Question

Show that under the transformation (11.1), all points $\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$ on a given straight line through the origin go into points $\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$ on another straight line through the origin. Hint: Solve (11.1) for $\mathit{x}$ and $\mathit{y}$ in terms of $\mathit{X}$ and $\mathit{Y}$and substitute into the equation $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{x}$ to get an equation $\mathit{Y}\mathbf{}\mathbf{=}\mathbf{}\mathit{k}\mathit{X}$, where $\mathit{k}$ is a constant. Further hint: If $\mathbf{\text{R}}\mathbf{=}\mathbf{\text{Mr}}$, then $\mathbf{\text{r}}\mathbf{=}{\mathbf{\text{M}}}^{\mathbf{-}\mathbf{1}}\mathbf{\text{R}}$.

Short answer

The equation of a straight line $Y=kX$ for, $k=\frac{2(m-1)}{(5-2m)}.$

Step-by-step solution

Given information

The transformation is given below.

$\left(\begin{array}{c}X\\ Y\end{array}\right)=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

Transformation Matrix

When a matrix turns one vector into another vector through the process of matrix multiplication, it is known as a transformation matrix.

Calculate the inverse of the transformation matrix

Transformation given,

$\left(\begin{array}{c}X\\ Y\end{array}\right)=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

From this, we obtain the two equations,

$$\begin{gathered} X=5x-2y \\ Y=-2x+2y \end{gathered}$$

Solve these two equations for $x$and $y$by adding these two,

$$\begin{gathered} x=\frac{X+Y}{3} \\ y=\frac{2X+5Y}{6} \end{gathered}$$

Find the inverse of the transformation matrix, that is,

$M^{-1}{\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)}^{-1}=\frac{1}{6}\left(\begin{array}{cc}2& 2\\ 2& 5\end{array}\right)$

$C_{11}=m_{22},{\text{C}}_{12}=-m_{21},C_{21}=-m_{12}\text{and}{C}_{22}=m_{11}\text{.}$

Insert the values into the equation of straight line

Insert this result into the equation of a straight line $y=mx$

$$\begin{gathered} \frac{2X+5Y}{6}=m\frac{X+Y}{3} \\ 2X+5Y=2m(X+Y) \\ Y(5-2m)=X(2m-2) \\ Y=\frac{2(m-1)}{(5-2m)}X \end{gathered}$$

From this, we can see that the equation of a straight line $Y=kX$and $k=\frac{2(m-1)}{(5-2m)}.$