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Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

In exercises 1 through 10, find all solutions of the linear systems using elimination. Then check your solutions7. $\left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}x+3y+4z=3\phantom{\rule{0ex}{0ex}}x+4y+5z=4\right|$

The system of equation has no solution.

See the step by step solution

Step 1:Transforming the system

To get the solution, we will transform the value ofx,y,z.

$\left|x+2y+3z=1\text{}......\left(1\right)\phantom{\rule{0ex}{0ex}}x+3y+4z=3\text{}......\left(2\right)\phantom{\rule{0ex}{0ex}}x+4y+5z=4\text{}......\left(3\right)\right|$Into the form $\left|x=....\phantom{\rule{0ex}{0ex}}y=....\phantom{\rule{0ex}{0ex}}z=...\phantom{\rule{0ex}{0ex}}\right|$

Step 2: Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 2 and 3. First of all subtract the equation 1 from equation 2 and then subtract the first equation from equation 3.

$\begin{array}{rcl}\left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}x-x+3y-2y+4z-3z=3-1\phantom{\rule{0ex}{0ex}}x-x+4y-2y+5z-3z=4-1\right|& =& \left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}y+z=2\phantom{\rule{0ex}{0ex}}2y+2z=3\right|\end{array}$

Now multiply equation 2 by 2 and then subtract equation 2 from equation 3.

$\begin{array}{rcl}\left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}2y+2z=4\phantom{\rule{0ex}{0ex}}2y+2z=3\right|& =& \left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}0=1\phantom{\rule{0ex}{0ex}}2y+2z=3\right|\end{array}$

For whatever the value of the value 0 will never equal to 1. Thus, the given system of equations is inconsistent.

Hence, the given system of the equation has no solution.