Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

In Exercises 1 through 12, find all solutions of the equations
with paper and pencil using Gauss–Jordan elimination.
Show all your work.
$| \begin{matrix} x_{1} + 2 x_{3} + 4 x_{4} = - 8 \\ x_{2} - 3 x - x_{4} = 6 \\ 3 x_{1} + 4 x_{2} - 6 x_{3} + 8 x_{4} = 0 \\ - x_{2} + 3 x_{3} + 4 x_{4} = - 12 \end{matrix} |$

For the given system of equation we have free variable is $x_{3}$ and $x_{1} = - 2 x_{3}, x_{2} = 4 + 3 x_{3}, x_{4} = - 2$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Augmented matrix

First of all we will make the augmented matrix to the corresponding system of equations.

Solving system of equations using gauss Jordan’s elimination method.

$| \begin{matrix} x_{1} + 2 x_{3} + 4 x_{4} = - 8 \\ x_{2} - 3 x - x_{4} = 6 \\ 3 x_{1} + 4 x_{2} - 6 x_{3} + 8 x_{4} = 0 \\ - x_{2} + 3 x_{3} + 4 x_{4} = - 12 \end{matrix} |$

Corresponding Augmented matrix

$[\begin{matrix} 1 & 0 & 2 & 4 \\ 0 & 1 & - 3 & - 1 \\ 3 & 4 & - 6 & 8 \\ 0 & - 1 & 3 & 4 \end{matrix} \begin{matrix} - 8 \\ 6 \\ 0 \\ - 12 \end{matrix}]$

Step 2: Solving the Augmented matrix

$[\begin{matrix} 1 & 0 & 2 & 4 \\ 0 & 1 & - 3 & - 1 \\ 3 & 4 & - 6 & 8 \\ 0 & - 1 & 3 & 4 \end{matrix} \begin{matrix} - 8 \\ 6 \\ 0 \\ - 12 \end{matrix}] R_{3} \to R_{3} - 3 R_{1}$

Proceeding for solving the augmented matrix

$[\begin{matrix} 1 & 0 & 2 & 4 \\ 0 & 1 & - 3 & - 1 \\ 0 & 4 & - 12 & - 4 \\ 0 & - 1 & 3 & 4 \end{matrix} \begin{matrix} - 8 \\ 6 \\ 24 \\ - 12 \end{matrix}] R_{3} \to \frac{1}{4} R_{3} R_{4} \to R_{4} + R_{2} R_{3} \to R_{3} - R_{2} [\begin{matrix} 1 & 0 & 2 & 4 \\ 0 & 1 & - 3 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 \end{matrix} \begin{matrix} - 8 \\ 6 \\ 0 \\ - 6 \end{matrix}]$

$R_{4} \to \frac{1}{3} R_{4} R_{1} \to R_{1} - 4 R_{4} R_{2} \to R_{2} + R_{4} [\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \begin{matrix} 0 \\ 4 \\ 0 \\ - 2 \end{matrix}]$

Corresponding system of equations

$|\begin{matrix} x_{1} + 2 x_{3} = 0 \\ x_{2} - 3 x_{3} = 4 \\ x_{4} = - 2 \end{matrix}| = |\begin{matrix} x_{1} = 2 x_{3} \\ x_{2} = 4 + 3 x_{3} \\ x_{4} = - 2 \end{matrix}|$

Here, we have $x_{3}$ is free variable.

Step 3: Values of variables

Since for free variables we can let any value for that.

We have free variable $x_{3}$ .

Let the value of $x_{3} = 1$

$\Rightarrow x_{1} = - 2, x_{2} = 7, x_{4} = - 2$

Step 4: Final answer

For the given system of equation is free variable $x_{3}$ and $x_{1} = - 2 x_{3}, x_{2} = 4 + 3 x_{3}, x_{4} = - 2$ .

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Augmented matrix

Step 2: Solving the Augmented matrix

Step 3: Values of variables

Step 4: Final answer

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Linear Algebra With Applications

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Short Answer

In Exercises 1 through 12, find all solutions of the equationswith paper and pencil using Gauss–Jordan elimination.Show all your work. |x1+2x3+4x4=-8x2-3x-x4=63x1+4x2-6x3+8x4=0 -x2+3x3+4x4=-12|

Step by Step Solution

Step 1: Augmented matrix

Step 2: Solving the Augmented matrix

Step 3: Values of variables

Step 4: Final answer

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