Answered: Solve the system by the method of…

Intermediate Algebra

19th Edition

ISBN:9780998625720

Author:Lynn Marecek

Publisher:Lynn Marecek

Chapter4: Systems Of Linear Equations

Section4.4: Solve Systems Of Equations With Three Variables

Problem 4.64TI: Solve the system by elimination: {4x+y+z=12x2y+z=22x+3yz=1 .

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Question

Solve the system by the method of Gaussian elimination:

x + y - z = -1

2x + 3y - z = -2

-3x - 2y + 2z = -3

Expert Solution

Given,

$x + y - z = - 1 2 x + 3 y - z = - 2 - 3 x - 2 y + 2 z = - 3$

The augmented matrix is,

$[\begin{matrix} 1 & 1 & - 1 & - 1 \\ 2 & 3 & - 1 & - 2 \\ - 3 & - 2 & 2 & - 3 \end{matrix}]$

Now apply $R_{2} \to R_{2} - 2 R_{1} & R_{3} \to R_{3} + 3 R_{1}$ , we get

$[\begin{matrix} 1 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & - 1 & - 6 \end{matrix}]$

Now apply $R_{3} \to R_{3} - R_{2}$ , we get

$[\begin{matrix} 1 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & - 2 & - 6 \end{matrix}]$

Now apply $R_{3} \to - \frac{1}{2} R_{3}$ , we get

$[\begin{matrix} 1 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 3 \end{matrix}]$

Now apply $R_{2} \to R_{2} - R_{3}$ , we get

$[\begin{matrix} 1 & 1 & - 1 & - 1 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 3 \end{matrix}]$

Now apply $R_{1} \to R_{1} + R_{3}$ , we get

$[\begin{matrix} 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 3 \end{matrix}]$

Now apply $R_{1} \to R_{1} - R_{2}$ , we get

$[\begin{matrix} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 3 \end{matrix}]$

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