# projective matrix [duplicate]

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I am quiet new in projective transformation and I need to know and understand how can find the projective transformation matrix which transforms the set of points {{0,0,1},{0,1,0},{1,3,8},{1,4,6}} to the set {{1,0,0},{0,1,0},{0,0,1},{1,1,1}}. So, if I have the transformation mapping f defined by f:A={{0,0,1},{0,1,0},{1,3,8},{1,4,6}} ----> B={{1,0,0},{0,1,0},{0,0,1},{1,1,1}}, how can find the matrix transformation which maps A to B.

## marked as duplicate by amd, user223391, zz20s, Claude Leibovici, LeucippusDec 24 '17 at 6:37

• Have you tried using a general 3 by 3 Matrix $P$? Then just try plug in your vectors $x$ and equate $Ax=b$. Try to find equations for the coefficients. – MrYouMath Sep 16 '15 at 11:08
• I am really confused about the matrix P, any explanation please. – nancy Sep 16 '15 at 11:15

You want to find a matrix map from A to B.

We define an arbitrary Matrix

$$P=\begin{bmatrix}p_{11} & p_{12} &p_{13}\\p_{21} & p_{22} &p_{23}\\p_{31} & p_{32} &p_{33}\end{bmatrix}$$ Now you calculate the map for $x_1=(0,0,1)^T$ to $y_1=(1,0,0)^T$.

$$Px_1= \begin{bmatrix}p_{11} & p_{12} &p_{13}\\p_{21} & p_{22} &p_{23}\\p_{31} & p_{32} &p_{33}\end{bmatrix} \cdot \begin{bmatrix}0 \\0\\1\end{bmatrix}=y_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$$ From this you will get a system of equations. And we can conclude that: $p_{13}=1, p_{23} =0$ and $p_{33}=0$.

Do the same with $x_2=(0,1,0)^T,x_3=(1,3,8)^T$ and $x_4=(1,4,6)^T$.

I leave the rest to you.

• for Px2 I have got p12=0, p22=1, and p32=0 but about Px3 I have got – nancy Sep 16 '15 at 11:51
• p11= p21=p31=1, p12=p22=p32=1, and p13=p23=p33=11 Right? in this case how can prove Px3=[0 0 1]^T – nancy Sep 16 '15 at 11:53
• Sorry for that I wrote anther numbers , p11= p21=p31=1, p12=p22=p32=3, and p13=p23=p33=8 Right? in this case how can prove Px3=[0 0 1]^T – nancy Sep 16 '15 at 11:55
• Sorry, but it is really hard to follow you. Please use p_{11}, with a dollar sign in front and at the end, for p11 etc. Also i don't understand your last question. You have found 6 coefficients form $x_1$ and $x_2$. The remaining coefficients can be obtained by solving the system of equations that you get from $x_3$ and $x_4$. – MrYouMath Sep 16 '15 at 11:58
• Sorry I couldn't get $P*X_[3}$ and $P*X_[4}$ – nancy Sep 16 '15 at 14:54