To determine To Solve: System of equations using matrices (row operation). If the system has no solution, say that it is inconsistent. Answer Solution: The solution is ( 13 9 , 7 18 , 19 18 ) Explanation Given: 2 x + 3 y − z = 3 x − y − z = 0 − x + y + z = 0 x + y + 3 z = 5 Calculation: Write the given equation in matrices. 2 x + 3 y − z = 3 x − y − z = 0 − x + y + z = 0 x + y + 3 z = 5 [ | 2 3 − 1 1 − 1 − 1 − 1 1 1 1 1 3 | 3 0 0 5 ] R 1 = R 1 9 [ | 2 9 1 3 − 1 9 1 − 1 − 1 − 1 1 1 1 1 3 | 1 3 0 0 5 ] R 4 = R 4 2 [ | 2 9 1 3 − 1 9 1 − 1 − 1 − 1 1 1 1 2 1 2 3 2 | 1 3 0 0 5 2 ] R 4 = R 4 − R 1 [ | 2 9 1 3 − 1 9 1 − 1 − 1 − 1 1 1 5 18 1 6 29 18 | 1 3 0 0 13 6 ] R 2 = R 2 3 [ | 2 9 1 3 − 1 9 1 3 − 1 3 − 1 3 − 1 1 1 5 18 1 6 29 18 | 1 3 0 0 13 6 ] R 1 = R 1 + R 2 [ | 5 9 0 − 4 9 1 3 − 1 3 − 1 3 − 1 1 1 5 18 1 6 29 18 | 1 3 0 0 13 6 ] R 3 = R 3 6 [ | 5 9 0 − 4 9 1 3 − 1 3 − 1 3 − 1 6 1 6 1 6 5 18 1 6 29 18 | 1 3 0 0 13 6 ] R 4 = R 4 − R 3 [ | 5 9 0 − 4 9 1 3 − 1 3 − 1 3 − 1 6 1 6 1 6 4 9 0 13 9 | 1 3 0 0 13 6 ] R 1 = 13 R 1 R 4 = 4 R 4 [ | 65 9 0 − 52 9 1 3 − 1 3 − 1 3 − 1 6 1 6 1 6 16 9 0 52 9 | 13 3 0 0 52 6 ] R 1 = R 1 + R 4 [ | 81 9 0 0 1 3 − 1 3 − 1 3 − 1 6 1 6 1 6 16 9 0 52 9 | 78 6 0 0 52 6 ] We have 81 9 x = 78 6 x = 13 9 16 9 x + 52 9 z = 52 6 16 9 ( 13 9 ) + 52 9 z = 52 6 z = 19 18 x − y − z = 0 13 9 − y − 19 18 = 0 y = 7 18 The solution is ( 13 9 , 7 18 , 19 18 ) .

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ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Textbook Question

Chapter 11.2, Problem 69AYU

In problems 39-74, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

${\begin{cases} 2 x + 3 y - z = 3 \\ x - y - z = 0 \\ - x + y + z = 0 \\ x + v + 3 z = 5 \end{cases}$

Expert Solution & Answer

To determine

To Solve: System of equations using matrices (row operation). If the system has no solution, say that it is inconsistent.

Answer to Problem 69AYU

Solution:

The solution is $(\frac{13}{9}, \frac{7}{18}, \frac{19}{18})$

Explanation of Solution

Given:

$\begin{array}{l} 2 x + 3 y - z = 3 \\ x - y - z = 0 \\ - x + y + z = 0 \\ x + y + 3 z = 5 \end{array}$

Calculation:

Write the given equation in matrices.

$\begin{array}{l} 2 x + 3 y - z = 3 \\ x - y - z = 0 \\ - x + y + z = 0 \\ x + y + 3 z = 5 \end{array}$

$\begin{array}{l} [| \begin{array}{r} 2 & 3 & - 1 \\ 1 & - 1 & - 1 \\ - 1 & 1 & 1 \\ 1 & 1 & 3 \end{array} | \begin{matrix} 3 \\ 0 \\ 0 \\ 5 \end{matrix}] \\ R_{1} = \frac{R_{1}}{9} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{2}{9} & \frac{1}{3} & \frac{- 1}{9} \\ 1 & - 1 & - 1 \\ - 1 & 1 & 1 \\ 1 & 1 & 3 \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ 5 \end{matrix}] \\ R_{4} = \frac{R_{4}}{2} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{2}{9} & \frac{1}{3} & \frac{- 1}{9} \\ 1 & - 1 & - 1 \\ - 1 & 1 & 1 \\ \frac{1}{2} & \frac{1}{2} & \frac{3}{2} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{5}{2} \end{matrix}] \\ R_{4} = R_{4} - R_{1} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{2}{9} & \frac{1}{3} & \frac{- 1}{9} \\ 1 & - 1 & - 1 \\ - 1 & 1 & 1 \\ \frac{5}{18} & \frac{1}{6} & \frac{29}{18} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{13}{6} \end{matrix}] \\ R_{2} = \frac{R_{2}}{3} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{2}{9} & \frac{1}{3} & \frac{- 1}{9} \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ - 1 & 1 & 1 \\ \frac{5}{18} & \frac{1}{6} & \frac{29}{18} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{13}{6} \end{matrix}] \\ R_{1} = R_{1} + R_{2} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{5}{9} & 0 & \frac{- 4}{9} \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ - 1 & 1 & 1 \\ \frac{5}{18} & \frac{1}{6} & \frac{29}{18} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{13}{6} \end{matrix}] \\ R_{3} = \frac{R_{3}}{6} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{5}{9} & 0 & \frac{- 4}{9} \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ \frac{- 1}{6} & \frac{1}{6} & \frac{1}{6} \\ \frac{5}{18} & \frac{1}{6} & \frac{29}{18} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{13}{6} \end{matrix}] \\ R_{4} = R_{4} - R_{3} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{5}{9} & 0 & \frac{- 4}{9} \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ \frac{- 1}{6} & \frac{1}{6} & \frac{1}{6} \\ \frac{4}{9} & 0 & \frac{13}{9} \end{array} | \begin{matrix} \frac{1}{3} \\ 0 \\ 0 \\ \frac{13}{6} \end{matrix}] \\ R_{1} = 13 R_{1} \\ R_{4} = 4 R_{4} \end{array}$

$\begin{array}{l} [| \begin{array}{r} \frac{65}{9} & 0 & \frac{- 52}{9} \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ \frac{- 1}{6} & \frac{1}{6} & \frac{1}{6} \\ \frac{16}{9} & 0 & \frac{52}{9} \end{array} | \begin{matrix} \frac{13}{3} \\ 0 \\ 0 \\ \frac{52}{6} \end{matrix}] \\ R_{1} = R_{1} + R_{4} \end{array}$

$[| \begin{array}{r} \frac{81}{9} & 0 & 0 \\ \frac{1}{3} & \frac{- 1}{3} & \frac{- 1}{3} \\ \frac{- 1}{6} & \frac{1}{6} & \frac{1}{6} \\ \frac{16}{9} & 0 & \frac{52}{9} \end{array} | \begin{matrix} \frac{78}{6} \\ 0 \\ 0 \\ \frac{52}{6} \end{matrix}]$

We have $\frac{81}{9} x = \frac{78}{6}$

$\begin{array}{l} x = \frac{13}{9} \\ \frac{16}{9} x + \frac{52}{9} z = \frac{52}{6} \\ \frac{16}{9 (\frac{13}{9})} + \frac{52}{9} z = \frac{52}{6} \\ z = \frac{19}{18} \end{array}$

$\begin{array}{l} x - y - z = 0 \\ \frac{13}{9} - y - \frac{19}{18} = 0 \\ y = \frac{7}{18} \end{array}$

The solution is $(\frac{13}{9}, \frac{7}{18}, \frac{19}{18})$ .

Chapter 11.2, Problem 68AYU

Chapter 11.2, Problem 70AYU

Chapter 11 Solutions

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In problems 39-74, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. { 2 x + 3 y − z = 3 x − y − z = 0 − x + y + z = 0 x + v + 3 z = 5