Systems of Linear Equations Exercises Solutions#
Solution to Exercise 2.1
(a)
\[\begin{split} \begin{align*}
\det\left(\begin{matrix}-4 & 2\\3 & 4\end{matrix}\right) &= -22, \\
\operatorname{adj}\left(\begin{matrix}-4 & 2\\3 & 4\end{matrix}\right) &= \left(\begin{matrix}4 & -2\\-3 & -4\end{matrix}\right), \\
\therefore A^{-1} &= - \frac{1}{22}\left(\begin{matrix}4 & -2\\-3 & -4\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{11} & \frac{1}{11}\\\frac{3}{22} & \frac{2}{11}\end{matrix}\right)
\end{align*} \end{split}\]
therefore
\[\begin{split}\begin{align*}
\vec{x} &= \left(\begin{matrix}- \frac{2}{11} & \frac{1}{11}\\\frac{3}{22} & \frac{2}{11}\end{matrix}\right)
\left(\begin{matrix}-22\\11\end{matrix}\right) =
\left(\begin{matrix}5\\-1\end{matrix}\right)
\end{align*} \end{split}\]
so the solution is \(x_1 = 5\) and \(x_2 = -1\).
(b)
\[\begin{split} \begin{align*}
\det\left(\begin{matrix}-4 & 2\\-1 & -3\end{matrix}\right) &= 14, \\
\operatorname{adj}\left(\begin{matrix}-4 & 2\\-1 & -3\end{matrix}\right) &= \left(\begin{matrix}-3 & -2\\1 & -4\end{matrix}\right), \\
\therefore A^{-1} &= \frac{1}{14}\left(\begin{matrix}-3 & -2\\1 & -4\end{matrix}\right) = \left(\begin{matrix}- \frac{3}{14} & - \frac{1}{7}\\\frac{1}{14} & - \frac{2}{7}\end{matrix}\right)
\end{align*} \end{split}\]
therefore
\[\begin{split}\begin{align*}
\vec{x} &= \left(\begin{matrix}- \frac{3}{14} & - \frac{1}{7}\\\frac{1}{14} & - \frac{2}{7}\end{matrix}\right)
\left(\begin{matrix}6\\-2\end{matrix}\right) =
\left(\begin{matrix}-1\\1\end{matrix}\right)
\end{align*} \end{split}\]
so the solution is \(x_1 = -1\) and \(x_2 = 1\).
(c)
\[\begin{split} \begin{align*}
\det\left(\begin{matrix}-4 & -4 & -2\\3 & 0 & 4\\1 & 0 & 0\end{matrix}\right) &= -16, \\
\operatorname{adj}\left(\begin{matrix}-4 & -4 & -2\\3 & 0 & 4\\1 & 0 & 0\end{matrix}\right) &= \left(\begin{matrix}0 & 0 & -16\\4 & 2 & 10\\0 & -4 & 12\end{matrix}\right), \\
\therefore A^{-1} &= - \frac{1}{16}\left(\begin{matrix}0 & 0 & -16\\4 & 2 & 10\\0 & -4 & 12\end{matrix}\right) = \left(\begin{matrix}0 & 0 & 1\\- \frac{1}{4} & - \frac{1}{8} & - \frac{5}{8}\\0 & \frac{1}{4} & - \frac{3}{4}\end{matrix}\right)
\end{align*} \end{split}\]
therefore
\[\begin{split}\begin{align*}
\vec{x} &= \left(\begin{matrix}0 & 0 & 1\\- \frac{1}{4} & - \frac{1}{8} & - \frac{5}{8}\\0 & \frac{1}{4} & - \frac{3}{4}\end{matrix}\right)
\left(\begin{matrix}16\\-8\\0\end{matrix}\right) =
\left(\begin{matrix}0\\-3\\-2\end{matrix}\right)
\end{align*} \end{split}\]
so the solution is \(x_1 = 0\), \(x_2 = -3\) and \(x_3 = -2\).
(d)
\[\begin{split} \begin{align*}
\det\left(\begin{matrix}4 & 0 & -4\\4 & -1 & 1\\3 & 1 & 2\end{matrix}\right) &= -40, \\
\operatorname{adj}\left(\begin{matrix}4 & 0 & -4\\4 & -1 & 1\\3 & 1 & 2\end{matrix}\right) &= \left(\begin{matrix}-3 & -4 & -4\\-5 & 20 & -20\\7 & -4 & -4\end{matrix}\right), \\
\therefore A^{-1} &= - \frac{1}{40}\left(\begin{matrix}-3 & -4 & -4\\-5 & 20 & -20\\7 & -4 & -4\end{matrix}\right) = \left(\begin{matrix}\frac{3}{40} & \frac{1}{10} & \frac{1}{10}\\\frac{1}{8} & - \frac{1}{2} & \frac{1}{2}\\- \frac{7}{40} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right)
\end{align*} \end{split}\]
therefore
\[\begin{split}\begin{align*}
\vec{x} &= \left(\begin{matrix}\frac{3}{40} & \frac{1}{10} & \frac{1}{10}\\\frac{1}{8} & - \frac{1}{2} & \frac{1}{2}\\- \frac{7}{40} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right)
\left(\begin{matrix}8\\-4\\-12\end{matrix}\right) =
\left(\begin{matrix}-1\\-3\\-3\end{matrix}\right)
\end{align*} \end{split}\]
so the solution is \(x_1 = -1\), \(x_2 = -3\) and \(x_3 = -3\).
Solution to Exercise 2.2
(a)
\[\begin{split} \begin{align*}
x_{1} &= \frac{\det\left(\begin{matrix}-20 & 4\\-5 & 1\end{matrix}\right)}{\det\left(\begin{matrix}1 & 4\\-4 & 1\end{matrix}\right)}
= \frac{0}{17} = 0, \\
x_{2} &= \frac{\det\left(\begin{matrix}1 & -20\\-4 & -5\end{matrix}\right)}{\det\left(\begin{matrix}1 & 4\\-4 & 1\end{matrix}\right)}
= \frac{-85}{17} = -5.
\end{align*} \end{split}\]
(b)
\[\begin{split} \begin{align*}
x_{1} &= \frac{\det\left(\begin{matrix}4 & 1\\12 & 4\end{matrix}\right)}{\det\left(\begin{matrix}1 & 1\\0 & 4\end{matrix}\right)}
= \frac{4}{4} = 1, \\
x_{2} &= \frac{\det\left(\begin{matrix}1 & 4\\0 & 12\end{matrix}\right)}{\det\left(\begin{matrix}1 & 1\\0 & 4\end{matrix}\right)}
= \frac{12}{4} = 3.
\end{align*} \end{split}\]
(c)
\[\begin{split} \begin{align*}
x_{1} &= \frac{\det\left(\begin{matrix}21 & -4 & -4\\8 & -1 & -1\\-14 & -1 & 3\end{matrix}\right)}{\det\left(\begin{matrix}3 & -4 & -4\\-2 & -1 & -1\\4 & -1 & 3\end{matrix}\right)}
= \frac{44}{-44} = -1, \\
x_{2} &= \frac{\det\left(\begin{matrix}3 & 21 & -4\\-2 & 8 & -1\\4 & -14 & 3\end{matrix}\right)}{\det\left(\begin{matrix}3 & -4 & -4\\-2 & -1 & -1\\4 & -1 & 3\end{matrix}\right)}
= \frac{88}{-44} = -2, \\
x_{3} &= \frac{\det\left(\begin{matrix}3 & -4 & 21\\-2 & -1 & 8\\4 & -1 & -14\end{matrix}\right)}{\det\left(\begin{matrix}3 & -4 & -4\\-2 & -1 & -1\\4 & -1 & 3\end{matrix}\right)}
= \frac{176}{-44} = -4.
\end{align*} \end{split}\]
(d)
\[\begin{split} \begin{align*}
x_{1} &= \frac{\det\left(\begin{matrix}5 & 4 & 1\\-1 & 1 & 1\\-14 & 4 & 2\end{matrix}\right)}{\det\left(\begin{matrix}4 & 4 & 1\\-2 & 1 & 1\\-5 & 4 & 2\end{matrix}\right)}
= \frac{-48}{-15} = \frac{16}{5}, \\
x_{2} &= \frac{\det\left(\begin{matrix}4 & 5 & 1\\-2 & -1 & 1\\-5 & -14 & 2\end{matrix}\right)}{\det\left(\begin{matrix}4 & 4 & 1\\-2 & 1 & 1\\-5 & 4 & 2\end{matrix}\right)}
= \frac{66}{-15} = - \frac{22}{5}, \\
x_{3} &= \frac{\det\left(\begin{matrix}4 & 4 & 5\\-2 & 1 & -1\\-5 & 4 & -14\end{matrix}\right)}{\det\left(\begin{matrix}4 & 4 & 1\\-2 & 1 & 1\\-5 & 4 & 2\end{matrix}\right)}
= \frac{-147}{-15} = \frac{49}{5}.
\end{align*} \end{split}\]
Solution to Exercise 2.3
(a)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cc|c}
-1 & 3 & -2 \\
-2 & 1 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 2 R_{1} \end{array} &
\longrightarrow
& \left( \begin{array}{cc|c}
-1 & 3 & -2 \\
0 & -5 & 5 \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{2} &= - \frac{1}{5} \left( 5 \right) = -1, \\
x_{1} &= - \left( -2 - 3 \left( -1 \right) \right) = -1.
\end{align*} \end{split}\]
(b)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
3 & 1 & 2 & 11 \\
4 & 0 & -4 & -4 \\
4 & -2 & 1 & 13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - \frac{4}{3} R_{1} \\ R_{3} - \frac{4}{3} R_{1} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & 1 & 2 & 11 \\
0 & - \frac{4}{3} & - \frac{20}{3} & - \frac{56}{3} \\
0 & - \frac{10}{3} & - \frac{5}{3} & - \frac{5}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{5}{2} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
3 & 1 & 2 & 11 \\
0 & - \frac{4}{3} & - \frac{20}{3} & - \frac{56}{3} \\
0 & 0 & 15 & 45 \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= \frac{1}{15} \left( 45 \right) = 3, \\
x_{2} &= - \frac{1}{\frac{4}{3}} \left( - \frac{56}{3} - \left( - \frac{20}{3} \right) \left( 3 \right) \right) = -1, \\
x_{1} &= \frac{1}{3} \left( 11 - 1 \left( -1 \right) - 2 \left( 3 \right) \right) = 2.
\end{align*} \end{split}\]
(c)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -17 \\
2 & -2 & -3 & -14 \\
3 & -1 & 4 & -13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} + 2 R_{1}\\ R_{3} + 3 R_{1}\end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -17 \\
0 & -12 & -7 & -48 \\
0 & -16 & -2 & -64 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{4}{3} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -17 \\
0 & -12 & -7 & -48 \\
0 & 0 & \frac{22}{3} & 0 \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= \frac{1}{\frac{22}{3}} \left( 0 \right) = 0, \\
x_{2} &= - \frac{1}{12} \left( -48 - \left( -7 \right) \left( 0 \right) \right) = 4, \\
x_{1} &= - \left( -17 - \left( -5 \right) \left( 4 \right) - \left( -2 \right) \left( 0 \right) \right) = -3.
\end{align*} \end{split}\]
(d)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -26 \\
2 & -2 & -3 & -19 \\
3 & -1 & -4 & -20 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} + 2 R_{1}\\ R_{3} + 3 R_{1}\end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -26 \\
0 & -12 & -7 & -71 \\
0 & -16 & -10 & -98 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{4}{3} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -26 \\
0 & -12 & -7 & -71 \\
0 & 0 & - \frac{2}{3} & - \frac{10}{3} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{3} &= - \frac{1}{\frac{2}{3}} \left( - \frac{10}{3} \right) = 5, \\
x_{2} &= - \frac{1}{12} \left( -71 - \left( -7 \right) \left( 5 \right) \right) = 3, \\
x_{1} &= - \left( -26 - \left( -5 \right) \left( 3 \right) - \left( -2 \right) \left( 5 \right) \right) = 1.
\end{align*} \end{split}\]
(e)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
2 & 3 & 3 & -3 & 11 \\
-2 & 2 & -5 & -4 & -21 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{2}{3} R_{1} \\ R_{4} + \frac{2}{3} R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
0 & \frac{19}{3} & \frac{17}{3} & - \frac{7}{3} & - \frac{23}{3} \\
0 & - \frac{4}{3} & - \frac{23}{3} & - \frac{14}{3} & - \frac{7}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} + \frac{19}{12} R_{2}\\ R_{4} - \frac{1}{3} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
0 & 0 & \frac{125}{12} & - \frac{26}{3} & \frac{229}{4} \\
0 & 0 & - \frac{26}{3} & - \frac{10}{3} & -16 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_{4} + \frac{104}{125} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
3 & -5 & -4 & -1 & 28 \\
0 & -4 & 3 & -4 & 41 \\
0 & 0 & \frac{125}{12} & - \frac{26}{3} & \frac{229}{4} \\
0 & 0 & 0 & - \frac{1318}{125} & \frac{3954}{125} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{4} &= - \frac{1}{\frac{1318}{125}} \left( \frac{3954}{125} \right) = -3, \\
x_{3} &= \frac{1}{\frac{125}{12}} \left( \frac{229}{4} - \left( - \frac{26}{3} \right) \left( -3 \right) \right) = 3, \\
x_{2} &= - \frac{1}{4} \left( 41 - 3 \left( 3 \right) - \left( -4 \right) \left( -3 \right) \right) = -5, \\
x_{1} &= \frac{1}{3} \left( 28 - \left( -5 \right) \left( -5 \right) - \left( -4 \right) \left( 3 \right) - \left( -1 \right) \left( -3 \right) \right) = 4.
\end{align*} \end{split}\]
(f)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cccc|c}
2 & -3 & -3 & 4 & -1 \\
4 & -5 & 1 & -5 & 42 \\
3 & 3 & -1 & -5 & 20 \\
1 & 0 & 1 & 3 & -4 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} - \frac{3}{2} R_{1} \\ R_{4} - \frac{1}{2} R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
2 & -3 & -3 & 4 & -1 \\
0 & 1 & 7 & -13 & 44 \\
0 & \frac{15}{2} & \frac{7}{2} & -11 & \frac{43}{2} \\
0 & \frac{3}{2} & \frac{5}{2} & 1 & - \frac{7}{2} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{15}{2} R_{2} \\ R_{4} - \frac{3}{2} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
2 & -3 & -3 & 4 & -1 \\
0 & 1 & 7 & -13 & 44 \\
0 & 0 & -49 & \frac{173}{2} & - \frac{617}{2} \\
0 & 0 & -8 & \frac{41}{2} & - \frac{139}{2} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ R_{4} - \frac{8}{49} R_{3} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
2 & -3 & -3 & 4 & -1 \\
0 & 1 & 7 & -13 & 44 \\
0 & 0 & -49 & \frac{173}{2} & - \frac{617}{2} \\
0 & 0 & 0 & \frac{625}{98} & - \frac{1875}{98} \\
\end{array} \right)
\end{align*} \end{split}\]
Solving by back substitution gives
\[\begin{split} \begin{align*}
x_{4} &= \frac{1}{\frac{625}{98}} \left( - \frac{1875}{98} \right) = -3, \\
x_{3} &= - \frac{1}{49} \left( - \frac{617}{2} - \frac{173}{2} \left( -3 \right) \right) = 1, \\
x_{2} &= 44 - 7 \left( 1 \right) - \left( -13 \right) \left( -3 \right) = -2, \\
x_{1} &= \frac{1}{2} \left( -1 - \left( -3 \right) \left( -2 \right) - \left( -3 \right) \left( 1 \right) - 4 \left( -3 \right) \right) = 4.
\end{align*} \end{split}\]
Solution to Exercise 2.4
(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
\[\begin{split} \begin{align*}
x_{4} &= \frac{1}{\frac{625}{98}} \left( - \frac{1875}{98} \right) = -3, \\
x_{3} &= - \frac{1}{\frac{98}{27}} \left( - \frac{617}{27} - \frac{173}{27} \left( -3 \right) \right) = 1, \\
x_{2} &= \frac{1}{\frac{27}{4}} \left( - \frac{23}{2} - \left( - \frac{7}{4} \right) \left( 1 \right) - \left( - \frac{5}{4} \right) \left( -3 \right) \right) = -2, \\
x_{1} &= \frac{1}{4} \left( 42 - \left( -5 \right) \left( -2 \right) - 1 \left( 1 \right) - \left( -5 \right) \left( -3 \right) \right) = 4.
\end{align*} \end{split}\]
Solution to Exercise 2.5
(a)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cc|c}
-1 & 3 & -2 \\
-2 & 1 & 1 \\
\end{array} \right)
\begin{array}{l} -1 R_{1}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{cc|c}
1 & -3 & 2 \\
-2 & 1 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} + 2 R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|c}
1 & -3 & 2 \\
0 & -5 & 5 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{1}{5} R_{2}\end{array} &
\longrightarrow
& \left( \begin{array}{cc|c}
1 & -3 & 2 \\
0 & 1 & -1 \\
\end{array} \right)
\begin{array}{l} R_{1} + 3 R_{2}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|c}
1 & 0 & -1 \\
0 & 1 & -1 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = -1\) and \(x_{2} = -1\).
(b)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
3 & 1 & 2 & 11 \\
4 & 0 & -4 & -4 \\
4 & -2 & 1 & 13 \\
\end{array} \right)
\begin{array}{l} \frac{1}{3} R_{1}\\ \phantom{x} \\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & \frac{1}{3} & \frac{2}{3} & \frac{11}{3} \\
4 & 0 & -4 & -4 \\
4 & -2 & 1 & 13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 4 R_{1} \\ R_{3} - 4 R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & \frac{1}{3} & \frac{2}{3} & \frac{11}{3} \\
0 & - \frac{4}{3} & - \frac{20}{3} & - \frac{56}{3} \\
0 & - \frac{10}{3} & - \frac{5}{3} & - \frac{5}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{3}{4} R_{2}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & \frac{1}{3} & \frac{2}{3} & \frac{11}{3} \\
0 & 1 & 5 & 14 \\
0 & - \frac{10}{3} & - \frac{5}{3} & - \frac{5}{3} \\
\end{array} \right)
\begin{array}{l} R_{1} - \frac{1}{3} R_{2} \\ \phantom{x} \\ R_{3} + \frac{10}{3} R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & -1 & -1 \\
0 & 1 & 5 & 14 \\
0 & 0 & 15 & 45 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \frac{1}{15} R_{3}\end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & -1 & -1 \\
0 & 1 & 5 & 14 \\
0 & 0 & 1 & 3 \\
\end{array} \right)
\begin{array}{l} R_{1} + R_{3} \\ R_{2} - 5 R_{3} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = 2\), \(x_{2} = -1\) and \(x_{3} = 3\).
(c)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -17 \\
2 & -2 & -3 & -14 \\
3 & -1 & 4 & -13 \\
\end{array} \right)
\begin{array}{l} -1 R_{1}\\ \phantom{x} \\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 17 \\
2 & -2 & -3 & -14 \\
3 & -1 & 4 & -13 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} - 3 R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 17 \\
0 & -12 & -7 & -48 \\
0 & -16 & -2 & -64 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{1}{12} R_{2}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 17 \\
0 & 1 & \frac{7}{12} & 4 \\
0 & -16 & -2 & -64 \\
\end{array} \right)
\begin{array}{l} R_{1} - 5 R_{2} \\ \phantom{x} \\ R_{3} + 16 R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & - \frac{11}{12} & -3 \\
0 & 1 & \frac{7}{12} & 4 \\
0 & 0 & \frac{22}{3} & 0 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \frac{3}{22} R_{3}\end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & - \frac{11}{12} & -3 \\
0 & 1 & \frac{7}{12} & 4 \\
0 & 0 & 1 & 0 \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{11}{12} R_{3}\\ R_{2} - \frac{7}{12} R_{3} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & 0 & -3 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 0 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = -3\), \(x_{2} = 4\) and \(x_{3} = 0\).
(d)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
-1 & -5 & -2 & -26 \\
2 & -2 & -3 & -19 \\
3 & -1 & -4 & -20 \\
\end{array} \right)
\begin{array}{l} -1 R_{1}\\ \phantom{x} \\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 26 \\
2 & -2 & -3 & -19 \\
3 & -1 & -4 & -20 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} - 3 R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 26 \\
0 & -12 & -7 & -71 \\
0 & -16 & -10 & -98 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{1}{12} R_{2}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 5 & 2 & 26 \\
0 & 1 & \frac{7}{12} & \frac{71}{12} \\
0 & -16 & -10 & -98 \\
\end{array} \right)
\begin{array}{l} R_{1} - 5 R_{2} \\ \phantom{x} \\ R_{3} + 16 R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & - \frac{11}{12} & - \frac{43}{12} \\
0 & 1 & \frac{7}{12} & \frac{71}{12} \\
0 & 0 & - \frac{2}{3} & - \frac{10}{3} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ - \frac{3}{2} R_{3}\end{array} &
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & - \frac{11}{12} & - \frac{43}{12} \\
0 & 1 & \frac{7}{12} & \frac{71}{12} \\
0 & 0 & 1 & 5 \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{11}{12} R_{3}\\ R_{2} - \frac{7}{12} R_{3} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 5 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = 1\), \(x_{2} = 3\) and \(x_{3} = 5\).
(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} \frac{1}{3} R_{1}\\ \phantom{x} \\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & - \frac{5}{3} & - \frac{4}{3} & - \frac{1}{3} & \frac{28}{3} \\
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} - 2 R_{1} \\ R_{4} + 2 R_{1}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & - \frac{5}{3} & - \frac{4}{3} & - \frac{1}{3} & \frac{28}{3} \\
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} \\ - \frac{1}{4} R_{2}\\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & - \frac{5}{3} & - \frac{4}{3} & - \frac{1}{3} & \frac{28}{3} \\
0 & 1 & - \frac{3}{4} & 1 & - \frac{41}{4} \\
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} R_{1} + \frac{5}{3} R_{2}\\ \phantom{x} \\ R_{3} - \frac{19}{3} R_{2} \\ R_{4} + \frac{4}{3} R_{2}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & - \frac{31}{12} & \frac{4}{3} & - \frac{31}{4} \\
0 & 1 & - \frac{3}{4} & 1 & - \frac{41}{4} \\
0 & 0 & \frac{125}{12} & - \frac{26}{3} & \frac{229}{4} \\
0 & 0 & - \frac{26}{3} & - \frac{10}{3} & -16 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \frac{12}{125} R_{3}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & - \frac{31}{12} & \frac{4}{3} & - \frac{31}{4} \\
0 & 1 & - \frac{3}{4} & 1 & - \frac{41}{4} \\
0 & 0 & 1 & - \frac{104}{125} & \frac{687}{125} \\
0 & 0 & - \frac{26}{3} & - \frac{10}{3} & -16 \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{31}{12} R_{3}\\ R_{2} + \frac{3}{4} R_{3}\\ \phantom{x} \\ R_{4} + \frac{26}{3} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & - \frac{102}{125} & \frac{806}{125} \\
0 & 1 & 0 & \frac{47}{125} & - \frac{766}{125} \\
0 & 0 & 1 & - \frac{104}{125} & \frac{687}{125} \\
0 & 0 & 0 & - \frac{1318}{125} & \frac{3954}{125} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ - \frac{125}{1318} R_{4}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & - \frac{102}{125} & \frac{806}{125} \\
0 & 1 & 0 & \frac{47}{125} & - \frac{766}{125} \\
0 & 0 & 1 & - \frac{104}{125} & \frac{687}{125} \\
0 & 0 & 0 & 1 & -3 \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{102}{125} R_{4}\\ R_{2} - \frac{47}{125} R_{4} \\ R_{3} + \frac{104}{125} R_{4}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & 0 & 4 \\
0 & 1 & 0 & 0 & -5 \\
0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 1 & -3 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = 4\), \(x_{2} = -5\), \(x_{3} = 3\) and \(x_{4} = -3\).
(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} \frac{1}{2} R_{1}\\ \phantom{x} \\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & - \frac{3}{2} & - \frac{3}{2} & 2 & - \frac{1}{2} \\
4 & -5 & 1 & -5 & 42 \\
3 & 3 & -1 & -5 & 20 \\
1 & 0 & 1 & 3 & -4 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 4 R_{1} \\ R_{3} - 3 R_{1} \\ R_{4} - R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & - \frac{3}{2} & - \frac{3}{2} & 2 & - \frac{1}{2} \\
0 & 1 & 7 & -13 & 44 \\
0 & \frac{15}{2} & \frac{7}{2} & -11 & \frac{43}{2} \\
0 & \frac{3}{2} & \frac{5}{2} & 1 & - \frac{7}{2} \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{3}{2} R_{2}\\ \phantom{x} \\ R_{3} - \frac{15}{2} R_{2} \\ R_{4} - \frac{3}{2} R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 9 & - \frac{35}{2} & \frac{131}{2} \\
0 & 1 & 7 & -13 & 44 \\
0 & 0 & -49 & \frac{173}{2} & - \frac{617}{2} \\
0 & 0 & -8 & \frac{41}{2} & - \frac{139}{2} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ - \frac{1}{49} R_{3}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 9 & - \frac{35}{2} & \frac{131}{2} \\
0 & 1 & 7 & -13 & 44 \\
0 & 0 & 1 & - \frac{173}{98} & \frac{617}{98} \\
0 & 0 & -8 & \frac{41}{2} & - \frac{139}{2} \\
\end{array} \right)
\begin{array}{l} R_{1} - 9 R_{3} \\ R_{2} - 7 R_{3} \\ \phantom{x} \\ R_{4} + 8 R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & - \frac{79}{49} & \frac{433}{49} \\
0 & 1 & 0 & - \frac{9}{14} & - \frac{1}{14} \\
0 & 0 & 1 & - \frac{173}{98} & \frac{617}{98} \\
0 & 0 & 0 & \frac{625}{98} & - \frac{1875}{98} \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \phantom{x} \\ \frac{98}{625} R_{4}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & - \frac{79}{49} & \frac{433}{49} \\
0 & 1 & 0 & - \frac{9}{14} & - \frac{1}{14} \\
0 & 0 & 1 & - \frac{173}{98} & \frac{617}{98} \\
0 & 0 & 0 & 1 & -3 \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{79}{49} R_{4}\\ R_{2} + \frac{9}{14} R_{4}\\ R_{3} + \frac{173}{98} R_{4}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cccc|c}
1 & 0 & 0 & 0 & 4 \\
0 & 1 & 0 & 0 & -2 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & -3 \\
\end{array} \right)
\end{align*} \end{split}\]
So the solution is \(x_{1} = 4\), \(x_{2} = -2\), \(x_{3} = 1\) and \(x_{4} = -3\).
Solution to Exercise 2.6
(a)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cc|cc}
-4 & 2 & 1 & 0 \\
3 & 4 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} - \frac{1}{4} R_{1}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
3 & 4 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 3 R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
0 & \frac{11}{2} & \frac{3}{4} & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \frac{2}{11} R_{2}\end{array} &
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
0 & 1 & \frac{3}{22} & \frac{2}{11} \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{1}{2} R_{2}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & 0 & - \frac{2}{11} & \frac{1}{11} \\
0 & 1 & \frac{3}{22} & \frac{2}{11} \\
\end{array} \right)
\end{align*} \end{split}\]
So \(A^{-1} = \left(\begin{matrix}- \frac{2}{11} & \frac{1}{11}\\\frac{3}{22} & \frac{2}{11}\end{matrix}\right)\). Checking this is correct
\[\begin{split} \begin{align*}
\left(\begin{matrix}- \frac{2}{11} & \frac{1}{11}\\\frac{3}{22} & \frac{2}{11}\end{matrix}\right)\left(\begin{matrix}-4 & 2\\3 & 4\end{matrix}\right) = \left(\begin{matrix}1 & 0\\0 & 1\end{matrix}\right) \quad \checkmark\end{align*} \end{split}\]
(b)
\[\begin{split} \begin{align*}
& \left( \begin{array}{cc|cc}
-4 & 2 & 1 & 0 \\
-1 & -3 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} - \frac{1}{4} R_{1}\\ \phantom{x} \end{array} &
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
-1 & -3 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} + R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
0 & - \frac{7}{2} & - \frac{1}{4} & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{2}{7} R_{2}\end{array} &
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & - \frac{1}{2} & - \frac{1}{4} & 0 \\
0 & 1 & \frac{1}{14} & - \frac{2}{7} \\
\end{array} \right)
\begin{array}{l} R_{1} + \frac{1}{2} R_{2}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{cc|cc}
1 & 0 & - \frac{3}{14} & - \frac{1}{7} \\
0 & 1 & \frac{1}{14} & - \frac{2}{7} \\
\end{array} \right)
\end{align*} \end{split}\]
So \(A^{-1} = \left(\begin{matrix}- \frac{3}{14} & - \frac{1}{7}\\\frac{1}{14} & - \frac{2}{7}\end{matrix}\right)\). Checking this is correct
\[\begin{split} \begin{align*}
\left(\begin{matrix}- \frac{3}{14} & - \frac{1}{7}\\\frac{1}{14} & - \frac{2}{7}\end{matrix}\right)\left(\begin{matrix}-4 & 2\\-1 & -3\end{matrix}\right) = \left(\begin{matrix}1 & 0\\0 & 1\end{matrix}\right) \quad \checkmark\end{align*} \end{split}\]
(c)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|ccc}
-4 & -4 & -2 & 1 & 0 & 0 \\
3 & 0 & 4 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} - \frac{1}{4} R_{1}\\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 1 & \frac{1}{2} & - \frac{1}{4} & 0 & 0 \\
3 & 0 & 4 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 3 R_{1} \\ R_{3} - R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 1 & \frac{1}{2} & - \frac{1}{4} & 0 & 0 \\
0 & -3 & \frac{5}{2} & \frac{3}{4} & 1 & 0 \\
0 & -1 & - \frac{1}{2} & \frac{1}{4} & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ - \frac{1}{3} R_{2}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 1 & \frac{1}{2} & - \frac{1}{4} & 0 & 0 \\
0 & 1 & - \frac{5}{6} & - \frac{1}{4} & - \frac{1}{3} & 0 \\
0 & -1 & - \frac{1}{2} & \frac{1}{4} & 0 & 1 \\
\end{array} \right)
\begin{array}{l} R_{1} - R_{2} \\ \phantom{x} \\ R_{3} + R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & \frac{4}{3} & 0 & \frac{1}{3} & 0 \\
0 & 1 & - \frac{5}{6} & - \frac{1}{4} & - \frac{1}{3} & 0 \\
0 & 0 & - \frac{4}{3} & 0 & - \frac{1}{3} & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ - \frac{3}{4} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & \frac{4}{3} & 0 & \frac{1}{3} & 0 \\
0 & 1 & - \frac{5}{6} & - \frac{1}{4} & - \frac{1}{3} & 0 \\
0 & 0 & 1 & 0 & \frac{1}{4} & - \frac{3}{4} \\
\end{array} \right)
\begin{array}{l} R_{1} - \frac{4}{3} R_{3} \\ R_{2} + \frac{5}{6} R_{3}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & - \frac{1}{4} & - \frac{1}{8} & - \frac{5}{8} \\
0 & 0 & 1 & 0 & \frac{1}{4} & - \frac{3}{4} \\
\end{array} \right)
\end{align*} \end{split}\]
So \(A^{-1} = \left(\begin{matrix}0 & 0 & 1\\- \frac{1}{4} & - \frac{1}{8} & - \frac{5}{8}\\0 & \frac{1}{4} & - \frac{3}{4}\end{matrix}\right)\). Checking this is correct
\[\begin{split} \begin{align*}
\left(\begin{matrix}0 & 0 & 1\\- \frac{1}{4} & - \frac{1}{8} & - \frac{5}{8}\\0 & \frac{1}{4} & - \frac{3}{4}\end{matrix}\right)\left(\begin{matrix}-4 & -4 & -2\\3 & 0 & 4\\1 & 0 & 0\end{matrix}\right) = \left(\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right) \quad \checkmark\end{align*} \end{split}\]
(d)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|ccc}
4 & 0 & -4 & 1 & 0 & 0 \\
4 & -1 & 1 & 0 & 1 & 0 \\
3 & 1 & 2 & 0 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \frac{1}{4} R_{1}\\ \phantom{x} \\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & -1 & \frac{1}{4} & 0 & 0 \\
4 & -1 & 1 & 0 & 1 & 0 \\
3 & 1 & 2 & 0 & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ R_{2} - 4 R_{1} \\ R_{3} - 3 R_{1} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & -1 & \frac{1}{4} & 0 & 0 \\
0 & -1 & 5 & -1 & 1 & 0 \\
0 & 1 & 5 & - \frac{3}{4} & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ -1 R_{2}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & -1 & \frac{1}{4} & 0 & 0 \\
0 & 1 & -5 & 1 & -1 & 0 \\
0 & 1 & 5 & - \frac{3}{4} & 0 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ R_{3} - R_{2} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & -1 & \frac{1}{4} & 0 & 0 \\
0 & 1 & -5 & 1 & -1 & 0 \\
0 & 0 & 10 & - \frac{7}{4} & 1 & 1 \\
\end{array} \right)
\begin{array}{l} \phantom{x} \\ \phantom{x} \\ \frac{1}{10} R_{3}\end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & -1 & \frac{1}{4} & 0 & 0 \\
0 & 1 & -5 & 1 & -1 & 0 \\
0 & 0 & 1 & - \frac{7}{40} & \frac{1}{10} & \frac{1}{10} \\
\end{array} \right)
\begin{array}{l} R_{1} + R_{3} \\ R_{2} + 5 R_{3}\\ \phantom{x} \end{array} \\ \\
\longrightarrow
& \left( \begin{array}{ccc|ccc}
1 & 0 & 0 & \frac{3}{40} & \frac{1}{10} & \frac{1}{10} \\
0 & 1 & 0 & \frac{1}{8} & - \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 1 & - \frac{7}{40} & \frac{1}{10} & \frac{1}{10} \\
\end{array} \right)
\end{align*} \end{split}\]
So \(A^{-1} = \left(\begin{matrix}\frac{3}{40} & \frac{1}{10} & \frac{1}{10}\\\frac{1}{8} & - \frac{1}{2} & \frac{1}{2}\\- \frac{7}{40} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right)\). Checking this is correct
\[\begin{split} \begin{align*}
\left(\begin{matrix}\frac{3}{40} & \frac{1}{10} & \frac{1}{10}\\\frac{1}{8} & - \frac{1}{2} & \frac{1}{2}\\- \frac{7}{40} & \frac{1}{10} & \frac{1}{10}\end{matrix}\right)\left(\begin{matrix}4 & 0 & -4\\4 & -1 & 1\\3 & 1 & 2\end{matrix}\right) = \left(\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right) \quad \checkmark\end{align*} \end{split}\]
Solution to Exercise 2.7
(a)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
1 & -1 & 2 & 2 \\
2 & 1 & 4 & 7 \\
4 & 1 & 1 & 4 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} - 4 R_{1} \end{matrix} &
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & -1 & 2 & 2 \\
0 & 3 & 0 & 3 \\
0 & 5 & -7 & -4 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ \phantom{x} \\ R_{3} - \frac{5}{3} R_{2} \end{matrix} \\ \\
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & -1 & 2 & 2 \\
0 & 3 & 0 & 3 \\
0 & 0 & -7 & -9 \\
\end{array} \right)
\end{align*} \end{split}\]
therefore \(\operatorname{rank}(A) = 3\) and \(\operatorname{rank}(A \mid \vec{b}) = 3\) so this system has a unique solution
\[\begin{split} \begin{align*}
x_{3} &= \frac{1}{-7} \left( -9\right) = \frac{9}{7}, \\
x_{2} &= \frac{1}{3} \left( 3 - 0 \left( \frac{9}{7} \right)\right) = 1, \\
x_{1} &= \frac{1}{1} \left( 2 - \left( -1 \right) \left( 1 \right) - 2 \left( \frac{9}{7} \right)\right) = \frac{3}{7}.
\end{align*} \end{split}\]
(b)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
1 & -1 & 2 & 3 \\
2 & -3 & 7 & 4 \\
-1 & 3 & -8 & 1 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} + R_{1} \end{matrix} &
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & -1 & 2 & 3 \\
0 & -1 & 3 & -2 \\
0 & 2 & -6 & 4 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ \phantom{x} \\ R_{3} + 2 R_{2} \end{matrix} \\ \\
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & -1 & 2 & 3 \\
0 & -1 & 3 & -2 \\
0 & 0 & 0 & 0 \\
\end{array} \right)
\end{align*} \end{split}\]
therefore \(\operatorname{rank}(A) = 2\) and \(\operatorname{rank}(A \mid \vec{b}) = 2\). Since \(\operatorname{rank}(A) = \operatorname{rank}(A \mid \vec{b}) < 3\) then this system has infintely many solutons. Let \(r = x_3\)
\[\begin{split} \begin{align*}
x_2 &= \frac{1}{-1}(-2 - 3r) = 2 + 3r, \\
x_1 &= 3 + 2 + 3r - 2r = 5 + r.
\end{align*} \end{split}\]
(c)
\[\begin{split} \begin{align*}
& \left( \begin{array}{ccc|c}
1 & 1 & -2 & 1 \\
2 & -1 & 1 & 9 \\
1 & 4 & -7 & 2 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ R_{2} - 2 R_{1} \\ R_{3} - R_{1} \end{matrix} &
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & 1 & -2 & 1 \\
0 & -3 & 5 & 7 \\
0 & 3 & -5 & 1 \\
\end{array} \right)
\begin{matrix} \phantom{x} \\ \phantom{x} \\ R_{3} + R_{2} \end{matrix} \\ \\
\longrightarrow &
\left( \begin{array}{ccc|c}
1 & 1 & -2 & 1 \\
0 & -3 & 5 & 7 \\
0 & 0 & 0 & 8 \\
\end{array} \right)
\end{align*} \end{split}\]
therefore \(\operatorname{rank}(A) = 2\) and \(\operatorname{rank}(A \mid \vec{b}) = 3\). Since \(\operatorname{rank}(A) < \operatorname{rank}(A \mid \vec{b})\) then this is an inconsistent system and does not have a solution.