Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - Chapter Review - Review Exercises - Page 794: 19

Answer

Singular.

Work Step by Step

We know that for a matrix \[ \left[\begin{array}{rr} a & b \\ c &d \\ \end{array} \right] \] the determinant $D$ is given by the formula $D=ad-bc.$ We also know that if $D=0$, then the matrix is singular. The given matrix has $D=(4\cdot2)-[(-8)\cdot(-1)]\\ D=8-8\\ D=0$ The square matrix has a determinant of zero therefore it is singular.
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