1. Equivalent transformations by matrices

I have done the first part. I can't think of a way to do the second part other than by making guesses. The answers are y=2x and y=-0.5x.

2. I have done it now. The second part is done by starting similarly to the first part.

3. Yes. To do the first part, look at $\begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ 3x\end{pmatrix}$ $= \begin{pmatrix}2x+ 3x \\ 2x- 6x\end{pmatrix}= \begin{pmatrix}5x \\ -4x\end{pmatrix}$.

So x'= 5x and y'= -4x. x= x'/5 so y'= -4(x'/5)= (-4/5)x'.

Any "line through the origin" is of the form y= ax with slope a. The line at right angles to that is y= (-1/a)x so you want to find a such that
$\begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ ax\end{pmatrix}$ $=\begin{pmatrix}2x+ ax \\ 2x- 2ax\end{pmatrix}= \begin{pmatrix}(2+a)x \\ (2- 2a)x\end{pmatrix}= \begin{pmatrix}x' \\ -(1/a)x'\end{pmatrix}$

That is, we must have x'= (2+a)x and y'= (-1/a)x'= (2- 2a)x. Dividing one equation by the other so that we eliminate x' on both sides of the equation, $-\frac{1}{a}= \frac{2+ a}{2+ 2a}$. Multiply on both sides by (2+ 2a)a to get rid of the fractions.

$2+ 2a= (2+ a)a= 2a+ a^2$ which reduces to $a^2= 2$.

Ouch, I divided the two equations incorrectly. Dividing x'= (2+ a)x by (-1/a)x'= (2- 2a)x gives $\frac{x'}{-\frac{1}{a}x'}= \frac{(2+a)x}{(2- 2a)x}$ or $-a= \frac{2+a}{2- 2a}$. Multiplying both sides by 2- 2a gives -(2- 2a)a= -2a+ 2a^2= 2+ a. $2a^2- 3a2 2= (2a+ 1)(a- 2)= 0$ which has roots a= 2 and a= -1/2. the lines, y= ax, are y= 2x and y= (-1/2)x as you said.

4. I used a slightly different method, still observing that multiplied gradients of perpendicular lines equals -1. I think you might have a mistake. You haven't finished with the equations of lines required.

5. Could you also fix the missing equals sign that is missing from the second row of matrix multiplications? Thanks.