Every vector $\mathbf{v}$ can be expressed uniquely in the form $\mathbf{a} + \mathbf{b},$ where $\mathbf{a}$ is a scalar multiple of $\begin{pmatrix} 2 \\ -1 \end{pmatrix},$ and $\mathbf{b}$ is a scalar multiple of $\begin{pmatrix} 3 \\ 1 \end{pmatrix}.$ Find the matrix $\mathbf{P}$ such that $\mathbf{P} \mathbf{v} = \mathbf{a}$ for all vectors $\mathbf{v}.$
I'd like help deriving $\mathbf{P}$, but I don't know how to do it. Any help would be much appreciated!
On
We let $v = x\begin{pmatrix} 2 \\ -1 \end{pmatrix} + y \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2x + 3y \\ -x +y \\ \end{pmatrix}$. Then we want to find a $2 \times 2 $ matrix $P$ such that $P \begin{pmatrix} 2x + 3y \\ -x +y \\ \end{pmatrix} = \begin{pmatrix} 2x \\ -x \\ \end{pmatrix}$. Then we let $P = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$. Then we have $\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} 2x + 3y \\ -x +y \\ \end{pmatrix} = \begin{pmatrix} 2x \\ -x \\ \end{pmatrix}$. Then we have $a(2x+3y) + b(-x+y) = 2ax + 3ay - bx + by = 2x$. Then this implies that $2a-b = 2$, $3a+b = 0$. Then we have $a = \frac{2}{5}$, $b = \frac{-6}{5}$. We also have $c(2x+3y) + d(-x+y) = 2cx + 3cy - dx + dy = -x$. Then we have $2c-d = -1$, $3c+d = 0$. Thus, $c = \frac{-1}{5}$, $d = \frac{3}{5}$. Thus, the matrix $P = \begin{pmatrix} \frac{2}{5} & \frac{-6}{5} \\ \frac{-1}{5} & \frac{3}{5} \\ \end{pmatrix}$
You can take the scalars of a and b into the vector and add the two of them together to get vector v=$(\begin{bmatrix} 2a+3b \cr -a+b \end{bmatrix})$ . To get P from Pv=a refer to this: How to find matrix $A$ given $Ax=b$. Also $det(A)$ & $sum(A)$ are known.