Question

A transformation is described by the equation $AX+B={{X}^{'}}$ , where $A=\left( \begin{matrix} 0 & -3 \\\ 1 & 0 \\\ \end{matrix} \right)$ and $B=\left( \begin{matrix} -3 \\\ 2 \\\ \end{matrix} \right)$ . Find the image of the straight line with equation $y=-2x+6$ under the transformation?

Answer 1

The above mentioned problem is a very simple example of transformation matrix. In such type of problems, we proceed by assuming that the matrix value of $X$ is taken to be as, $X=\left( \begin{matrix} x \\\ y \\\ \end{matrix} \right)$ , and also, we assume the matrix value of ${{X}^{'}}$ , to be as ${{X}^{'}}=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$ . The given equation $y=-2x+6$ is assumed to be of the matrix $X=\left( \begin{matrix} x \\\ y \\\ \end{matrix} \right)$ and the equation we are required to find, the image of the given straight line is assumed to be of the matrix ${{X}^{'}}=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$ . Using the relation of the problem, we find a relation between ${{X}^{'}}$ and $X$ .

Complete step by step answer:
Now starting off with the solution, we can firstly try to find out the relation between ${{X}^{'}}$ and $X$ . Evaluating the transformation by putting in the values of $A=\left( \begin{matrix} 0 & -3 \\\ 1 & 0 \\\ \end{matrix} \right)$ and $B=\left( \begin{matrix} -3 \\\ 2 \\\ \end{matrix} \right)$ we get,

0 & -3 \\\ 1 & 0 \\\ \end{matrix} \right)\left( \begin{matrix} x \\\ y \\\ \end{matrix} \right)+\left( \begin{matrix} -3 \\\ 2 \\\ \end{matrix} \right)=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$$ Now, we do matrix multiplication, to find the relation between $${{X}^{'}}$$ and $$X$$. We evaluate it as, $$\Rightarrow \left( \begin{matrix} 0x-3y \\\ 1x+0y \\\ \end{matrix} \right)+\left( \begin{matrix} -3 \\\ 2 \\\ \end{matrix} \right)=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$$ Multiplying the inner terms and adding the required terms we get, $$\Rightarrow \left( \begin{matrix} 3y \\\ 1x \\\ \end{matrix} \right)+\left( \begin{matrix} -3 \\\ 2 \\\ \end{matrix} \right)=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$$ Now, adding the first two matrices, we get, $$\Rightarrow \left( \begin{matrix} 3y-3 \\\ x-2 \\\ \end{matrix} \right)=\left( \begin{matrix} {{x}^{'}} \\\ {{y}^{'}} \\\ \end{matrix} \right)$$ Now, comparing the above two matrices we can write, $$\Rightarrow 3y-3={{x}^{'}}$$ and $$x-2={{y}^{'}}$$ Now, representing the above two found equations as a function of $$x$$ and $$y$$ respectively, we get, $$\Rightarrow y=\dfrac{{{x}^{'}}+3}{3}$$ and $$x={{y}^{'}}+2$$ , Now putting these values of $$x$$ and $$y$$, in equation $$y=-2x+6$$ we get, $$\Rightarrow \dfrac{{{x}^{'}}+3}{3}=-2\left( {{y}^{'}}+2 \right)+6$$ Now evaluating this we get, $$\begin{aligned} & \Rightarrow \dfrac{{{x}^{'}}+3}{3}=-2{{y}^{'}}-4+6 \\\ & \Rightarrow \dfrac{{{x}^{'}}+3}{3}=-2{{y}^{'}}+2 \\\ & \Rightarrow {{x}^{'}}+3=-6{{y}^{'}}+6 \\\ & \Rightarrow {{x}^{'}}+-6{{y}^{'}}=3 \\\ \end{aligned}$$ Thus the equation $${{x}^{'}}+-6{{y}^{'}}=3$$ is thus our required transformation equation for this problem. **Note:** For these types of problems we must remember all the properties of matrices. We must also very carefully find the transformation equation to find the respective relations. After finding out the relations between $${{X}^{'}}$$ and $$X$$ , we just substitute one relation in the other, to find our required answer. We must also remember that the two equations of the straight line, one of which is the transformation of the other, lie in the same plane.