Module 1: Now You Try III

1. Compute the following linear combination of vectors:

$3 \begin{pmatrix} -1 \\ 2 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ -2 \end{pmatrix}.$

Solution:

$3 \begin{pmatrix} -1 \\ 2 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \end{pmatrix} + \begin{pmatrix} -10 \\ 10 \end{pmatrix} = \begin{pmatrix} -13 \\ 16 \end{pmatrix}$

2. Show that there does not exist any number $t$ so that

$t \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 5 \end{pmatrix}.$

Solution:

Note that: $t \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} t \\ 2t \\ 3t \end{pmatrix}.$ By comparing the first components we see that $t=2$. By comparing the third components we see $t=5/2$. Since these disagree we see that there are no solutions.

3. Let $\vec{u},\vec{v},\vec{x},\vec{y}$ all be vectors and that$2\vec{u}+\vec{v} = \vec{x}, \quad \vec{u}-\vec{v} = \vec{y}.$Express $\vec{u}$ and $\vec{v}$ in terms of $\vec{x}$ and $\vec{y}$.

Solution:

Note that by summing the equations together we have:$3 \vec{u} = \vec{x} + \vec{y} \Rightarrow \vec{u} = \frac{1}{3} \vec{x} + \frac{1}{3} \vec{y}.$Substituting the latter equality into either of the original equations gives$\vec{v} = \frac{1}{3} \vec{x} - \frac{2}{3} \vec{y}.$