Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.3 - An Example of Gaussian Elimination - Problem Set - Page 16: 8

Answer

$k=0$, there is 1 solution $k=3$, there is 0 solution $k=-3,$ there are $\infty$ solution

Work Step by Step

$kx+3y = 6$ $3x+ky = -6$ They can be converted in the following form $ \begin{bmatrix} k && 3 && 6\\ 3 && k && -6 \end{bmatrix} $ This elimination breaks down when k = 0 $\sim \begin{bmatrix} 0 && 3 && 6\\ 3 && 0 && -6 \end{bmatrix} $ But it can be fixed by a row exchange $\sim \begin{bmatrix} 3 && 0 && 6\\ 0 && 3 && -6 \end{bmatrix} $ which on solving gives $3x = 6, 3y = -6$ $x = 2, y = -2$ Hence, one solution For $ k\neq 0$ $\sim \begin{bmatrix} k && 3 && 6\\ 3 && k && -6 \end{bmatrix} $ subtract $3/k$ times the first equation from the second $\sim \begin{bmatrix} k && 3 && 6\\ 0 && k-\frac{9}{k} && -6-\frac{18}{k} \end{bmatrix} $ $\sim \begin{bmatrix} k && 3 && 6\\ 0 && \frac{k^2-9}{k} && -6-\frac{18}{k} \end{bmatrix} $ The elimination here breaks down if $k = -3,3$ For $k = 3$ $\sim \begin{bmatrix} 3 && 3 && 6\\ 0 && 0 && -12 \end{bmatrix} $ The second row gives $0\cdot x+0\cdot y = -12$ which is false Hence, there are 0 solutions for $k = 3$ For $k = -3$ $\sim \begin{bmatrix} -3 && 3 && 6\\ 0 && 0 && 0 \end{bmatrix} $ The first row gives $-3\cdot x+3\cdot y = 6$ which has infinitely many solutions.
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