, $1\leq i\leq m$,
 <alignmath>
-\matrixentry{A}{ij}
+<![CDATA[&\matrixentry{A}{ij}\\]]>
 <![CDATA[&=]]>
 0\matrixentry{A}{i1}+
 \cdots+
 1\matrixentry{A}{ij}+
 0\matrixentry{A}{i,j+1}+
 \cdots+
-0\matrixentry{A}{in}\\
+0\matrixentry{A}{in}
+<![CDATA[&&]]>\text{<acroref type="theorem" acro="PCNA" />}\\
 <![CDATA[&=]]>
 \matrixentry{A}{i1}\vectorentry{\vect{e}_j}{1}+
 \matrixentry{A}{i2}\vectorentry{\vect{e}_j}{2}+
 0\matrixentry{B}{in}
 <![CDATA[&&]]>\text{<acroref type="definition" acro="SUV" />}\\
 <![CDATA[&=\matrixentry{B}{ij}]]>
+<![CDATA[&&]]>\text{<acroref type="theorem" acro="PCNA" />}
 </alignmath></p>
 
 <p>So by <acroref type="definition" acro="ME" /> the matrices $A$ and $B$ are equal, as desired.</p>
 <definition acro="MM" index="matrix multiplication">
 <title>Matrix Multiplication</title>
 <indexlocation index="vector!product with matrix" />
-<p>Suppose $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix with columns $\vectorlist{B}{p}$.  Then the <define>matrix product</define> of $A$ with $B$ is the $m\times p$ matrix where column $i$ is the matrix-vector product $A\vect{B}_i$.  Symbolically,
+<p>Suppose $A$ is an $m\times n$ matrix and $\vectorlist{B}{p}$ are the columns of an $n\times p$ matrix $B$.  Then the <define>matrix product</define> of $A$ with $B$ is the $m\times p$ matrix where column $i$ is the matrix-vector product $A\vect{B}_i$.  Symbolically,
 <equation>
 AB=A\matrixcolumns{B}{p}=\left[A\vect{B}_1|A\vect{B}_2|A\vect{B}_3|\ldots|A\vect{B}_p\right].
 </equation>
 <![CDATA[&=]]>
 \sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{B}{kj}
 </alignmath></p>
-
 </statement>
 
 <proof>
-<p>Denote the columns of  $A$ as the vectors $\vectorlist{A}{n}$ and the columns of  $B$ as the vectors $\vectorlist{B}{p}$.  Then for $1\leq i\leq m$, $1\leq j\leq p$,
+<p>Let the vectors $\vectorlist{A}{n}$ denote the columns of $A$ and let the vectors $\vectorlist{B}{p}$ denote the columns of $B$.  Then for $1\leq i\leq m$, $1\leq j\leq p$,
 <alignmath>
 \matrixentry{AB}{ij}
 <![CDATA[&=\vectorentry{A\vect{B}_j}{i}&&]]>\text{<acroref type="definition" acro="MM" />}\\
 <![CDATA[&=\vectorentry{]]>
 \vectorentry{\vect{B}_j}{1}\vect{A}_1+
 \vectorentry{\vect{B}_j}{2}\vect{A}_2+
-\vectorentry{\vect{B}_j}{3}\vect{A}_3+
 \cdots+
 \vectorentry{\vect{B}_j}{n}\vect{A}_n
 <![CDATA[}{i}&&]]>\text{<acroref type="definition" acro="MVP" />}\\
 <![CDATA[&=]]>
 \vectorentry{\vectorentry{\vect{B}_j}{1}\vect{A}_1}{i}+
 \vectorentry{\vectorentry{\vect{B}_j}{2}\vect{A}_2}{i}+
-\vectorentry{\vectorentry{\vect{B}_j}{3}\vect{A}_3}{i}+
 \cdots+
 \vectorentry{\vectorentry{\vect{B}_j}{n}\vect{A}_n}{i}
 <![CDATA[&&]]>\text{<acroref type="definition" acro="CVA" />}\\
 <![CDATA[&=]]>
 \vectorentry{\vect{B}_j}{1}\vectorentry{\vect{A}_1}{i}+
 \vectorentry{\vect{B}_j}{2}\vectorentry{\vect{A}_2}{i}+
-\vectorentry{\vect{B}_j}{3}\vectorentry{\vect{A}_3}{i}+
 \cdots+
 \vectorentry{\vect{B}_j}{n}\vectorentry{\vect{A}_n}{i}
 <![CDATA[&&]]>\text{<acroref type="definition" acro="CVSM" />}\\
 <![CDATA[&=]]>
 \matrixentry{B}{1j}\matrixentry{A}{i1}+
 \matrixentry{B}{2j}\matrixentry{A}{i2}+
-\matrixentry{B}{3j}\matrixentry{A}{i3}+
 \cdots+
 \matrixentry{B}{nj}\matrixentry{A}{in}
 <![CDATA[&&]]>\text{<acroref type="definition" acro="M" />}\\
 <![CDATA[&=]]>
 \matrixentry{A}{i1}\matrixentry{B}{1j}+
 \matrixentry{A}{i2}\matrixentry{B}{2j}+
-\matrixentry{A}{i3}\matrixentry{B}{3j}+
 \cdots+
 \matrixentry{A}{in}\matrixentry{B}{nj}
 <![CDATA[&&]]>\text{<acroref type="property" acro="CMCN" />}\\

src/section-MO.xml

 \end{bmatrix}
 </alignmath>
 then
-<equation>
-A+B=
+<alignmath>
+<![CDATA[A+B&=]]>
 \begin{bmatrix}
 <![CDATA[2&-3&4\\]]>
 <![CDATA[1&0&-7]]>
 \begin{bmatrix}
 <![CDATA[6&2&-4\\]]>
 <![CDATA[3&5&2]]>
-\end{bmatrix}
-=
+\end{bmatrix}\\
+<![CDATA[&=]]>
 \begin{bmatrix}
 <![CDATA[2+6&-3+2&4+(-4)\\]]>
 <![CDATA[1+3&0+5&-7+2]]>
 <![CDATA[8&-1&0\\]]>
 <![CDATA[4&5&-5]]>
 \end{bmatrix}
-</equation>
+</alignmath>
 </p>
 
 </example>
 <![CDATA[$B = \begin{bmatrix} 3 & 2 & 1 \\ -2 & -6 & 5\end{bmatrix}$ and]]>
 <![CDATA[$C = \begin{bmatrix} 2 & 4 \\ 4 & 0 \\ -2 & 2\end{bmatrix}$.]]>
 Let $\alpha = 4$ and $\beta = 1/2$.
-Perform the following calculations:
-<ol><li> $A + B$
-</li><li> $A + C$
-</li><li> $\transpose{B} + C$
-</li><li> $A + \transpose{B}$
-</li><li> $\beta C$
-</li><li> $4A - 3B$
-</li><li> $\transpose{A} + \alpha C$
-</li><li> $A + B - \transpose{C}$
-</li><li> $4A + 2B - 5\transpose{C}$
-</li></ol>
+Perform the following calculations:\quad
+(1)~$A + B$,\quad
+(2)~$A + C$,\quad
+(3)~$\transpose{B} + C$,\quad
+(4)~$A + \transpose{B}$,\quad
+(5)~$\beta C$,\quad
+(6)~$4A - 3B$,\quad
+(7)~$\transpose{A} + \alpha C$,\quad
+(8)~$A + B - \transpose{C}$,\quad
+(9)~$4A + 2B - 5\transpose{C}$
 </problem>
 <solution contributor="chrisblack"><ol>
      <li> <![CDATA[$A + B = \begin{bmatrix} 4 &  6 & -2 \\ 4 & -3 & 5 \end{bmatrix}$]]>
 </exercise>
 
 <exercise type="M" number="22" rough="Stray lin ind set in M_23">
-<problem contributor="robertbeezer">Determine if the set
+<problem contributor="robertbeezer">Determine if the set $S$ below is linearly independent in $M_{2,3}$.
 <equation>
-S=\set{
+\set{
 <![CDATA[\begin{bmatrix} -2 & 3 & 4 \\ -1 & 3 & -2 \end{bmatrix},\,]]>
 <![CDATA[\begin{bmatrix} 4 & -2 & 2 \\ 0 & -1 & 1 \end{bmatrix},\,]]>
 <![CDATA[\begin{bmatrix}-1 & -2 & -2 \\ 2 & 2 & 2 \end{bmatrix},\,]]>
 <![CDATA[\begin{bmatrix}-1 & 2 & -2 \\ 0 & -1 & -2 \end{bmatrix}]]>
 }
 </equation>
-is linearly independent in $M_{2,3}$.
+
 </problem>
 <solution contributor="chrisblack">Suppose that there exist constants $a_1$, $a_2$, $a_3$, $a_4$, and $a_5$ so that
 <alignmath>