Projektion med matris

Vilken matris, P, ger projektionen av en punkt i linjen x+2y=0

Så här har jag gjort:

$$\begin{gathered} \stackrel{\rightharpoonup }{n}=(1,2) \\ (x_2,y_2)=\left(x_1{y}_1\right)-t(1,2) \\ \left\{\begin{array}{l}{x}_2=x_1-t\\ y_2=y_1-2t\end{array}\right. \\ (x_1-t)+2(y_1-2t)=0 \\ {x}_1-t+2y_1-4t=0 \\ {x}_1+2y_1-5t=0 \\ 5t=x_1+2y_1 \\ t=\frac{x_1+2y_1}{5} \\ \left\{\begin{array}{l}{x}_2=x_1-\frac{x_1+2y_1}{5}\\ y_2=y_1-2\frac{x_1+2y_1}{5}\end{array}\right.\iff \left\{\begin{array}{l}{x}_2=\frac{5x_1-x_1+2y_1}{5}\\ y_2=\frac{5y_1-2x_1+4y_1}{5}\end{array}\right.\iff \left\{\begin{array}{l}{x}_2=\frac{4x_1+2y_1}{5}\\ y_2=\frac{9y_1-2x_1}{5}\end{array}\iff \left\{\begin{array}{l}{x}_2=\frac{4}{5}{x}_1+\frac{2}{5}{y}_1\\ y_2=-\frac{2}{5}{x}_1+\frac{9}{5}{y}_1\end{array}\right.\right. \\ \iff \left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}4& 2\\ -2& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right] \\ \mathit{P}=\frac{1}{5}\left[\begin{array}{cc}4& 2\\ -2& 4\end{array}\right] \end{gathered}$$

Men i facit står det

$\mathit{P}=\frac{1}{5}\left[\begin{array}{cc}4& -2\\ -2& 1\end{array}\right]$

Vad gör jag fel????