1. ## Spans and vectors

Is the vector

$\displaystyle \mathbf{b}=\begin{pmatrix}-2\\ -6\\ -4\end{pmatrix}\in \mbox{span}(\mathbf{v_1, v_2, v_3, v_4})$ where:

$\displaystyle \mathbf{v_1}=\begin{pmatrix}1\\ 3\\ 0\end{pmatrix}$, $\displaystyle \mathbf{v_2}=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}$, $\displaystyle \mathbf{v_3}=\begin{pmatrix}-1\\ 0\\ -1\end{pmatrix}$, $\displaystyle \mathbf{v_4}=\begin{pmatrix}1\\ -2\\ 1\end{pmatrix}$

Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations?

2) Is the set of vectors $\displaystyle \mathbf{v_1}=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}$, $\displaystyle \mathbf{v_2}=\begin{pmatrix}1\\ 1\\ -1\end{pmatrix}$, $\displaystyle \mathbf{v_3}=\begin{pmatrix}-1\\ 0\\ 5\end{pmatrix}$ a spanning set of $\displaystyle \mathbb{R}^3$?

Do we just make this a matrix with say $\displaystyle \mathbf{x}=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}$ as our solution and see if we can solve the matrix?

2. So your first question is asking whether there exist scalars $\displaystyle a_{1},a_{2},a_{3},a_{4}$ such that

$\displaystyle \mathbf{b}=a_{1}\mathbf{v}_{1}+a_{2}\mathbf{v}_{2} +a_{3}\mathbf{v}_{3}+a_{4}\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $\displaystyle A\mathbf{x}=\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no.

Your second question, again, is asking whether there exist scalars $\displaystyle c_{1}, c_{2}, c_{3}$ such that $\displaystyle c_{1}\mathbf{v}_{1}+c_{2}\mathbf{v}_{2}+c_{3}\math bf{v}_{3}=\mathbf{x}$ for your arbitrary $\displaystyle \mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.