Problem:
Add the two matrices
$\left(
\begin{array}{cc}
 7 & -4 \\
 \frac{3}{2} & -6 \\
 -\frac{15}{2} & -6 \\
\end{array}
\right)$ and
$\left(
\begin{array}{cc}
 -6 & -3 \\
 \frac{7}{2} & -2 \\
 -\frac{3}{2} & 4 \\
\end{array}
\right)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Simplify the following}: \\
 \left(
\begin{array}{cc}
 7 & -4 \\
 \frac{3}{2} & -6 \\
 -\frac{15}{2} & -6 \\
\end{array}
\right)+\left(
\begin{array}{cc}
 -6 & -3 \\
 \frac{7}{2} & -2 \\
 -\frac{3}{2} & 4 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \left(
\begin{array}{cc}
 7 & -4 \\
 \frac{3}{2} & -6 \\
 -\frac{15}{2} & -6 \\
\end{array}
\right)+\left(
\begin{array}{cc}
 -6 & -3 \\
 \frac{7}{2} & -2 \\
 -\frac{3}{2} & 4 \\
\end{array}
\right)=\left(
\begin{array}{cc}
 7-6 & -4-3 \\
 \frac{3}{2}+\frac{7}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right): \\
 \left(
\begin{array}{cc}
 7-6 & -4-3 \\
 \frac{3}{2}+\frac{7}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 7-6=1: \\
 \left(
\begin{array}{cc}
 \fbox{$1$} & -4-3 \\
 \frac{3}{2}+\frac{7}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -4-3=-(4+3): \\
 \left(
\begin{array}{cc}
 1 & \fbox{$-(4+3)$} \\
 \frac{3}{2}+\frac{7}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 4+3=7: \\
 \left(
\begin{array}{cc}
 1 & -\fbox{$7$} \\
 \frac{3}{2}+\frac{7}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \frac{3}{2}+\frac{7}{2} \text{= }\frac{3+7}{2}: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 \fbox{$\frac{3+7}{2}$} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 3+7=10: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 \frac{\fbox{$10$}}{2} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{gcd }\text{of }\text{10 }\text{and }2 \text{is }2, \text{so }\frac{10}{2}=\frac{2\times 5}{2\times 1}=\frac{2}{2}\times 5=5: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 \fbox{$5$} & -6-2 \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -6-2=-(6+2): \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & \fbox{$-(6+2)$} \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 6+2=8: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -\fbox{$8$} \\
 -\frac{15}{2}-\frac{3}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{15}{2}-\frac{3}{2} \text{= }\frac{-15-3}{2}: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -8 \\
 \fbox{$\frac{-15-3}{2}$} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -15-3=-(15+3): \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -8 \\
 \frac{\fbox{$-(15+3)$}}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 15+3=18: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -8 \\
 \frac{-\fbox{$18$}}{2} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \frac{18}{2}=\frac{2\times 9}{2}=9: \\
 \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -8 \\
 -\fbox{$9$} & -6+4 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -6+4=-2: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \left(
\begin{array}{cc}
 1 & -7 \\
 5 & -8 \\
 -9 & \fbox{$-2$} \\
\end{array}
\right) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}
{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/add_w_steps/362.txt'}