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: 83


This system won't have a unique solution for $$k\in\{-2,1\}$$

Work Step by Step

To analyze this system easier we will transform it to row echelon form doing the steps described below: Step 1: Subtract the first equation from the second to eliminate $x$: $$x-x+ky-y+z-kz = 2-3$$ which becomes $$(k-1)y-(k-1)z=-1$$ Our system now reads \begin{align*} x+y+kz=&3\\ (k-1)y-(k-1)z=-1\\ kx+y+z=1 \end{align*} Step 2: Subtract the first equation multiplied by $k$ from the third one to loose $x$: $$kx-kx+y-ky+z-k^2z=1-3k$$ and this becomes $$-(k-1)y -(k-1)(k+1)z=1-3k$$ where we used the fact that $1-k^2 = (1-k)(1+k) = -(k-1)(k+1)$. Multiplying by $-1$ we get $$(k-1)y+(k-1)(k+1)z=3k-1$$ Now our system reads \begin{align*} x+y+kz=&3\\ (k-1)y-(k-1)z=&-1\\ (k-1)y+(k-1)(k+1)z=&3k-1 \end{align*} Step 3: Now subtract the second equation from the third to eliminate $y$: $$(k-1)y-(k-1)y+(k-1)(k+1)z-(-(k-1)z)=3k-1-(-1)$$ and this becomes $$(k+2)(k-1)z=3k$$ Our system reads: \begin{align*} x+y+kz=&3\\ (k-1)y-(k-1)z=&-1\\ (k+2)(k-1)z=3k. \end{align*} Step 4. Note that if we take $k=1$ the second equation will say $$(1-1)z-(1-1)y=-1\Rightarrow 0=-1$$ which is impossible so then we will have no solutions. Also if we take $k=-2$ the last equation will read: $$(-2+2)(-2-1)z=3\times(-2)\Rightarrow 0=-6$$ which also impossible and we will have no solutions. For any other value of $k$ except these two we can work our way backwards and solve first for $z$ then for $y$ and then for $x$ and find a unique solution. In the problem we are required NOT to have a unique solution and that means we should choose $k\in\{-2,1\}$
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