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: 82

Answer

For ANY $k\neq 0$ we will have exactly one solution and it will be given by $$x= -\frac{23}{9},\quad y=-\frac{16}{9};\quad z=\frac{13}{3}.$$

Work Step by Step

Firstly we will transform this system to Row-Echelon form in order to analyze it easier following the steps below: Step 1: Change the order of the equations as follows: \begin{align*} x+y+z=&0\\ 2x-y+z=&0\\ kx+2ky+3kz=&4k \end{align*} Step 2: Subtract from the second equation the first one multiplied by $2$ to eliminate $x$: $$2x-2x-y-2y+z-2z=1-2\times0$$ which becomes $$-3y-z=1$$ and multiplied by $-1$ for nicer looks: $$3y+z=-1$$ Our system is now: \begin{align*} x+y+z=&0\\ 3y+z=&-1\\ kx+2ky+3kz=&4k \end{align*} Step 3: Subtract from the third equation the first one multiplied by $k$ to loose $x$: $$ kx-kx+2ky-ky+3kz-kz=4k-k\times0 $$ which becomes $$ky+2kz =4k.$$ Now our system reads \begin{align*} x+y+z=&0\\ 3y+z=&-1\\ ky+2kz=&4k \end{align*} Step 4: Subtract from the third equation the second one multiplied by $\frac{k}{3}$ to loose $y$: $$ky-\frac{k}{3}3y+2kz-\frac{k}{3}z = 4k-\frac{k}{3}(-1)$$ which becomes $$\frac{5}{3}kz=\frac{13}{3}k$$ Multiply by $3$ for nicer looks $$5kz=13k.$$ Now our system reads: \begin{align*} x+y+z=&0\\ 3y+z=&-1\\ 3kz=&13k \end{align*} Step 5: Notice that for $k=0$ the third equation won't be an actual equation but will just say $0=0$ and in this cas we would have infinitely many solutions depending on one parameter. For any $k\neq$ we will have exactly one solution because then it is "legal" (because we know $k$ is not zero) to divide the last equation with $k$ and that gives $$3z=13\Rightarrow z=\frac{13}{3}.$$ Now we can put this into the second equation to get $$3y+\frac{13}{3} = -1\Rightarrow 3y = -1-\frac{13}{3}\Rightarrow 3y= -\frac{16}{3}$$ and that gives the solution for $y$: $$y=-\frac{16}{9}.$$ Finally we can put the calculated values for $y$ and $z$ into the first equation and find $x$: $$x-\frac{16}{9}+\frac{13}{3} = 0\Rightarrow x +\frac{23}{9} = 0$$ and this gives $$x=-\frac{23}{9}$$ So for ANY $k\neq 0$ we will have exactly one solution and it will be given by $$x= -\frac{23}{9},\quad y=-\frac{16}{9};\quad z=\frac{13}{3}.$$
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