Answer Solution: The solution is ( 1 , 2 , 0 , 1 ) . Explanation Given: x + y + z + w = 4 − x + 2 y + z = 0 2 x + 3 y + z − w = 6 − 2 x + y − 2 z + 2 w = − 1 Calculation: Write the given equation in matrices. x + y + z + w = 4 − x + 2 y + z = 0 2 x + 3 y + z − w = 6 − 2 x + y − 2 z + 2 w = − 1 [ | 1 1 1 1 − 1 2 1 0 2 3 1 − 1 − 2 1 − 2 2 | 4 0 6 − 1 ] R 3 = R 3 − R 2 [ | 1 1 1 1 − 1 2 1 0 3 1 0 − 1 − 2 1 − 2 2 | 4 0 6 − 1 ] R 4 = R 4 + R 3 [ | 1 1 1 1 − 1 2 1 0 3 1 0 − 1 1 2 − 2 1 | 4 0 6 5 ] R 4 = − R 1 + R 4 [ | 1 1 1 1 − 1 2 1 0 3 1 0 − 1 0 1 − 3 0 | 4 0 6 1 ] R 1 = R 1 − R 2 [ | 2 − 1 0 1 − 1 2 1 0 3 1 0 − 1 0 1 − 3 0 | 4 0 6 1 ] R 3 = R 1 + R 3 [ | 2 − 1 0 1 − 1 2 1 0 5 0 0 0 0 1 − 3 0 | 4 0 10 1 ] 5 x = 10 x = 2 − x + 2 y + z = 0 2 y + z = 2 -----(1) y − 3 z = 1 -----(1) Solving equation 1 and 2 we have z = 0 2 y + 0 = 2 y = 1 Substituting the values of x , y and z we have 2 x − y + w = 4 w = 1 The solution is ( 2 , 1 , 0 , 1 ) .

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Question

Chapter 11.2, Problem 64AYU

To determine

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

Expert Solution & Answer

Answer to Problem 64AYU

Solution:

The solution is $(1, 2, 0, 1)$ .

Explanation of Solution

Given:

$\begin{array}{l} x + y + z + w = 4 \\ - x + 2 y + z = 0 \\ 2 x + 3 y + z - w = 6 \\ - 2 x + y - 2 z + 2 w = - 1 \end{array}$

Calculation:

Write the given equation in matrices.

$\begin{array}{l} x + y + z + w = 4 \\ - x + 2 y + z = 0 \\ 2 x + 3 y + z - w = 6 \\ - 2 x + y - 2 z + 2 w = - 1 \end{array}$

$\begin{array}{l} [| \begin{matrix} 1 & 1 & 1 & 1 \\ - 1 & 2 & 1 & 0 \\ 2 & 3 & 1 & - 1 \\ - 2 & 1 & - 2 & 2 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 6 \\ - 1 \end{matrix}] \\ R_{3} = R_{3} - R_{2} \end{array}$

$\begin{array}{l} [| \begin{matrix} 1 & 1 & 1 & 1 \\ - 1 & 2 & 1 & 0 \\ 3 & 1 & 0 & - 1 \\ - 2 & 1 & - 2 & 2 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 6 \\ - 1 \end{matrix}] \\ R_{4} = R_{4} + R_{3} \end{array}$

$\begin{array}{l} [| \begin{matrix} 1 & 1 & 1 & 1 \\ - 1 & 2 & 1 & 0 \\ 3 & 1 & 0 & - 1 \\ 1 & 2 & - 2 & 1 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 6 \\ 5 \end{matrix}] \\ R_{4} = - R_{1} + R_{4} \end{array}$

$\begin{array}{l} [| \begin{matrix} 1 & 1 & 1 & 1 \\ - 1 & 2 & 1 & 0 \\ 3 & 1 & 0 & - 1 \\ 0 & 1 & - 3 & 0 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 6 \\ 1 \end{matrix}] \\ R_{1} = R_{1} - R_{2} \end{array}$

$\begin{array}{l} [| \begin{matrix} 2 & - 1 & 0 & 1 \\ - 1 & 2 & 1 & 0 \\ 3 & 1 & 0 & - 1 \\ 0 & 1 & - 3 & 0 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 6 \\ 1 \end{matrix}] \\ R_{3} = R_{1} + R_{3} \end{array}$

$[| \begin{matrix} 2 & - 1 & 0 & 1 \\ - 1 & 2 & 1 & 0 \\ 5 & 0 & 0 & 0 \\ 0 & 1 & - 3 & 0 \end{matrix} | \begin{matrix} 4 \\ 0 \\ 10 \\ 1 \end{matrix}]$

$\begin{array}{l} 5 x = 10 \\ x = 2 \\ - x + 2 y + z = 0 \end{array}$

$2 y + z = 2$ -----(1)

$y - 3 z = 1$ -----(1)

Solving equation 1 and 2 we have

$\begin{array}{l} z = 0 \\ 2 y + 0 = 2 \\ y = 1 \end{array}$

Substituting the values of $x$ , $y$ and $z$ we have

$\begin{array}{l} 2 x - y + w = 4 \\ w = 1 \end{array}$

The solution is $(2, 1, 0, 1)$ .

Chapter 11.2, Problem 63AYU

Chapter 11.2, Problem 65AYU

Chapter 11 Solutions

Precalculus Enhanced with Graphing Utilities

Ch. 11.1 - Prob. 1AYU Ch. 11.1 - Prob. 2AYU Ch. 11.1 - Prob. 3AYU Ch. 11.1 - Prob. 4AYU Ch. 11.1 - Prob. 5AYU Ch. 11.1 - Prob. 6AYU Ch. 11.1 - Prob. 7AYU Ch. 11.1 - Prob. 8AYU Ch. 11.1 - Prob. 9AYU Ch. 11.1 - Prob. 10AYU

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

Solution Summary: The author explains how to solve a system of equations using matrices. If the system has no solution, it is inconsistent.