(a)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cc|c}
-1 & 3 & -2 \\
-2 & 1 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{1} \leftrightarrow R_{2} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{cc|c}
-2 & 1 & 1 \\
-1 & 3 & -2 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{1}{2} R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|c}
-2 & 1 & 1 \\
0 & \frac{5}{2} & - \frac{5}{2} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{2} &= \frac{1}{\frac{5}{2}} \left( - \frac{5}{2} \right) = -1, \\
x_{1} &= - \frac{1}{2} \left( 1 - 1 \left( -1 \right) \right) = -1.
\end{align*} \end{split}\]
(b)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
3 & 1 & 2 & 11 \\
4 & 0 & -4 & -4 \\
4 & -2 & 1 & 13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{1} \leftrightarrow R_{2} \\ \\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
4 & 0 & -4 & -4 \\
3 & 1 & 2 & 11 \\
4 & -2 & 1 & 13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{3}{4} R_{1} \\ R_{3} - R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
4 & 0 & -4 & -4 \\
0 & 1 & 5 & 14 \\
0 & -2 & 5 & 17 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{2} \leftrightarrow R_{3} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
4 & 0 & -4 & -4 \\
0 & -2 & 5 & 17 \\
0 & 1 & 5 & 14 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} + \frac{1}{2} R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
4 & 0 & -4 & -4 \\
0 & -2 & 5 & 17 \\
0 & 0 & \frac{15}{2} & \frac{45}{2} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= \frac{1}{\frac{15}{2}} \left( \frac{45}{2} \right) = 3, \\
x_{2} &= - \frac{1}{2} \left( 17 - 5 \left( 3 \right) \right) = -1, \\
x_{1} &= \frac{1}{4} \left( -4 - 0 \left( -1 \right) - \left( -4 \right) \left( 3 \right) \right) = 2.
\end{align*} \end{split}\]
(c)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -17 \\
2 & -2 & -3 & -14 \\
3 & -1 & 4 & -13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{1} \leftrightarrow R_{3} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & 4 & -13 \\
2 & -2 & -3 & -14 \\
-1 & -5 & -2 & -17 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{2}{3} R_{1} \\ R_{3} + \frac{1}{3} R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & 4 & -13 \\
0 & - \frac{4}{3} & - \frac{17}{3} & - \frac{16}{3} \\
0 & - \frac{16}{3} & - \frac{2}{3} & - \frac{64}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{2} \leftrightarrow R_{3} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & 4 & -13 \\
0 & - \frac{16}{3} & - \frac{2}{3} & - \frac{64}{3} \\
0 & - \frac{4}{3} & - \frac{17}{3} & - \frac{16}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{1}{4} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & 4 & -13 \\
0 & - \frac{16}{3} & - \frac{2}{3} & - \frac{64}{3} \\
0 & 0 & - \frac{11}{2} & 0 \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= - \frac{1}{\frac{11}{2}} \left( 0 \right) = 0, \\
x_{2} &= - \frac{1}{\frac{16}{3}} \left( - \frac{64}{3} - \left( - \frac{2}{3} \right) \left( 0 \right) \right) = 4, \\
x_{1} &= \frac{1}{3} \left( -13 - \left( -1 \right) \left( 4 \right) - 4 \left( 0 \right) \right) = -3.
\end{align*} \end{split}\]
(d)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -26 \\
2 & -2 & -3 & -19 \\
3 & -1 & -4 & -20 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{1} \leftrightarrow R_{3} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & -4 & -20 \\
2 & -2 & -3 & -19 \\
-1 & -5 & -2 & -26 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{2}{3} R_{1} \\ R_{3} + \frac{1}{3} R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & -4 & -20 \\
0 & - \frac{4}{3} & - \frac{1}{3} & - \frac{17}{3} \\
0 & - \frac{16}{3} & - \frac{10}{3} & - \frac{98}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{2} \leftrightarrow R_{3} \\ \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & -4 & -20 \\
0 & - \frac{16}{3} & - \frac{10}{3} & - \frac{98}{3} \\
0 & - \frac{4}{3} & - \frac{1}{3} & - \frac{17}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{1}{4} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & -1 & -4 & -20 \\
0 & - \frac{16}{3} & - \frac{10}{3} & - \frac{98}{3} \\
0 & 0 & \frac{1}{2} & \frac{5}{2} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= \frac{1}{\frac{1}{2}} \left( \frac{5}{2} \right) = 5, \\
x_{2} &= - \frac{1}{\frac{16}{3}} \left( - \frac{98}{3} - \left( - \frac{10}{3} \right) \left( 5 \right) \right) = 3, \\
x_{1} &= \frac{1}{3} \left( -20 - \left( -1 \right) \left( 3 \right) - \left( -4 \right) \left( 5 \right) \right) = 1.
\end{align*} \end{split}\]
(e)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
2 & 3 & 3 & -3 & 11 \\
-2 & 2 & -5 & -4 & -21 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{2}{3} R_{1} \\ R_{4} + \frac{2}{3} R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
0 & \frac{19}{3} & \frac{17}{3} & - \frac{7}{3} & - \frac{23}{3} \\
0 & - \frac{4}{3} & - \frac{23}{3} & - \frac{14}{3} & - \frac{7}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{2} \leftrightarrow R_{3} \\ \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & \frac{19}{3} & \frac{17}{3} & - \frac{7}{3} & - \frac{23}{3} \\
0 & -4 & 3 & -4 & 41 \\
0 & - \frac{4}{3} & - \frac{23}{3} & - \frac{14}{3} & - \frac{7}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} + \frac{12}{19} R_{2}\\ R_{4} + \frac{4}{19} R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & \frac{19}{3} & \frac{17}{3} & - \frac{7}{3} & - \frac{23}{3} \\
0 & 0 & \frac{125}{19} & - \frac{104}{19} & \frac{687}{19} \\
0 & 0 & - \frac{123}{19} & - \frac{98}{19} & - \frac{75}{19} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_{4} + \frac{123}{125} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & \frac{19}{3} & \frac{17}{3} & - \frac{7}{3} & - \frac{23}{3} \\
0 & 0 & \frac{125}{19} & - \frac{104}{19} & \frac{687}{19} \\
0 & 0 & 0 & - \frac{1318}{125} & \frac{3954}{125} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{4} &= - \frac{1}{\frac{1318}{125}} \left( \frac{3954}{125} \right) = -3, \\
x_{3} &= \frac{1}{\frac{125}{19}} \left( \frac{687}{19} - \left( - \frac{104}{19} \right) \left( -3 \right) \right) = 3, \\
x_{2} &= \frac{1}{\frac{19}{3}} \left( - \frac{23}{3} - \frac{17}{3} \left( 3 \right) - \left( - \frac{7}{3} \right) \left( -3 \right) \right) = -5, \\
x_{1} &= \frac{1}{3} \left( 28 - \left( -5 \right) \left( -5 \right) - \left( -4 \right) \left( 3 \right) - \left( -1 \right) \left( -3 \right) \right) = 4.
\end{align*} \end{split}\]
(f)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cccc|c}
2 & -3 & -3 & 4 & -1 \\
4 & -5 & 1 & -5 & 42 \\
3 & 3 & -1 & -5 & 20 \\
1 & 0 & 1 & 3 & -4 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{1} \leftrightarrow R_{2} \\ \\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
4 & -5 & 1 & -5 & 42 \\
2 & -3 & -3 & 4 & -1 \\
3 & 3 & -1 & -5 & 20 \\
1 & 0 & 1 & 3 & -4 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{1}{2} R_{1} \\ R_{3} - \frac{3}{4} R_{1} \\ R_{4} - \frac{1}{4} R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
4 & -5 & 1 & -5 & 42 \\
0 & - \frac{1}{2} & - \frac{7}{2} & \frac{13}{2} & -22 \\
0 & \frac{27}{4} & - \frac{7}{4} & - \frac{5}{4} & - \frac{23}{2} \\
0 & \frac{5}{4} & \frac{3}{4} & \frac{17}{4} & - \frac{29}{2} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{2} \leftrightarrow R_{3} \\ \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
4 & -5 & 1 & -5 & 42 \\
0 & \frac{27}{4} & - \frac{7}{4} & - \frac{5}{4} & - \frac{23}{2} \\
0 & - \frac{1}{2} & - \frac{7}{2} & \frac{13}{2} & -22 \\
0 & \frac{5}{4} & \frac{3}{4} & \frac{17}{4} & - \frac{29}{2} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} + \frac{2}{27} R_{2}\\ R_{4} - \frac{5}{27} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
4 & -5 & 1 & -5 & 42 \\
0 & \frac{27}{4} & - \frac{7}{4} & - \frac{5}{4} & - \frac{23}{2} \\
0 & 0 & - \frac{98}{27} & \frac{173}{27} & - \frac{617}{27} \\
0 & 0 & \frac{29}{27} & \frac{121}{27} & - \frac{334}{27} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_{4} + \frac{29}{98} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
4 & -5 & 1 & -5 & 42 \\
0 & \frac{27}{4} & - \frac{7}{4} & - \frac{5}{4} & - \frac{23}{2} \\
0 & 0 & - \frac{98}{27} & \frac{173}{27} & - \frac{617}{27} \\
0 & 0 & 0 & \frac{625}{98} & - \frac{1875}{98} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{4} &= \frac{1}{\frac{625}{98}} \left( - \frac{1875}{98} \right) = -3, \\
x_{3} &= - \frac{1}{\frac{98}{27}} \left( - \frac{617}{27} - \frac{173}{27} \left( -3 \right) \right) = 1, \\
x_{2} &= \frac{1}{\frac{27}{4}} \left( - \frac{23}{2} - \left( - \frac{7}{4} \right) \left( 1 \right) - \left( - \frac{5}{4} \right) \left( -3 \right) \right) = -2, \\
x_{1} &= \frac{1}{4} \left( 42 - \left( -5 \right) \left( -2 \right) - 1 \left( 1 \right) - \left( -5 \right) \left( -3 \right) \right) = 4.
\end{align*} \end{split}\]