Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 9 - Systems and Matrices - 9.8 Matrix Inverses - 9.8 Exercises - Page 942: 19

Answer

\[\left[ {\begin{array}{*{20}{c}} 2&1 \\ { - 3/2}&{ - 1/2} \end{array}} \right]\]

Work Step by Step

\[\begin{gathered} \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 2} \\ 3&4 \end{array}} \right] \hfill \\ {\text{Let a matrix A}} = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right],{\text{ its inverse is: }}{{\text{A}}^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right] \hfill \\ {\text{Then}} \hfill \\ {{\text{A}}^{ - 1}} = \frac{1}{{\left( { - 1} \right)\left( 4 \right) - \left( { - 2} \right)\left( 3 \right)}}\left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 3}&{ - 1} \end{array}} \right] \hfill \\ {\text{Simplifying}} \hfill \\ {{\text{A}}^{ - 1}} = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 3}&{ - 1} \end{array}} \right] \hfill \\ {{\text{A}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 2&1 \\ { - 3/2}&{ - 1/2} \end{array}} \right] \hfill \\ \end{gathered} \]
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