Problem:
Subtract the second matrix from the first:
$\left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)$
$\left(
\begin{array}{ccc}
 6 & -\frac{11}{3} & -\frac{17}{3} \\
 \frac{26}{3} & 8 & \frac{11}{3} \\
\end{array}
\right)$
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Simplify the following}: \\
 \left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)-\left(
\begin{array}{ccc}
 6 & -\frac{11}{3} & -\frac{17}{3} \\
 \frac{26}{3} & 8 & \frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 -1\left(
\begin{array}{ccc}
 6 & -\frac{11}{3} & -\frac{17}{3} \\
 \frac{26}{3} & 8 & \frac{11}{3} \\
\end{array}
\right)=\left(
\begin{array}{ccc}
 -6 & -\frac{-11}{3} & -\frac{-17}{3} \\
 -\frac{26}{3} & -8 & -\frac{11}{3} \\
\end{array}
\right): \\
 \left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)+\fbox{$\left(
\begin{array}{ccc}
 -6 & -\frac{-11}{3} & -\frac{-17}{3} \\
 -\frac{26}{3} & -8 & -\frac{11}{3} \\
\end{array}
\right)$} \\
\end{array}
 \\
 
\begin{array}{l}
 -(-11) \text{= }11: \\
 \left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)+\left(
\begin{array}{ccc}
 -6 & \frac{\fbox{$11$}}{3} & -\frac{-17}{3} \\
 -\frac{26}{3} & -8 & -\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -(-17) \text{= }17: \\
 \left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)+\left(
\begin{array}{ccc}
 -6 & \frac{11}{3} & \frac{\fbox{$17$}}{3} \\
 -\frac{26}{3} & -8 & -\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \left(
\begin{array}{ccc}
 -\frac{11}{3} & -2 & -\frac{2}{3} \\
 -\frac{25}{3} & \frac{10}{3} & -\frac{23}{3} \\
\end{array}
\right)+\left(
\begin{array}{ccc}
 -6 & \frac{11}{3} & \frac{17}{3} \\
 -\frac{26}{3} & -8 & -\frac{11}{3} \\
\end{array}
\right)=\left(
\begin{array}{ccc}
 -\frac{11}{3}-6 & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right): \\
 \left(
\begin{array}{ccc}
 -\frac{11}{3}-6 & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Put }-\frac{11}{3}-6 \text{over }\text{the }\text{common }\text{denominator }3. -\frac{11}{3}-6 \text{= }\frac{-11}{3}+\frac{3 (-6)}{3}: \\
 \left(
\begin{array}{ccc}
 \fbox{$\frac{-11}{3}+\frac{3 (-6)}{3}$} & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 3 (-6) \text{= }-18: \\
 \left(
\begin{array}{ccc}
 \frac{-11}{3}+\frac{\fbox{$-18$}}{3} & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{11}{3}-\frac{18}{3} \text{= }\frac{-11-18}{3}: \\
 \left(
\begin{array}{ccc}
 \fbox{$\frac{-11-18}{3}$} & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -11-18=-(11+18): \\
 \left(
\begin{array}{ccc}
 \frac{\fbox{$-(11+18)$}}{3} & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{c}
 
\begin{array}{ccc}
\hline
 \text{} & 1 & 8 \\
\hline
 + & 1 & 1 \\
 \text{} & 2 & 9 \\
\end{array}
 \\
\end{array}
: \\
 \left(
\begin{array}{ccc}
 \frac{-\fbox{$29$}}{3} & -2+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Put }-2+\frac{11}{3} \text{over }\text{the }\text{common }\text{denominator }3. -2+\frac{11}{3} \text{= }\frac{3 (-2)}{3}+\frac{11}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \fbox{$\frac{3 (-2)}{3}+\frac{11}{3}$} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 3 (-2) \text{= }-6: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{\fbox{$-6$}}{3}+\frac{11}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{6}{3}+\frac{11}{3} \text{= }\frac{-6+11}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \fbox{$\frac{-6+11}{3}$} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -6+11=5: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{\fbox{$5$}}{3} & -\frac{2}{3}+\frac{17}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{2}{3}+\frac{17}{3} \text{= }\frac{-2+17}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & \fbox{$\frac{-2+17}{3}$} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -2+17=15: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & \frac{\fbox{$15$}}{3} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{gcd }\text{of }\text{15 }\text{and }3 \text{is }3, \text{so }\frac{15}{3}=\frac{3\times 5}{3\times 1}=\frac{3}{3}\times 5=5: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & \fbox{$5$} \\
 -\frac{25}{3}-\frac{26}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{25}{3}-\frac{26}{3} \text{= }\frac{-25-26}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 \fbox{$\frac{-25-26}{3}$} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -25-26=-(25+26): \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 \frac{\fbox{$-(25+26)$}}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{c}
 
\begin{array}{ccc}
\hline
 \text{} & 1 & \text{} \\
\hline
 \text{} & 2 & 6 \\
 + & 2 & 5 \\
 \text{} & 5 & 1 \\
\end{array}
 \\
\end{array}
: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 \frac{-\fbox{$51$}}{3} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{c}
 
\begin{array}{c}
 \text{} \\
 3 \\
\end{array}
\nthickspace \nthickspace \nthickspace 
\begin{array}{lll}
 \text{} &  &  \\
 \text{} &  & 1 \\
 - &  & \text{} \\
 \text{} & 2 & 1 \\
 - &  &  \\
 \text{} & \text{} & 0 \\
\end{array}
 \\
\end{array}
: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -\fbox{$17$} & \frac{10}{3}-8 & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Put }\frac{10}{3}-8 \text{over }\text{the }\text{common }\text{denominator }3. \frac{10}{3}-8 \text{= }\frac{10}{3}+\frac{3 (-8)}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & \fbox{$\frac{10}{3}+\frac{3 (-8)}{3}$} & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 3 (-8) \text{= }-24: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & \frac{10}{3}+\frac{\fbox{$-24$}}{3} & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \frac{10}{3}-\frac{24}{3} \text{= }\frac{10-24}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & \fbox{$\frac{10-24}{3}$} & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 10-24 \text{= }-(24-10): \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & \frac{\fbox{$-(24-10)$}}{3} & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{c}
 
\begin{array}{ccc}
 \text{} & 2 & 4 \\
 - & 1 & 0 \\
\hline
 \text{} & 1 & 4 \\
\end{array}
 \\
\end{array}
: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & \frac{-\fbox{$14$}}{3} & -\frac{23}{3}-\frac{11}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -\frac{23}{3}-\frac{11}{3} \text{= }\frac{-23-11}{3}: \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & -\frac{14}{3} & \fbox{$\frac{-23-11}{3}$} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 -23-11=-(23+11): \\
 \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & -\frac{14}{3} & \frac{\fbox{$-(23+11)$}}{3} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{c}
 
\begin{array}{ccc}
\hline
 \text{} & 2 & 3 \\
\hline
 + & 1 & 1 \\
 \text{} & 3 & 4 \\
\end{array}
 \\
\end{array}
: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \left(
\begin{array}{ccc}
 -\frac{29}{3} & \frac{5}{3} & 5 \\
 -17 & -\frac{14}{3} & \frac{-\fbox{$34$}}{3} \\
\end{array}
\right) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}
{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/subtract_w_steps/523.txt'}