Update chapter4.tex

Fixed all warnings arising from the addition of hyperref. Compiling with pdflatex should give a virtually identical log as the one without the addition of the package hyperref.

Author: pabloocal

Chapters/chapter4.tex

@@ -2698,7 +2698,7 @@ \subsection{Adding a multiple of one row to another doesn't change
 Here we used linearity in a row and the fact that the determinant of a
 matrix with two rows the same is zero.
 
-\subsection{The determinant of $QA$}
+\subsection{The determinant of \texorpdfstring{$QA$}{QA}}
 \label{sec:dettheory4}
 
 To begin, we compute the determinants of the elementary
@@ -2733,8 +2733,8 @@ \subsection{The determinant of $QA$}
 \det(Q_1Q_2Q_3\cdots Q_kA)=\det(Q_1)\det(Q_2)\cdots\det(Q_k)\det(A).
 \]
 
-\subsection{The determinant of $A$ is zero exactly when $A$ is 
-not invertible}
+\subsection{The determinant of \texorpdfstring{$A$}{A} is zero exactly when 
+\texorpdfstring{$A$}{A} is not invertible}
 \label{sec:dettheory5}
 
 Recall that if $R$ denotes the reduced form of $A$, obtained by
@@ -2756,7 +2756,9 @@ \subsection{The determinant of $A$ is zero exactly when $A$ is
 other words, we can take $R=I$. Then $\det(R)=1$. Each
 $\det(Q_i^{-1})$ is non-zero too, so $\det(A)\ne 0$.
 
-\subsection{The product formula: $\det(AB)=\det(A)\det(B)$}
+\subsection{The product formula: 
+\texorpdfstring{$\det(AB)=\det(A)\det(B)$}{determinant of product equals 
+product of determinants}}
 \label{sec:dettheory6}
 
 If either $A$ or $B$ is non-invertible, then $AB$ is non-invertible