## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - 4.1 Vectors in Rn - 4.1 Exercises: 23

#### Answer

$\mathbf{z}=(\frac{7}{2},3,\frac{5}{2})$

#### Work Step by Step

$\mathbf{u}=(1,2,3)\quad\mathbf{v}=(2,2-1)\quad\mathbf{w}=(4,0,-4)$ Find $\mathbf{z}$ if $2\mathbf{z}-3\mathbf{u}=\mathbf{w}$ We can solve for $\mathbf{z}$: $2\mathbf{z}=\mathbf{w}+3\mathbf{u}$ $\mathbf{z}=\frac{1}{2}\mathbf{w}+\frac{3}{2}\mathbf{u}$. All we need to do now is plug in the vectors $\mathbf{w}$ and $\mathbf{u}$: $\mathbf{z}=\frac{1}{2}(4,0,-4)+\frac{3}{2}(1,2,3)$ $\mathbf{z}=(2,0,-2)+(\frac{3}{2},3,\frac{9}{2})=(\frac{7}{2},3,\frac{5}{2})$

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.