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\begin{document}
\centerline{Math 201 lecture for Wednesday, Week 8}
\bigskip

\centerline{\textbf{Permutation expansion of the determinant}}  
\medskip

\textbf{Definition.} A {\em permutation} of a set~$X$ is a bijective mapping of~$X$
to itself.  If~$\sigma$ and~$\tau$ are permutations of~$X$, then so is their
composition~$\sigma\circ\tau$.  The collection of all permutations of~$X$ along
with the binary operation~$\circ$ given by composition of functions is called the {\em
symmetric group on~$X$}.  For each nonnegative integer~$n$, let~$[n]:=\left\{
1,\dots,n \right\}$.  The symmetric group on~$[n]$ is called the {\em symmetric
group of degree~$n$} and denoted by~$\sn_n$.
\medskip

\textbf{Example.} Here are six elements of~$\sn_3$:
\begin{center}
  \begin{tikzpicture}
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l)--(1r);
      \draw[->] (2l)--(2r);
      \draw[->] (3l)--(3r);
    \end{scope}
    \begin{scope}[xshift=1.4in,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (2r);
      \draw[->] (2l) to[out=0,in=180] (1r);
      \draw[->] (3l) to[out=0,in=180] (3r);
    \end{scope}
    \begin{scope}[xshift=2.8in,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=-20,in=180] (3r);
      \draw[->] (2l) to[out=0,in=180] (2r);
      \draw[->] (3l) to[out=0,in=200] (1r);
    \end{scope}

    \begin{scope}[xshift=0,yshift=-1.0in]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (1r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (2r);
    \end{scope}
    \begin{scope}[xshift=1.4in,yshift=-1.0in]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (2r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (1r);
    \end{scope}
    \begin{scope}[xshift=2.8in,yshift=-1.0in]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (3r);
      \draw[->] (2l) to[out=0,in=180] (1r);
      \draw[->] (3l) to[out=0,in=180] (2r);
    \end{scope}
  \end{tikzpicture}.
\end{center}
\medskip

\textbf{Note.} Define the {\em factorial} of a natural number as follows: $0!=1$,
and for each integer~$n>0$, recursively define~$n!=n(n-1)!$.
Thus,~$1!=1$,~$2!=2\cdot 1$,~$3!=3\cdot 2\cdot1=6$,~$4!=4\cdot3\cdot2\cdot1=24$,
etc.  Then the number of elements of~$\sn_n=n!$ since we can uniquely determine
every permutation~$\sigma$ by first choosing one of~$n$ values for~$\sigma(1)$,
then any of the remaining values~$n-1$ for~$\sigma(2)$, then one of the
remaining~$n-2$ values of~$\sigma(3)$, etc.
\medskip

\textbf{Definition.} Let~$\sigma\in\sn_n$.  The {\em permutation matrix
corresponding to~$\sigma$} is the~$n\times n$ matrix~$P_{\sigma}$
whose~$i$-th row is~$e_{\sigma(i)}$.  Another way of saying this is that,~$P_{\sigma}$ is
obtained by permuting the columns of the identity matrix,~$I_n$, according
to~$\sigma$: put~$e_j$ in column~$\sigma(j)$.
\medskip

\textbf{Example.} Let~$\sigma\in\sn_3$ be defined by~$\sigma(1)=2$, $\sigma(2)=3$,
and~$\sigma(3)=1$.  Then
\[
  P_{\sigma}=
  \left(\begin{array}{ccc}
      0&1&0\\
      0&0&1\\
      1&0&0
  \end{array} \right).
\]
\medskip
\medskip

\textbf{Exercise.} Let~$\sigma,\tau\in\sn_n$ and let~$A$ be an~$n\times n$ matrix.
\begin{enumerate}
  \item If the rows of~$A$ are~$(r_1,\dots,r_n)$, then the~$i$-th row of
    $P_{\sigma}A$ is~$r_{\sigma(i)}$.  In other words, the multiplying on the
    left by~$P_{\sigma}$ permutes the rows of~$A$ in the same way that the rows
    of~$I_n$ are permuted to form~$P_{\sigma}$.  We leave it as an exercise to
    the reader to investigate the effect of multiplying~$A$ on the right
    by~$P$.
  \item $P_{\sigma}e_{\sigma(i)}=e_{i}$,
    and~$P_{\sigma}P_{\tau}=P_{\tau\circ\sigma}$.  (Note that the order
      of~$\sigma$ and~$\tau$ have switched.)
\end{enumerate}

\textbf{Rook placements.}  Permutation matrices are exactly those that have a
single~$1$ in each row and in each column.  Thus, if the~$1$s in a permutation
matrix were replaced by rooks in the game of chess, then no rook would be
attacking another.  We sometimes call a permutation matrix a {\em rook
placement}.
\medskip

\textbf{Definition.} The {\em sign} of~$\sigma\in\sn_n$ is
\[
  \mathrm{sign}(\sigma) = \det(P_{\sigma})=\pm1.
\]
A permutation is {\em even} if its sign is~$1$ and {\em odd} if its sign
is~$-1$.
\medskip

Every permutation matrix~$P_{\sigma}$ may be obtained from~$I_n$ through a sequence
of transpositions of columns, i.e., a sequence in which each step consists of
swapping two columns.  In the example above,~$P_{\sigma}$ is formed by permuting
columns~$1$,~$2$, and~$3$ of~$I_3$ as follows:
\begin{center}
  \begin{tikzpicture}
    \begin{scope}[xshift=0in,yshift=-0in]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (2r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (1r);
    \end{scope}
  \end{tikzpicture}.
\end{center}
This permutation could have been obtained from two transpositions:
\begin{center}
  \begin{tikzpicture}
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (1rr) at (3.4,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (2rr) at (3.4,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \node (3rr) at (3.4,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (1r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (2r);
      \draw[->] (1r) to[out=0,in=180] (2rr);
      \draw[->] (2r) to[out=0,in=180] (1rr);
      \draw[->] (3r) to[out=0,in=180] (3rr);
    \end{tikzpicture}.
\end{center}
Thus, every permutation matrix can be obtained as the product of permutation
matrices corresponding to transpositions.  Swapping two columns in a matrix
changes the sign of the determinant.  Therefore, even though a
permutation~$\sigma$ may be realized in different ways as sequences of
transpositions, the parity (evenness or oddness) of the number of
transpositions required is well-defined: the number is even
if~$\det(P_{\sigma})=1$ and odd if~$\det(P_{\sigma})=-1$.
\medskip

\textbf{Theorem.} Let~$A$ be an~$n\times n$ matrix.  Then
\[
  \det(A)=\sum_{\sigma\in\sn_n}\mathrm{sign}(\sigma)A_{1\sigma(1)}A_{2\sigma(2)}\cdots A_{n\sigma(n)}.
\]

\textbf{Example.} Consider the case~$n=3$.  Then
\[
A = 
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right).
\]
Each term $A_{1\sigma(1)}A_{2\sigma(2)}\cdots A_{n\sigma(n)}$ in the formula in
the theorem should be thought of as the product of the entries corresponding to
a rook placement.  The permutations, rook placements, and corresponding
summands appear in Figure~\ref{fig1}.\footnote{Note that the sign of the permutation is
also equal to~$(-1)^{c}$ where~$c$ is the number of times two arrows cross in
the diagram for the permutation in the left-most column.}
\begin{figure}
  \centering
  \begin{tikzpicture}
    % ()
    \begin{scope}[yshift=0in]
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (1r);
      \draw[->] (2l) to[out=0,in=180] (2r);
      \draw[->] (3l) to[out=0,in=180] (3r);
    \end{scope}
    \begin{scope}[xshift=2in,yshift=0]
      \node at (0,0) {$
	\displaystyle
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right) 
      $};
      \def\x{-1.2}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{-0.3}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{0.6}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
    \end{scope}
    \begin{scope}[xshift=3.5in,yshift=0]
      \node at (0,0) {$a_{11}a_{22}a_{33}$};
    \end{scope}
  \end{scope}
    % (123)
    \begin{scope}[yshift=-1in]
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (2r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (1r);
    \end{scope}
    \begin{scope}[xshift=2in,yshift=0]
      \node at (0,0) {$
	\displaystyle
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right) 
      $};
      \def\x{-1.2}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{-0.3}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{0.6}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
    \end{scope}
    \begin{scope}[xshift=3.5in,yshift=0]
      \node at (0,0) {$a_{12}a_{23}a_{31}$};
    \end{scope}
  \end{scope}
    % (132)
    \begin{scope}[yshift=-2in]
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (3r);
      \draw[->] (2l) to[out=0,in=180] (1r);
      \draw[->] (3l) to[out=0,in=180] (2r);
    \end{scope}
    \begin{scope}[xshift=2in,yshift=0]
      \node at (0,0) {$
	\displaystyle
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right) 
      $};
      \def\x{-1.2}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{-0.3}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{0.6}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
    \end{scope}
    \begin{scope}[xshift=3.5in,yshift=0]
      \node at (0,0) {$a_{13}a_{21}a_{32}$};
    \end{scope}
  \end{scope}
    % (12)
    \begin{scope}[yshift=-3in]
      \begin{scope}[xshift=0,yshift=0]
	\node (1l) at (0.6,0.6) {$1$};
	\node (1r) at (2,0.6) {$1$};
	\node (2l)  at (0.6,0) {$2$};
	\node (2r)  at (2,0) {$2$};
	\node (3l)  at (0.6,-0.6) {$3$};
	\node (3r)  at (2,-0.6) {$3$};
	\draw[->] (1l) to[out=0,in=180] (2r);
	\draw[->] (2l) to[out=0,in=180] (1r);
	\draw[->] (3l) to[out=0,in=180] (3r);
      \end{scope}
      \begin{scope}[xshift=2in,yshift=0]
	\node at (0,0) {$
	  \displaystyle
  \left(\begin{array}{ccc}
      a_{11}&a_{12}&a_{13}\\
      a_{21}&a_{22}&a_{23}\\
      a_{31}&a_{32}&a_{33}
  \end{array} \right) 
	$};
	\def\x{-1.2}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
	\def\x{-0.3}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
	\def\x{0.6}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \end{scope}
      \begin{scope}[xshift=3.5in,yshift=0]
	\node at (0,0) {$-a_{12}a_{21}a_{33}$};
      \end{scope}
    \end{scope}
    % (13)
    \begin{scope}[yshift=-4in]
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=-20,in=180] (3r);
      \draw[->] (2l) to[out=0,in=180] (2r);
      \draw[->] (3l) to[out=0,in=200] (1r);
    \end{scope}
    \begin{scope}[xshift=2in,yshift=0]
      \node at (0,0) {$
	\displaystyle
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right) 
      $};
      \def\x{-1.2}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{-0.3}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{0.6}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
    \end{scope}
    \begin{scope}[xshift=3.5in,yshift=0]
      \node at (0,0) {$-a_{13}a_{22}a_{31}$};
    \end{scope}
  \end{scope} 
    % (23)
    \begin{scope}[yshift=-5in]
    \begin{scope}[xshift=0,yshift=0]
      \node (1l) at (0.6,0.6) {$1$};
      \node (1r) at (2,0.6) {$1$};
      \node (2l)  at (0.6,0) {$2$};
      \node (2r)  at (2,0) {$2$};
      \node (3l)  at (0.6,-0.6) {$3$};
      \node (3r)  at (2,-0.6) {$3$};
      \draw[->] (1l) to[out=0,in=180] (1r);
      \draw[->] (2l) to[out=0,in=180] (3r);
      \draw[->] (3l) to[out=0,in=180] (2r);
    \end{scope}
    \begin{scope}[xshift=2in,yshift=0]
      \node at (0,0) {$
	\displaystyle
\left(\begin{array}{ccc}
    a_{11}&a_{12}&a_{13}\\
    a_{21}&a_{22}&a_{23}\\
    a_{31}&a_{32}&a_{33}
\end{array} \right) 
      $};
      \def\x{-1.2}; \def\y{0.7}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{-0.3}; \def\y{-0.3}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
      \def\x{0.6}; \def\y{0.2}; \fill[blue,opacity=0.15] (\x,\y) rectangle (\x+0.6,\y-0.45); 
    \end{scope}
    \begin{scope}[xshift=3.5in,yshift=0]
      \node at (0,0) {$-a_{11}a_{23}a_{32}$};
    \end{scope}
  \end{scope}
\end{tikzpicture}.
  \caption{Computing the determinant of the~$3\times 3$ matrix~$A=(a_{ij})$ via
    rook placements.  The determinant is the sum of the terms in the right-most
column.}\label{fig1}
\end{figure}

\medskip

\begin{proof}[Proof of permutation formula for the determinant.]
  We want to compute
  \[
    \det(A_{11}e_1+A_{12}e_2+\cdots+A_{1n}e_n,\quad\dots\quad,A_{n1}e_1+A_{n2}e_2+\cdots+A_{nn}e_n).
  \]
  Each of the~$n$ components in the above expression consists of~$n$ summands
  where each of the summands has the form~$a_{ij}e_j$. Using the multilinear
  properties of the determinant, when we expand the above express, we
  get~$n!$ terms, each of the form
  \[
    A_{1j_1}A_{2j_2}\cdots A_{nj_n}\det(e_{1j_1},e_{2j_2},\dots,e_{nj_n}).
  \]
  If any pair of these~$e_{kj_k}$ is the same, this term will evaluate
  to~$0$.  Thus, for the nonzero terms, $e_{1j_1},e_{2j_2},\dots,e_{nj_n}$ must be
  some permutation of~$e_1,\dots,e_n$.  We then have
  \[
    \det(e_{1j_1},e_{2j_2},\dots,e_{nj_n})=\pm 1,
  \]
  depending on the sign of the permutation~$\sigma$ defined by
  \[
    \sigma(1)=j_1,\ \sigma(2)=j_2,\ \dots\ ,\ \sigma(n)=j_n,
  \]
  and we can write
  \[
    A_{1j_1}A_{2j_2}\cdots A_{n_{j_n}}\det(e_{1j_1},e_{2j_2},\dots,e_{nj_n})
    =
    \mathrm{sign}(\sigma)A_{1\sigma(1)}A_{2\sigma(2)}\cdots A_{n\sigma(n)}.
  \]
\end{proof}
\bigskip

See the next pages for all of the details of the above proof in the case~$n=3$. 
\newpage

Let's look at the proof again in the case~$n=3$.  The~$i$-th row vector
of~$A$ is
\[
  r_i = a_{i1}e_1+a_{i2}e_2+a_{i3}e_3.
\]
To compute the determinant of~$A$ we start by expanding using multilinearity:
\begin{align*}
  \det(A)
  &= \det(r_1,r_2,r_3)\\[10pt]
  &=
  \det(a_{11}e_1+a_{12}e_2+a_{13}e_3,a_{21}e_1+a_{22}e_2+a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\[10pt]
  &= \det(a_{11}e_1,a_{21}e_1+a_{22}e_2+a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad+\det(a_{12}e_2,a_{21}e_1+a_{22}e_2+a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad+\det(a_{13}e_3,a_{21}e_1+a_{22}e_2+a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\[12pt]
  &= \det(a_{11}e_1,a_{21}e_1,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad+ \det(a_{11}e_1,a_{22}e_2,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad+ \det(a_{11}e_1,a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\[10pt]
  &\qquad+\det(a_{12}e_2,a_{21}e_1,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad+\det(a_{12}e_2,a_{22}e_2,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad\qquad+\det(a_{12}e_2,a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\[10pt]
  &\qquad\qquad+\det(a_{13}e_3,a_{21}e_1,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad\qquad+\det(a_{13}e_3,a_{22}e_2,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
  &\qquad\qquad\qquad\qquad+\det(a_{13}e_3,a_{23}e_3,a_{31}e_1+a_{32}e_2+a_{33}e_3)\\
\end{align*}
There is one more step to go in the complete expansion, at which point, we'll
have~$27$ terms.  For completeness, I'll list these all on the next page.
\newpage

\begin{align*}
  &= \det(a_{11}e_1,a_{21}e_1,a_{31}e_1)\\
  &\qquad+ \det(a_{11}e_1,a_{21}e_1,a_{32}e_2)\\
  &\qquad\qquad+ \det(a_{11}e_1,a_{21}e_1,a_{33}e_3)\\[12pt]
  &\qquad+ \det(a_{11}e_1,a_{22}e_2,a_{31}e_1)\\
  &\qquad\qquad+ \det(a_{11}e_1,a_{22}e_2,a_{32}e_2)\\
  &\qquad\qquad\qquad+ \det(a_{11}e_1,a_{22}e_2,a_{33}e_3)\\[12pt]
  &\qquad\qquad+ \det(a_{11}e_1,a_{23}e_3,a_{31}e_1)\\
  &\qquad\qquad\qquad+ \det(a_{11}e_1,a_{23}e_3,a_{32}e_2)\\
  &\qquad\qquad\qquad\qquad+ \det(a_{11}e_1,a_{23}e_3,a_{33}e_3)\\[20pt]
  % blah blah
  &+\det(a_{12}e_2,a_{21}e_1,a_{31}e_1)\\
  &\qquad+\det(a_{12}e_2,a_{21}e_1,a_{32}e_2)\\
  &\qquad\qquad+\det(a_{12}e_2,a_{21}e_1,a_{33}e_3)\\[12pt]
  %blah
  &\qquad+\det(a_{12}e_2,a_{22}e_2,a_{31}e_1)\\
  &\qquad\qquad+\det(a_{12}e_2,a_{22}e_2,a_{32}e_2)\\
  &\qquad\qquad\qquad+\det(a_{12}e_2,a_{22}e_2,a_{33}e_3)\\[12pt]
  %blah
  &\qquad\qquad+\det(a_{12}e_2,a_{23}e_3,a_{31}e_1)\\
  &\qquad\qquad\qquad+\det(a_{12}e_2,a_{23}e_3,a_{32}e_2)\\
  &\qquad\qquad\qquad\qquad+\det(a_{12}e_2,a_{23}e_3,a_{33}e_3)\\[20pt]
  &+\det(a_{13}e_3,a_{21}e_1,a_{31}e_1)\\
  &\qquad+\det(a_{13}e_3,a_{21}e_1,a_{32}e_2)\\
  &\qquad\qquad+\det(a_{13}e_3,a_{21}e_1,a_{33}e_3)\\[12pt]
  %blah
  &\qquad+\det(a_{13}e_3,a_{22}e_2,a_{31}e_1)\\
  &\qquad\qquad+\det(a_{13}e_3,a_{22}e_2,a_{32}e_2)\\
  &\qquad\qquad\qquad+\det(a_{13}e_3,a_{22}e_2,a_{33}e_3)\\[12pt]
  %blah
  &\qquad\qquad+\det(a_{13}e_3,a_{23}e_3,a_{31}e_1)\\
  &\qquad\qquad\qquad+\det(a_{13}e_3,a_{23}e_3,a_{32}e_2)\\
  &\qquad\qquad\qquad\qquad+\det(a_{13}e_3,a_{23}e_3,a_{33}e_3)
\end{align*}
\newpage

Use linearity to pull out the constants:
\begin{align*}
  &= a_{11}a_{21}a_{31}\,\det(e_1,e_1,e_1)\\
  &\qquad+ a_{11}a_{21}a_{32}\,\det(e_1,e_1,e_2)\\
  &\qquad\qquad+ a_{11}a_{21}a_{33}\,\det(e_1,e_1,e_3)\\[12pt]
  &\qquad+ a_{11}a_{22}a_{31}\,\det(e_1,e_2,e_1)\\
  &\qquad\qquad+ a_{11}a_{22}a_{32}\,\det(e_1,e_2,e_2)\\
  &\qquad\qquad\qquad+ a_{11}a_{22}a_{33}\,\det(e_1,e_2,e_3)\\[12pt]
  &\qquad\qquad+ a_{11}a_{23}a_{31}\,\det(e_1,e_3,e_1)\\
  &\qquad\qquad\qquad+ a_{11}a_{23}a_{32}\,\det(e_1,e_3,e_2)\\
  &\qquad\qquad\qquad\qquad+ a_{11}a_{23}a_{33}\,\det(e_1,e_3,e_3)\\[20pt]
  % blah blah
  &+a_{12}a_{21}a_{31}\,\det(e_2,e_1,e_1)\\
  &\qquad+a_{12}a_{21}a_{32}\,\det(e_2,e_1,e_2)\\
  &\qquad\qquad+a_{12}a_{21}a_{33}\,\det(e_2,e_1,e_3)\\[12pt]
  %blah
  &\qquad+a_{12}a_{22}a_{31}\,\det(e_2,e_2,e_1)\\
  &\qquad\qquad+a_{12}a_{22}a_{32}\,\det(e_2,e_2,e_2)\\
  &\qquad\qquad\qquad+a_{12}a_{22}a_{33}\,\det(e_2,e_2,e_3)\\[12pt]
  %blah
  &\qquad\qquad+a_{12}a_{23}a_{31}\,\det(e_2,e_3,e_1)\\
  &\qquad\qquad\qquad+a_{12}a_{23}a_{32}\,\det(e_2,e_3,e_2)\\
  &\qquad\qquad\qquad\qquad+a_{12}a_{23}a_{33}\,\det(e_2,e_3,e_3)\\[20pt]
  &+a_{13}a_{21}a_{31}\,\det(e_3,e_1,e_1)\\
  &\qquad+a_{13}a_{21}a_{32}\,\det(e_3,e_1,e_2)\\
  &\qquad\qquad+a_{13}a_{33}a_{21}\det(e_3,e_1,e_3)\\[12pt]
  %blah
  &\qquad+a_{13}a_{22}a_{31}\,\det(e_3,e_2,e_1)\\
  &\qquad\qquad+a_{13}a_{22}a_{32}\,\det(e_3,e_2,e_2)\\
  &\qquad\qquad\qquad+a_{13}a_{22}a_{33}\,\det(e_3,e_2,e_3)\\[12pt]
  %blah
  &\qquad\qquad+a_{13}a_{23}a_{31}\,\det(e_3,e_3,e_1)\\
  &\qquad\qquad\qquad+a_{13}a_{23}a_{32}\,\det(e_3,e_3,e_2)\\
  &\qquad\qquad\qquad\qquad+a_{13}a_{23}a_{33}\,\det(e_3,e_3,e_3)
\end{align*}
\newpage

Now we use the alternating property of the determinant.  If any row is repeated,
the determinant is~$0$.  Getting rid of those terms leaves:
\begin{align*}
  \det(A)
  &=a_{11}a_{22}a_{33}\,\det(e_1,e_2,e_3)\\
  &\hspace{2em}+a_{11}a_{23}a_{32}\,\det(e_1,e_3,e_2)\\
  &\hspace{4em}+a_{12}a_{21}a_{33}\,\det(e_2,e_1,e_3)\\
  &\hspace{6em}+a_{12}a_{23}a_{31}\,\det(e_2,e_3,e_1)\\
  &\hspace{8em}+a_{13}a_{21}a_{32}\,\det(e_3,e_1,e_2)\\
  &\hspace{10em}+a_{13}a_{22}a_{31}\,\det(e_3,e_2,e_1).
\end{align*}
Next notice that each determinant appearing above is the determinant of a permutation matrix.
For instance, the term
\[
  a_{12}a_{23}a_{31}\,\det(e_2,e_3,e_1)
\]
contains~$\det(e_2,e_3,e_1)$, which is the determinant of the permutation
matrix for the permutation~$\sigma(1)=2$,~$\sigma(2)=3$, and~$\sigma(3)=1$.
We have
\begin{align*}
  a_{12}a_{23}a_{31}\,\det(e_2,e_3,e_1)
  &=a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}\det(P_{\sigma})\\
      &=a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}\mathrm{sign}(\sigma).
\end{align*}
In this way, the six terms in the sum can be expressed as follows:
\begin{align*}
  \det(A)
  &=\sum_{\sigma\in\sn_3}\det(P_{\sigma})a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}\\[14pt]
  &=\sum_{\sigma\in\sn_3}\mathrm{sign}(P_{\sigma})a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}.
\end{align*}
\end{document}
