1. ## Transformations Involving Matrices

Hello, I am stuck on a question in my FP1 textbook, in the practice exam section. I'm doing the OCR board.

It tells me that there are two 2 x 2 matrices R and S represent the following transformations in the x-y place.

R: a rotation through angle θ anticlockwise about origin
S: a stretch with scale factor 2 parallel to the x-axis (with the y-axis invariant)

RSR-¹ is denoted by M

I have found that M is:

$\left(
\begin{array}{cc}
1 + cos^2 \theta & sin \theta cos \theta \\
sin \theta cos \theta & 1 + sin^2 \theta
\end{array}
\right)$

The next part says the the point P has the column matrix:

$\left(
\begin{array}{c}
x \\
y
\end{array}
\right)$

And that the point Q is the image of P under the transformation represented by M.

I have found that the column matrix of Q would be:

$
\left(
\begin{array}{c}
x + xcos^2 \theta + ysin \theta cos \theta \\
ysin \theta cos \theta + y+xsin^2 \theta
\end{array}
\right)$

But the next part asks me to show that the line joining P to Q makes an angle θ with the x-axis.

2. Hi

You have made a mistake on Q which is actually

$
\left(
\begin{array}{c}
x + xcos^2 \theta + ysin \theta cos \theta \\
xsin \theta cos \theta + y+ysin^2 \theta
\end{array}
\right)$

Then the column matrix of PQ is

$
\left(
\begin{array}{c}
xcos^2 \theta + ysin \theta cos \theta \\
xsin \theta cos \theta +ysin^2 \theta
\end{array}
\right) = (x \cos \theta + y \sin \theta) \left(
\begin{array}{c}
cos \theta \\
sin \theta
\end{array}
\right)$

3. Oh yes, sorry about that. >_<

But I am unsure how the column matrix of PQ proves that it makes an angle of θ? Please could you explain?

4. The column matrix of PQ is

$
\left(
\begin{array}{c}
xcos^2 \theta + ysin \theta cos \theta \\
xsin \theta cos \theta +ysin^2 \theta
\end{array}
\right) = (x \cos \theta + y \sin \theta) \left(
\begin{array}{c}
cos \theta \\
sin \theta
\end{array}
\right)$

$PQ^2 = (x \cos \theta + y \sin \theta)^2 (\cos^2\theta + \sin^2\theta) = (x \cos \theta + y \sin \theta)^2$

$||\overrightarrow{PQ}|| = PQ = |x \cos \theta + y \sin \theta|$

$\frac{\overrightarrow{PQ}}{||\overrightarrow{PQ}|| } = \pm \left(
\begin{array}{c}
cos \theta \\
sin \theta
\end{array}
\right)$

This shows that the angle between PQ and x-axis is θ or θ+pi
Therefore the angle between the line (PQ) and x-axis is θ