Precalculus Vector and Matrix problem

[MathGeek5321]

Every vector v can be expressed uniquely in the form a + b, where a is a scalar multiple of \begin{pmatrix} 2 \\ -1 \end{pmatrix} and \mathbf{b}$ is a scalar multiple of $\begin{pmatrix} 3 \\ 1 \end{pmatrix}. What is the matrix P such that
\[\mathbf{P} \mathbf{v} = \mathbf{a}\]for all vectors v?

 

[Steven G]

May I please ask what you expect as a result of your post above?

 

[pka]

Every vector v can be expressed uniquely in the form a + b, where a is a scalar multiple of \begin{pmatrix} 2 \\ -1 \end{pmatrix} and \mathbf{b}$ is a scalar multiple of $\begin{pmatrix} 3 \\ 1 \end{pmatrix}. What is the matrix P such that
\[\mathbf{P} \mathbf{v} = \mathbf{a}\]for all vectors v?

Here are the two given vectors: \(\vec{a}=\left( {\begin{array}{*{20}{c}}2 \\ { - 1}\end{array}} \right)~\&~\vec{b}=\left( {\begin{array}{*{20}{c}}3 \\ { 1 }\end{array}} \right)\)
Having taught vector analysis many times, I am with Jomo in find your post very confused.
I think that you mean that very vector \(\vec{v}=\left( {\begin{array}{*{20}{c}}p \\ { q}\end{array}} \right)\) can be written as \(\vec{v}=\alpha\vec{a}+\beta\vec{b}\).
If that is the question then \(\alpha=\frac{1}{5}p-\frac{3}{5}q\quad \&\quad\beta=\frac{1}{5}+\frac{2}{5}q\)
Reread the question and see what you can do with that.