Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.5 Exercises - Page 49: 36

Answer

\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

Work Step by Step

For $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} $ to be a solution, we can have $x_1 = x_3$ and $x_2$ to be a free variable. The matrix which matches these conditions is: \begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} Checking that $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} $ is a solution, we find: \begin{equation} \begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} * \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{equation} Thus, the matrix satisfied all of the problem conditions.
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