This is the best I can think: use a tikz matrix to create a matrix of math nodes (which you can include inside a math environment and delimit with brackets if you want), and then use the implicit naming of nodes to refer to individual cells of the matrix, as for example: m-1-1.north east to refer to the north east corner of the first element.
In order to avoid alignment problems, you have to ensure that all the nodes of that matrix have the same dimensions, by giving a minimum width and minimum height option. I'm not very satisfied with this solution, because it requires you to know the dimensions of the larger cell. However, appropiate values are not difficult to find by trial and error.
After some tries, my code is the following:
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{inputenc}
\usepackage{xcolor}
\usepackage{tikz}
\begin{document}
\thispagestyle{empty}
\usetikzlibrary{matrix}
\usetikzlibrary{calc,fit}
\tikzset{%
highlight1/.style={rectangle,rounded corners,color=red!,fill=red!15,draw,fill opacity=0.5,thick,inner sep=0pt}
}
\tikzset{%
highlight2/.style={rectangle,rounded corners,color=green!,fill=green!15,draw,fill opacity=0.5,thick,inner sep=0pt}
}
\begin{equation}
\renewcommand{\arraystretch}{1.5}
A_{L}=
\begin{tikzpicture}[baseline=(m.center)]
\matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
row sep=1mm, nodes={minimum width=3em, minimum height=1.6em}] {
-T^{1}_{11} & 0 & 0 & -T^{1}_{12} \\
-T^{2}_{12} & -T^{2}_{11} & 0 & 0 \\
0 & -T^{3}_{12} & |(r)| T^{3}_{11} & 0 \\
0 & 0 & 0 & 0 \\
};
\node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
\node[highlight1, fit=(m-3-1.north west) (m-4-4.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{A} \\
\phi_{B} \\
\phi_{C} \\
\phi_{D}
\end{array}\right]
\label{eq:ALphif}
\end{equation}
\begin{equation}\renewcommand{\arraystretch}{1.5}
B_{L}=
\begin{tikzpicture}[baseline=(m.center)]
\matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
row sep=1mm, nodes={minimum width=5.5em, minimum height=1.6em}] {
(T^{1}_{11}+T^{1}_{12}) & 0 & 0 \\
0 & (T^{2}_{11}+T^{2}_{12}) & 0 \\
0 & 0 & (T^{3}_{11}+T^{3}_{12}) \\
0 & 0 & 0 \\
};
\node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
\node[highlight1, fit=(m-3-1.north west) (m-4-3.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{1} \\
\phi_{2} \\
\phi_{3}
\end{array}\right]
\label{eq:BLphii}
\end{equation}
\begin{equation}
\renewcommand{\arraystretch}{1.5}
A_{R}=
\begin{tikzpicture}[baseline=(m.center)]
\matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
row sep=1mm, nodes={minimum width=2.5em, minimum height=1.6em}] {
T^{2}_{22} & T^{2}_{21} & 0 & 0 \\
0 & T^{3}_{22} & T^{3}_{21} & 0 \\
0 & 0 & 0 & 0 \\
T^{1}_{21} & 0 & 0 & T^{1}_{22}\\
};
\node[highlight2, fit=(m-1-1.north west) (m-2-2.south east)] {};
\node[highlight1, fit=(m-3-1.north west) (m-4-4.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{A} \\
\phi_{B} \\
\phi_{C} \\
\phi_{D}
\end{array}\right]
\label{eq:ARphif}
\end{equation}
\begin{equation}
\renewcommand{\arraystretch}{1.5}
B_{R}=
\begin{tikzpicture}[baseline=(m.center)]
\matrix (m) [matrix of math nodes, left delimiter={[}, right delimiter={]},
row sep=1mm, nodes={minimum width=6.5em, minimum height=1.6em}] {
0 & -(T^{2}_{21}+T^{2}_{22}) & 0 \\
0 & 0 & -(T^{3}_{21}+T^{3}_{22})\\
0 & 0 & 0 \\
-(T^{1}_{21}+T^{1}_{22}) & 0 & 0 \\
};
\node[highlight2, fit=(m-1-1.north west) (m-2-3.south east)] {};
\node[highlight1, fit=(m-3-1.north west) (m-4-3.south east)] {};
\end{tikzpicture}
\left[\begin{array}{c}
\phi_{1} \\
\phi_{2} \\
\phi_{3}
\end{array}\right]
\label{eq:BRphii}
\end{equation}
\end{document}
Which produces the following output:
Best Answer
(revised this answer completely after receiving further information from the OP)
I'm not sure if I understand that nature of your
arrayenvironments. It seems they all have five columns, and that the fourth column can have entries of widely varying widths. Suppose, furthermore, that the single widest entry in any of the arrays' fourth columns is the number "28978". If that's the case, it suffices to exchange the variable-widtyhccolumn type with a fixed-width column type, as is done in the code shown below.