Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Question: Solve the linear system
$| \begin{matrix} y + z = a \\ X & + z = b \\ x + y & = C \end{matrix} |$ , where a,b and c are arbitrary constants.

Answer:

The solution of the linear system is $|\begin{matrix} y + z = a \\ y + z = b \\ y + y = c \end{matrix}|, x = \frac{b + c - a}{2}, y = \frac{a + c - b}{2}, z = \frac{a + b - c}{2}$ ,

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Consider the system

Consider the linear system.

$\begin{matrix} y + z = a \\ y + z = b \\ y + y = c \end{matrix}$

The matrix form of the system is,

$[\begin{matrix} 0 & 1 & 1 & | & 1 \\ 1 & 0 & 1 & | & b \\ 1 & 1 & 0 & | & c \end{matrix}]$

Step 2: Compute the system.

Consider the matrix,

$[\begin{matrix} 0 & 1 & 1 & | & 1 \\ 1 & 0 & 1 & | & b \\ 1 & 1 & 0 & | & c \end{matrix}]$

Perform the row reduction operation.

$[\begin{matrix} 0 & 1 & 1 & | & 1 \\ 1 & 0 & 1 & | & b \\ 1 & 1 & 0 & | & c \end{matrix}] ~ [\begin{matrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & b \\ 1 & 1 & - 2 & c - b - a \end{matrix}] [\begin{matrix} 1 & 0 & 0 & \frac{b - a + c}{2} \\ 0 & 1 & 0 & \frac{a - b + c}{2} \\ 0 & 0 & 1 & \frac{a + b - c}{2} \end{matrix}]$

The values are, $x = \frac{b + c - a}{2}, y = \frac{a + c - b}{2}, z = \frac{a + b - c}{2}$

Hence, $|\begin{matrix} y + z = a \\ y + z = b \\ y + y = c \end{matrix}|,$ $, x = \frac{b + c - a}{2}, y = \frac{a + c - b}{2}, z = \frac{a + b - c}{2}$ is the solution of the linear system, .

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Linear Algebra With Applications

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Short Answer

Question: Solve the linear system
$| \begin{matrix} y + z = a \\ X & + z = b \\ x + y & = C \end{matrix} |$ , where a,b and c are arbitrary constants.

Step by Step Solution

Step 1: Consider the system

Step 2: Compute the system.

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Step by Step Solution

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Question: Solve the linear system
$| \begin{matrix} y + z = a \\ X & + z = b \\ x + y & = C \end{matrix} |$ , where a,b and c are arbitrary constants.