diff --git a/source/linear-algebra/source/05-GT/01.ptx b/source/linear-algebra/source/05-GT/01.ptx
index b6fd2316c..cb62fec92 100644
--- a/source/linear-algebra/source/05-GT/01.ptx
+++ b/source/linear-algebra/source/05-GT/01.ptx
@@ -158,7 +158,7 @@ What is the area of the transformed unit square?
   <me>B\vec e_1=\left[\begin{array}{cc} 2 &amp; 3 \\ 0 &amp; 4 \end{array}\right]\left[\begin{array}{c}1\\0\end{array}\right]
   =\left[\begin{array}{c}2\\0\end{array}\right]=2\vec e_1</me>
 
-  <me>
+  <me>\def\arraystretch{1.25}
     B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]
       =
     \left[\begin{array}{cc} 2 &amp; 3 \\ 0 &amp; 4 \end{array}\right]\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]
@@ -175,7 +175,7 @@ What is the area of the transformed unit square?
   \draw[thin,gray,&lt;-&gt;] (-4,0)-- (4,0);
   \draw[thin,gray,&lt;-&gt;] (0,-4)-- (0,4);
   \draw[thick,blue,-&gt;] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0);
-  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
+  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(\def\arraystretch{1.25} B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
   \draw[thick,red,-&gt;] (0,0) -- (1,0);
   \draw[thick,red,-&gt;] (0,0) -- (0.75,0.5);
   \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5);
@@ -228,7 +228,7 @@ What is the area of the transformed unit square?
   \draw[thin,gray,&lt;-&gt;] (-4,0)-- (4,0);
   \draw[thin,gray,&lt;-&gt;] (0,-4)-- (0,4);
   \draw[thick,blue,-&gt;] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0);
-  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
+  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(\def\arraystretch{1.25} B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
   \draw[thick,red,-&gt;] (0,0) -- (1,0);
   \draw[thick,red,-&gt;] (0,0) -- (0.75,0.5);
   \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5);
@@ -275,7 +275,7 @@ In order to figure out how to compute it, we first figure out the properties it
   \draw[thin,gray,&lt;-&gt;] (-4,0)-- (4,0);
   \draw[thin,gray,&lt;-&gt;] (0,-4)-- (0,4);
   \draw[thick,blue,-&gt;] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0);
-  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
+  \draw[thick,blue,-&gt;] (0,0) -- ++(3,2) node[above] {\(\def\arraystretch{1.25} B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)};
   \draw[thick,red,-&gt;] (0,0) -- (1,0);
   \draw[thick,red,-&gt;] (0,0) -- (0.75,0.5);
   \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5);
@@ -301,7 +301,7 @@ If <m>\det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)</m> is the
 area of resulting parallelogram, what is the value of <m>\det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)</m>?
         </p>
 <figure>
-	<image width="75%" xml:id="GT1-image-identity-determinant">
+	<image width="50%" xml:id="GT1-image-identity-determinant">
 		<latex-image>
 \begin{tikzpicture}
 \fill[red!50!white] (0,0) rectangle (1,1);
diff --git a/source/linear-algebra/source/05-GT/03.ptx b/source/linear-algebra/source/05-GT/03.ptx
index 1a06ceea5..f56ed8187 100644
--- a/source/linear-algebra/source/05-GT/03.ptx
+++ b/source/linear-algebra/source/05-GT/03.ptx
@@ -15,7 +15,7 @@
   <activity>
     <introduction>
       <p>
-        Let <m>R\colon\IR^2\to\IR^2</m> be the transformation given by rotating vectors about the origin through and angle of <m>45^\circ</m>, and let <m>S\colon\IR^2\to\IR^2</m> denote the transformation that reflects vectors about the line <m>x_1=x_2</m>.
+        Let <m>R\colon\IR^2\to\IR^2</m> be the transformation given by rotating vectors about the origin through an angle of <m>45^\circ</m>, and let <m>S\colon\IR^2\to\IR^2</m> denote the transformation that reflects vectors about the line <m>x_1=x_2</m>.
       </p>
     </introduction>
     <task>