Minor latex edits so codespell avoids 'te'

Author: cbm755

Chapters/chapter4.tex

@@ -3032,9 +3032,8 @@ \section{Solutions to Chapter Problems}
 0&1 + t + {{t^2}\over{2}} + {{t^3}\over{3!}} + \cdots\cr
 }\right] \\
 &=&\left[\matrix{
-e^t&
-te^t\cr
-0&e^t\cr
+  e^t & t e^t\cr
+  0   & e^t\cr
 }\right]
 \end{eqnarray*}
 \item[d,e)] We are looking for all matrices that satisfy $B^2=A$. Let 
@@ -4068,4 +4067,4 @@ \section{Solutions to Chapter Problems}
 \end{eqnarray*}
 
 
-\end{enumerate}
+\end{enumerate}

Chapters/chapter6.tex

@@ -1779,7 +1779,7 @@ \subsection{The matrix exponential and differential equations}
 cases. Consider the matrix $A=\left[\matrix{1&1\cr0&1}\right]$. This
 matrix does not have a basis of eigenvectors. So it cannot be
 diagonalized.  However, in a homework problem, you showed that $e^{tA}
-= \left[\matrix{e^t&te^t\cr 0&e^t}\right]$. Thus the solution to
+= \left[\matrix{e^t & t e^t\cr 0 & e^t}\right]$. Thus the solution to
 \[
 \yy'(t) = \left[\matrix{1&1\cr 0&1}\right]\yy(t)
 \]
@@ -1790,8 +1790,8 @@ \subsection{The matrix exponential and differential equations}
 is
 \[
 \yy(t) = e^{tA}\left[\matrix{2\cr 1\cr}\right]
-=\left[\matrix{e^t&te^t\cr 0&e^t}\right]\left[\matrix{2\cr 1\cr}\right]
-=\left[\matrix{2e^t+te^t\cr e^t\cr}\right]
+=\left[\matrix{e^t & t e^t\cr 0 & e^t}\right]\left[\matrix{2\cr 1\cr}\right]
+=\left[\matrix{2e^t + t e^t\cr e^t\cr}\right]
 \]
 Notice that this solution involves a power of $t$ in addition to
 exponentials.