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I am having trouble with this seemingly simple question: Write the standard coordinate form of an affine transformation in $\mathbb{A}^{2}(\mathbb{R})$ that maps the point (1, −2) to the point (0, 10), and the lines $10x_{1} − 4x_{2} = 1$ and $3x_{1} − 3x_{2} = −7$ to the lines $x_{1} − 2x_{2} = −3$ and $x_{1} − x_{2} = 6$, respectively.

Now as I understand for the point transformation $\begin{pmatrix}0\\10\\0\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12} & b_{1}\\a_{21} & a_{22} & b_{2}\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\-2\\0\end{pmatrix}$ and I need to find this matrix of transformation, this matrix equation gives me two equations with 6 unknowns. So I guess 4 more equations will be coming from the transformations of the two lines. Now here is where I am having trouble, I don't know how to get those equations. Any kind of hints would be really helpful. Thanks in advance.

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I think the added third coordinates should be $1$ instead of $0$ to obtain the translation effect of the matrix multiplication.

Write $x=(x_1,x_2,1)^T$.
If an expression $e(x)$ defines a given set $E$, in the sense that $e(x)$ is true iff $x\in E$, and $A$ is an invertible transformation, then the image of $E$ can be described by the expression $$f(y):=e(A^{-1}(y))$$ because $y\in A(E)\iff A^{-1}(y)\in E\iff e(A^{-1}(y))$.
Since we don't want to calculate the inverse of the unknown matrix, we can rewrite the above formula by the $x=A^{-1}(y)$ substitution as $$f(A(x))=e(x)\,.$$ So, we have the following connection between the first equations: $$(a_{11}x_1+a_{12}x_2+b_1)\ -\ 2(a_{21}x_1+a_{22}x_2+b_2)\ =\ -3\ \iff\ 10x_1-4x_2\,=\,1\,,$$ which means that the two equations are a scalar multiple of each other. This actually brings in one more variable and 3 equations.
And the same to be done with the other two lines.

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  • $\begingroup$ Thank you so much. It took me a while but I was able to solve it, thanks to your help. Just one simple question, how does the third coordinate being 1 instead of 0 give the translation effect of matrix multiplication? I feel like I have missed something. $\endgroup$ – Paul Dirac jr jr jr... Jan 20 at 12:01
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    $\begingroup$ If you multiply $\pmatrix{1&0&b_1\\ 0&1&b_2\\ 0&0&1}\pmatrix{x_1\\x_2\\1}$ you receive $\pmatrix{x_1+b_1\\x_2+b_2\\1}$, but the $b_i$'s would disappear if you used $0$ as third coordinate. $\endgroup$ – Berci Jan 20 at 12:25
  • $\begingroup$ Oh yes. Thank you so much for your help. $\endgroup$ – Paul Dirac jr jr jr... Jan 20 at 12:29
  • $\begingroup$ Your answer is very good and short. I have provided one which is longer but has, so I think, its own merits... $\endgroup$ – Jean Marie Jan 20 at 14:12
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Here is a different method. Let $A$ be the looked for matrix.

First of all, let us determine the intersection of the "initial" axes and of the "final" axes. It is rather easy to find that the first two axes intersect at $(14/9,35/9)$ and the second pair of axes at $(15,9)$. Therefore, $(15,9,1)$ is the image of $(14/9,35/9,1)$ by $A$.

Besides, directing vectors of the initial axes are

$$\binom{2}{5} \ \ \text{and} \ \ \binom{1}{1}$$

and directing vectors of the final axis are:

$$s\binom{2}{1} \ \ \text{and} \ \ t\binom{1}{1}$$

(parameters $s$ and $t$ are there in order to take into account the change of scale)

With all these informations, we can say that

$$A : \begin{pmatrix}2\\5\\0\end{pmatrix} \to \begin{pmatrix}2s\\1s\\0\end{pmatrix}, \ \begin{pmatrix}1\\1\\0\end{pmatrix} \to \begin{pmatrix}1t\\1t\\0\end{pmatrix}, \ \begin{pmatrix}14/9\\35/9\\1\end{pmatrix}\to \begin{pmatrix}15\\9\\1\end{pmatrix} $$

Otherwise said :

$$A \times \underbrace{\begin{pmatrix}2&1&14/9\\5&1&35/9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{initial affine frame}\\B\end{array}} = \underbrace{\begin{pmatrix}2s&1t&15\\1s&1t&9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{final affine frame}\\C\end{array}} $$

which is equivalent to :

$$A=C \times B^{-1} = \dfrac{1}{3}\begin{pmatrix}5t-2s&2s-2t&45-14s/3\\ 5t-s&s-2t& 27-7s/3\\ 0& 0& 3\end{pmatrix}\tag{1}$$

Now, the last step is expressing that

$$A\begin{pmatrix}1\\-2\\1\end{pmatrix}=\begin{pmatrix}0\\10\\1\end{pmatrix}$$

which yields a system of two equations in two unknowns $s$ and $t$, finally giving :

$$s=9 \ \ \text{and} \ \ t=17/3$$

that have to be plugged into (1) to give the final answer:

$$A=\begin{pmatrix} 31/9&20/9&1\\ 58/9&-7/9& 2\\0&0&1\end{pmatrix}$$

Remark: the interest of this method is that all the computations can be done using a Computer Algebra System ; here is what I have done with Matlab:

syms x1 x2 s t 
[X1,X2]=solve(10*x1-4*x2==0,...
               3*x1-3*x2==-7,...
               x1,x2)
B=[2, 1, X1;
   5, 1, X2;
   0, 0, 1];
[X1,X2]=solve(x1-2*x2==-3,...
              x1  -x2==6,...
              x1,x2)
C=[2*s, t, X1 ; 
   s,   t, X2;
   0,   0, 1];
A=C*inv(B);
[S,T]=solve(A*[1,-2,1]'==[0,10,1]',s,t);
subs(A,{s,t},{S,T})
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