tikzpictures underneath matrices

Question

I'm trying to construct a diagram of how to calculate a 3x3 matrix determinant via the Rule of Sarrus, and I've managed to come this far after lots of research, but I still would like to make the arrows apear underneath the numbers, and the curved arrow to have a circled minus sign in the middle, unfortunately I have yet to find a way of doing this, so if someone can point me in the right direction I would very much appreciate.

This is the code I'm using, and the result bellow.

%preamble
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{circuitikz}
\usepackage{tikz}
\newcommand\tikznode[1]{\tikz[overlay,remember picture] \node (#1) {};}
\tikzset{>=latex}
------------
%document
\begin{equation*}
    \left|\begin{matrix}
    \tikznode{11}2& \tikznode{12}7 & \tikznode{13}6\\ 
    \tikznode{21}9& \tikznode{22}5 & \tikznode{23}1\\ 
    \tikznode{31}4& \tikznode{32}3 & \tikznode{33}8
    \end{matrix}
    \middle|\;\begin{matrix}
    \tikznode{14}2& \tikznode{15}7\\ 
    \tikznode{24}9& \tikznode{25}5\\ 
    \tikznode{34}4& \tikznode{35}3
    \end{matrix}\right| \qquad
    \begin{matrix}
    4*5*6+3*1*2+8*9*7 = 630\tikznode{top}\\
     \\
     \\
     \\
    2*5*8+7*1*4+6*9*3 = 270\tikznode{btn}
    \end{matrix}\qquad \qquad
    270-630 = -360
    \begin{tikzpicture}[overlay, remember picture]
        \foreach \i in {1,2,3}
            \def\e{\the\numexpr\i+2}
            \draw[->,red,line width=1pt] ($(1\i.north east)+(-0.5,0.5)$) -- ($(3\e.south west)+(0.6,-0.2)$);
        \foreach \i in {3,4,5}
            \def\e{\the\numexpr\i-2}
            \draw[<-,blue,line width=1pt] ($(1\i.north west)+(0.7,0.5)$) -- ($(3\e.south east)+(-0.4,-0.2)$);
        \path[->,line width=1pt] ($(btn)+(0.25,0)$) edge[bend right=60] ($(top)+(0.25,0)$);
    \end{tikzpicture}
\end{equation*}

And the desired result is something along this lines (without the + and - signs ono top and bottom):

Answer 1

The pstricks way:

\documentclass{article}
\usepackage{pst-node}%

\begin{document}

\[ \renewcommand{\arraystretch}{1.25}
\begin{array}{|ccc|cc|}
\Rnode{a1}{2} &\Rnode{b1}{7} & \Rnode{c1}{6} & \Rnode{A1}{2} & \Rnode{B1}{7} \\
\Rnode{a2}{9} & \Rnode{b2}{5} & \Rnode{c2}{1} & \Rnode{A2}{9}& 5\\
\Rnode{a3}{4} & \Rnode{b3}{3} & \Rnode{c3}{8} & \Rnode{A3}{4} & \Rnode{B3}{3}
\end{array}
\psset{linecolor=red, nodesepA=0.5pt, nodesepB = 0.5pt, arrowinset=0.1}
\foreach \s/\t/\u in {a1/b2/c3,b1/c2/A3,c1/A2/B3} {\ncline{-}{\s}{\t}\ncline{->}{\t}{\u}}
 \psset{linecolor=blue}
\foreach \s/\t/\u in {a3/b2/c1,b3/c2/A1,c3/A2/B1} {\ncline{-}{\s}{\t}\ncline{->}{\t}{\u}}
\]

\end{document} 

Answer 2

Here is a solution with {NiceArray} of nicematrix.

\documentclass{article}
\usepackage{nicematrix}
\usepackage{tikz}

\begin{document}

\NiceMatrixOptions
  { pgf-node-code = \pgfsetfillcolor{white} \pgfusepathqfill }

\pgfset{nicematrix/cell-node/.append style = { inner sep = 1pt } }


$\begin{NiceArray}{|ccc|>{\color{gray}}c>{\color{gray}}c}
\CodeBefore [create-cell-nodes]
    \begin{tikzpicture} [shorten < = 2pt,shorten > = 2pt]
    \draw [red,->] (1-1) -- (3-3) ;
    \draw [red,->] (1-2) -- (3-4) ;
    \draw [red,->] (1-3) -- (3-5) ;
    \draw [blue,->] (3-1) -- (1-3) ;
    \draw [blue,->] (3-2) -- (1-4) ;
    \draw [blue,->] (3-3) -- (1-5) ;
    \end{tikzpicture}
\Body
    a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\[2mm]
    a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\[2mm]
    a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \\
\end{NiceArray}$

\end{document}

You need several compilations (because of the PGF/Tikz nodes).

Answer 3

the red and blue arrows are essentially a play of cardinal points around each number in the matrix-- each cardinal point of the compass acts as an anchor for the start or end of the arrow -- the idea is to offset each arrow tip by a small amount plus or minus so that the arrow does not collide with the number in the matrix -- for example -- \draw[->,red, line width=1pt] ($(1\i.north east)+(0.1,-0.1)$) -- ($(3\e.north)+(0,0)$); -- means start the arrow from 11.north east which is the top right corner of the number 2 (first row and first column of matrix) -- offset the start point of the arrow 0.1 cm in x direction, and 0.1cm down in y direction -- the end of the arrow is at 33.north -- which is the number 8 (third row and third column of matrix) and offset the arrow tip by (0,0)

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{circuitikz}
\usepackage{tikz}
\newcommand\tikznode[1]{\tikz[overlay,remember picture] \node (#1) {};}
\tikzset{>=latex}
%------------
%document
\begin{document}
    

\begin{equation*}
    \left|\begin{matrix}
        \tikznode{11}2& \tikznode{12}7 & \tikznode{13}6\\ 
        \tikznode{21}9& \tikznode{22}5 & \tikznode{23}1\\ 
        \tikznode{31}4& \tikznode{32}3 & \tikznode{33}8
    \end{matrix}
    \middle|\;\begin{matrix}
        \tikznode{14}2& \tikznode{15}7\\ 
        \tikznode{24}9& \tikznode{25}5\\ 
        \tikznode{34}4& \tikznode{35}3
    \end{matrix}\right| \qquad
    \begin{matrix}
        4*5*6+3*1*2+8*9*7 = 630\tikznode{top}\\
        \\
        \\
        \\
        2*5*8+7*1*4+6*9*3 = 270\tikznode{btn}
    \end{matrix}\qquad \qquad
    270-630 = -360
    \begin{tikzpicture}[overlay, remember picture]
        \foreach \i in {1,2,3}
        \def\e{\the\numexpr\i+2}
        \draw[->,red,line width=1pt] ($(1\i.north east)+(0.1,-0.1)$) -- ($(3\e.north)+(0,0)$);
        \foreach \i in {3,4,5}
        \def\e{\the\numexpr\i-2}
        \draw[<-,blue,line width=1pt] ($(1\i.south)+(0,0.1)$) -- ($(3\e.north east)+(0.1,0.1)$);
        \path[->,line width=1pt] ($(btn)+(0.25,0)$) edge[bend right=60]node[fill=white, circle,draw](){-} ($(top)+(0.25,0)$);
    \end{tikzpicture}
\end{equation*}
\end{document}