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: 61


The solution to this system is unique and it reads: $$x=0;\quad y=0;\quad z=0.$$

Work Step by Step

This system always has a solution $$x=0;\quad y=0;\quad z=0;$$ because in that case each term on the left side of the equations is something times zero which will always add up to zero, and that is exactly equal to the right side. To check if there are other solutions we have to transform this system to Row-Echelon form by doing the following steps: Step 1: From the last equation subtract the first one term by term to eliminate $x$: $$4x-4x + 2y-3y+19z-17z=0$$ which then becomes $$-y+2z=0.$$ Now our system reads: \begin{align*} 4x+3y+17z=&0\\ 5x+4y+22z=&0\\ -y+2z=&0 \end{align*} Step 2: We now want to eliminate $x$ from the second equation as well. To do so we will subtract from it the first equation multiplied by $\frac{5}{4}$: $$5x-\frac{5}{4}\times 4x + 4y - \frac{5}{4}\times 3y + 22z-\frac{5}{4}\times 17 z = 0$$ which simplifies to $$\frac{1}{4}y +\frac{3}{4}z = 0.$$ To loose the fractions and make this equation look nicer we can multiply it by $4$ which then gives $$y+3z=0.$$ Now the system reads \begin{align*} 4x+3y+17z=&0\\ y+3z=&0\\ -y+2z=&0 \end{align*} Step 3: To eliminate y from the third equation just add the second one to it: $$-y+y+2z+3z=0$$ and this yields $$5z=0$$ which means that $$z=0$$ Step 4: Now going back to the second equation we substitute the calculated value for $z$ (zero) and get $$-y+2\times 0 = 0$$ which gives $$y=0.$$ To continue substitute $y=0$ and $z=0$ to the first equation to obtain $$4x+3\times0+17\times0=0$$ and that gives $$x=0$$ Step 5: Transforming to Row-Echelon form showed us that the solution to this system is unique and it is: $$x=0;\quad y=0;\quad z=0.$$
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