@@ -2698,7 +2698,7 @@ \subsection{Adding a multiple of one row to another doesn't change
26982698Here we used linearity in a row and the fact that the determinant of a
26992699matrix with two rows the same is zero.
27002700
2701- \subsection {The determinant of $ QA$ }
2701+ \subsection {The determinant of \texorpdfstring { $ QA$ }{QA} }
27022702\label {sec:dettheory4 }
27032703
27042704To begin, we compute the determinants of the elementary
@@ -2733,8 +2733,8 @@ \subsection{The determinant of $QA$}
27332733\det (Q_1Q_2Q_3\cdots Q_kA)=\det (Q_1)\det (Q_2)\cdots\det (Q_k)\det (A).
27342734\]
27352735
2736- \subsection {The determinant of $ A$ is zero exactly when $ A $ is
2737- not invertible }
2736+ \subsection {The determinant of \texorpdfstring { $ A$ }{A} is zero exactly when
2737+ \texorpdfstring { $ A $ }{A} is not invertible }
27382738\label {sec:dettheory5 }
27392739
27402740Recall 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
27562756other words, we can take $ R=I$ . Then $ \det (R)=1 $ . Each
27572757$ \det (Q_i^{-1})$ is non-zero too, so $ \det (A)\ne 0 $ .
27582758
2759- \subsection {The product formula: $ \det (AB)=\det (A)\det (B)$ }
2759+ \subsection {The product formula:
2760+ \texorpdfstring {$ \det (AB)=\det (A)\det (B)$ }{determinant of product equals
2761+ product of determinants} }
27602762\label {sec:dettheory6 }
27612763
27622764If either $ A$ or $ B$ is non-invertible, then $ AB$ is non-invertible
0 commit comments