Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 3 - Matrices - 3.1 Matrix Operations - Exercises 3.1 - Page 152: 4

Answer

$\begin{bmatrix}-3 & 2\\5 &2 \\4 & 3\end{bmatrix}$

Work Step by Step

We have $B^{T}=\begin{bmatrix}4 & 0\\-2 & 2 \\1 & 3\end{bmatrix}$ Now, $C-B^{T}=\begin{bmatrix}1 & 2\\3 & 4 \\5 & 6\end{bmatrix}-\begin{bmatrix}4 & 0\\-2 & 2 \\1 & 3\end{bmatrix}$ or, $=\begin{bmatrix}1-4 & 2-0\\3+2 &4- 2 \\5-1 & 6-3\end{bmatrix}$ After simplification, we obtain: $=\begin{bmatrix}-3 & 2\\5 &2 \\4 & 3\end{bmatrix}$
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.