Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises: 62


This system has a unique solution and it is $$x=0;\quad y=0;\quad z=0.$$

Work Step by Step

This system always has the solution $x=0;\quad y=0;\quad z=0$ because if you substitute this to each equation all of the terms on the left side will just be equal to zero and that will always exactly match the right side which is also zero. To check whether there are any other solutions we will transform this system to the Row-Echelon form following the steps below: Step 1: We will put the second equation first, the third equation on the second place, and the first equation at the last place. Also we will rewrite the equations in the order of appearing $z$ then $y$ and then $x$: \begin{align*} -z+3y+4x=&0\\ 3z+3y+8x=&0\\ 3y+2x=&0 \end{align*} Step 2: To eliminate $z$ from the second equation add to it term by term the first equation multiplied by $3$: $$3z+3\times (-z) + 3y + 3\times3y + 8x+ 3\times4x = 0$$ which becomes $$12y+20x=0.$$ To simplify what we got divide it by $4$ which gives: $$3y+5x=0.$$ Now our system reads: \begin{align*} -z+3y+4x=&0\\ 3y+5x=&0\\ 3y+2x=&0 \end{align*} Step 3: To eliminate $y$ form the last equation subtract from it the second one: $$3y-3y+2x-5x = 0$$ which yields $$-3x=0$$ and that means that $$x=0.$$ Step 4: Substitute $x=0$ to the second equation to get $$3y+5\times 0 = 0$$ which is just $$3y=0$$ and that means that $$y=0.$$ Step 5: Substitute $x=0$ and $y=0$ to the first equation: $$-z+3\times0+4\times0=0$$ which is just $$-z=0$$ and that means that $$z=0.$$ Step 6: Note that the transformation to Row-Echelon form showed that there are no other solutions apart from $$x=0;\quad y=0;\quad z=0$$ so it is unique
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