Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 2 - Matrices - Review Exercises - Page 98: 13

Answer

$$A^T=\left[\begin{array}{rrr}{1} & {3} & {-1} \end{array}\right].$$ $$A^TA=\left[\begin{array}{rrr}{11} \end{array}\right].$$ $$AA^T=\left[\begin{array}{rrr}{1} & {3} &{-1} \\ {3} & {9} &{-3}\\{-1}&{-3}&{1} \end{array}\right].$$

Work Step by Step

Let $A$ be given by $$A=\left[\begin{array}{rrr}{1} \\ {3} \\ {-1} \end{array}\right].$$ Now, we have $$A^T=\left[\begin{array}{rrr}{1} & {3} & {-1} \end{array}\right].$$ $$A^TA=\left[\begin{array}{rrr}{1} & {3} & {-1} \end{array}\right]\left[\begin{array}{rrr}{1} \\ {3} \\ {-1} \end{array}\right]=\left[\begin{array}{rrr}{11} \end{array}\right].$$ $$AA^T=\left[\begin{array}{rrr}{1} \\ {3} \\ {-1} \end{array}\right]\left[\begin{array}{rrr}{1} & {3} & {-1} \end{array}\right]=\left[\begin{array}{rrr}{1} & {3} &{-1} \\ {3} & {9} &{-3}\\{-1}&{-3}&{1} \end{array}\right].$$
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