Linear Algebra and Its Applications (5th Edition)

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

Chapter 2 - Matrix Algebra - 2.2 Exercises - Page 112: 13

Answer

$B=C$ This is not true in general when A is not invertible.

Work Step by Step

Let $AB=AC$ where $B$ and $C$ are $n$$\times$$p$ matrices and $A$ is invertible. $AB=AC$ $A^{-1}AB=A^{-1}AC$ $IB=IC$ $B=C$ Let $A=\begin{bmatrix}1&0\\0&0\\\end{bmatrix}$, $B=\begin{bmatrix}2&0\\0&0\\\end{bmatrix}$, and $C=\begin{bmatrix}2&0\\0&1\\\end{bmatrix}$. A is not invertible because $detA=0$. $AB=\begin{bmatrix}1&0\\0&0\\\end{bmatrix}\begin{bmatrix}2&0\\0&0\\\end{bmatrix}=\begin{bmatrix}2&0\\0&0\\\end{bmatrix}$ $AC=\begin{bmatrix}1&0\\0&0\\\end{bmatrix}\begin{bmatrix}2&0\\0&1\\\end{bmatrix}=\begin{bmatrix}2&0\\0&0\\\end{bmatrix}$ $AB=AC$ but $B$$\ne$$C$ so this is not true in general when $A$ is not invertible.
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