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 - Chapter Test - Page 796: 11

Answer

$\begin{bmatrix}2&-1\\-\frac{5}{2}&\frac{3}{2}\end{bmatrix}$

Work Step by Step

We are given the matrix: $A=\begin{bmatrix} 3&2\\5&4\end{bmatrix}$ Compute $det A$: $detA=\begin{vmatrix}3&2\\5&4\end{vmatrix}=3(4)-5(2)=2$ Because $detA\not=0$, the inverse of the matrix exists. We determine it using the formula: $A^{-1}=\dfrac{1}{det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$, where $detA=ad-bc$ and $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ $A^{-1}=\dfrac{1}{2}\begin{bmatrix}4&-2\\-5&3\end{bmatrix}=\begin{bmatrix}2&-1\\-\frac{5}{2}&\frac{3}{2}\end{bmatrix}$
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