Find $P\begin{pmatrix} 1 & 0 & 2 \\ 2 & 0 & 1\\ 4 & 0 & -1 \end{pmatrix} - R\begin{pmatrix} 1 & 1 & 2 \\ -1 & 1 & 0\\ 2 & 0 & 2 \end{pmatrix}$.
I know how to multiply matrices, I just have to find P and R. I was able to find that for P:
$\begin{pmatrix} a & b & c \\ d & e & f\\ g & h & i \end{pmatrix}\begin{pmatrix} 1 \\ 2\\ 4 \end{pmatrix}= \begin{pmatrix} a + 2b + 2c \\ d + 2e + 4f \\ g + 2h + 4i \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$
and
$\begin{pmatrix} a & b & c \\ d & e & f\\ g & h & i \end{pmatrix}\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}= \begin{pmatrix} -4a + c \\ -4d + f \\ -4g + i \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
(by multiplying the vector by itself it turns out to be the same, and perpendicular equals the zero vector).
but I need another equation to find P, and I don't know what. Please help! Thank you so much!
You don’t need to know $P$ and $R$ explicitly to solve this.
It appears that you know that $P$ is the identity map on its image, so you have the first column of $P\,\tiny{\begin{bmatrix}1&0&2\\2&0&1\\4&0&-1\end{bmatrix}}$. Any linear transformation maps zero to zero, so you have the second column of this product as well. For the third column, observe that $(2,1,-1)^T$ and $(1,2,4)^T$ are orthogonal. What is the kernel of the (presumably orthogonal) projection represented by $P$?
You can proceed in a similar way for the second term. Hints: What vectors are mapped to themselves by a reflection? What does a reflection do to vectors orthogonal to the reflector? Can you find a relationship among the columns of the matrix being multiplied by $R$?