Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 934: 84


The provided statement is False.

Work Step by Step

The inverse of a $2\times 2$ matrix is determined as below: $A=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]$ $B=\left[ \begin{matrix} e & f \\ g & h \\ \end{matrix} \right]$ Now, we will consider that the multiplication equals, $\left[ AB \right]=\left[ \begin{matrix} p & q \\ r & s \\ \end{matrix} \right]$ Then, the inverse is ${{\left[ AB \right]}^{-1}}=\frac{1}{ps-qr}\left[ \begin{matrix} s & -r \\ -q & p \\ \end{matrix} \right]$ Thus, the matrix is invertible. If $ad-bc\ne 0$, then it is not.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.