## Differential Equations and Linear Algebra (4th Edition)

1. $\vec{v}_1$=(6,2) 2. $\vec{v}_2$=(-3,6) 3. $\vec{v}_3$=(3,8)
1. Subtitute the components of the geometric $\vec{x}$ into the equation of $\vec{v}_1$. Then, each components of the geometric $\vec{v}_1$ is times with 2. $\vec{v}_1$=2$\vec{x}$=2(3,1)=(6,2). 2. Subtitute the components of the geometric $\vec{y}$ into the equation of $\vec{v}_2$. Then, each components of the geometric $\vec{v}_2$ is times with 3. $\vec{v}_2$=3$\vec{y}$=3(-1,2)=(-3,6). 3. $\vec{v}_3$ is the summation of $\vec{v}_1$ and $\vec{v}_2$. Therefore, using the previous answers, replaced $\vec{v}_3$ by $\vec{v}_3$= $\vec{v}_1$+ $\vec{v}_2$ =(6,2)+(-3,6)=(3,8).