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:
A new verion, after clarification. No additional packages used. Each column is a separate table. $'s left, however rather unneeded.
\documentclass{article}
\begin{document}
\def\barr{\begin{tabular}{l}}
\def\earr{\end{tabular}}
\begin{tabular}{|c|l|c|c|c|c|}
\hline
$\textbf{Q }$&
\barr
$\textbf{Header one}$
\earr &
\barr
$\textbf{Test}$\\
$\textbf{Text4} $\\
\earr &
\barr
$\textbf{Test1}$\\
$\textbf{Text5}$
\earr &
\barr
$\textbf{Test2}$\\
$\textbf{Text6}$
\earr &
\barr
$\textbf{Test3}$\\
$\textbf{Text7}$\\
$\textbf{Text8}$
\earr\\
\hline
\end{tabular}
\end{document}
Best Answer
This might be a starting point for you. I created a new column type
Pwhich makes it easier distributing the columns over the width of the textblock, because with this solution the widths of thencolumns should add up to1and not to1-n*\tabcolsep.