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: 80

Answer

For $k=1$ and $k=-1$ system won't have any solution.

Work Step by Step

Step 1: Multiply the first equation by $k$: \begin{align*} kx+k^2y=&2k\\ kx+y=&4 \end{align*} Step 2: Subtract these equations to eliminate $x$: $$kx-kx+k^2y-y=2k-4$$ which becomes $$y(k^2-1) = 2k-4$$ and then we get for $y$ $$y=\frac{2k-4}{k^2-1}$$ Step 3: Now only multiply the second equation by $k$: \begin{align*} x+ky=&2\\ k^2x+ky=&4k \end{align*} Step 4: Subtract these equation to eliminate $y$: $$x-k^2x +ky-ky = 2-4k$$ which becomes $$x(1-k^2) = 2-4k$$ and that gives $$x=\frac{4k-2}{k^2-1}$$ If the calculated values for $x$ and $y$ are valid expressions there will always be solutions to this system. So we require that they are not valid expressions i.e. that there are zeros in their denominators. This will happen if $k^2-1 = 0$ i.e. when $$k=\pm 1.$$ So for these two values for $k$ the system won't have any solutions.
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