## Elementary Linear Algebra 7th Edition

(a) ${u} \cdot {v}=-6$. (b) ${v} \cdot {v}=13$. (c) $\|u\|^{2}=25$. (d) $({u} \cdot {v})v=-6\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$. (e) ${u} \cdot {5v}=-30$.
Given ${u}=\left[\begin{array}{l}{3} \\ {4}\end{array}\right]$ and ${v}=\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$, then we have (a) ${u} \cdot {v}={u}^{T} {v}=[3 \ \ 4]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right]=[(3)(2)+(-3)(4)]=-6$. (b) ${v} \cdot {v}={v}^{T} {v}=[2 \ \ -3]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right]=[(2)(2)+(-3)(-3)]=13$. (c) $\|u\|^{2}={u} \cdot {u}={u}^{T} {u}=[3 \ \ 4]\left[\begin{array}{l}{3} \\ {4}\end{array}\right]=[(3)(3)+(4)(4)]=25$. (d) $({u} \cdot {v})v=({u}^{T} {v}) v=[3 \ \ 4]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right] \left[\begin{array}{l}{2}\\ {-3}\end{array}\right]=-6\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$. (e) ${u} \cdot {5v}={u}^{T} {5v}=[3 \ \ 4]\left[\begin{array}{l}{10} \\ {-15}\end{array}\right]=[(3)(10)+(4)(-15)]=-30$.