FP1 2016 6 Edexcel

(a)

We can see U is a rotation. Note that:

\begin{pmatrix}\cos \theta & - \sin\theta \\ \sin\theta & \cos \theta\end{pmatrix}

represents anti-clockwise rotation through an angle of \theta about the origin.

We see that \cos \theta = -\dfrac 1 {\sqrt 2} and \sin \theta = \dfrac 1 {\sqrt 2}, satisfied by \theta = \dfrac {3 \pi} 4.

So U is an anti-clockwise rotation through an angle of \dfrac {3 \pi} 4 (radians) about the origin.

(b)

We have:

\begin{pmatrix}-\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2}\end{pmatrix} \begin{pmatrix}p\\q\end{pmatrix} = \begin{pmatrix}-\frac 1 {\sqrt 2} p - \frac 1 {\sqrt 2} q \\ \frac 1 {\sqrt 2} p - \frac 1 {\sqrt 2} q\end{pmatrix} = \begin{pmatrix}6 \sqrt 2 \\ 3 \sqrt 2\end{pmatrix}

So:

-p - q = 6 \times \sqrt 2 \times \sqrt 2 = 12

and:

p - q = 3 \times \sqrt 2 \times \sqrt 2 = 6

So:

-2q = 18

giving:

q = -9

and:

p = -3

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Note that reflection in the line y = (\tan \theta)x can be represented by the matrix:

\begin{pmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{pmatrix}

Setting \theta = \dfrac \pi 4, we have:

Q = \begin{pmatrix}\cos \frac \pi 2 & \sin \frac \pi 2 \\ \sin \frac \pi 2 & -\cos \frac \pi 2\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}

(d)

Note that the transformation T = V \circ U is represented by the matrix QP. So we have:

\displaystyle R = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}-\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2}\end{pmatrix} = \begin{pmatrix}\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2} \\ -\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2}\end{pmatrix}

(e)

Either note that T is a reflection, (in y = -x) so is self-inverse, or that:

\begin{align*}R^{-1} &= \frac 1 {\frac 1 {\sqrt 2} \times \left(-\frac 1 {\sqrt 2}\right) - \left(-\frac 1 {\sqrt 2}\right) \times \left(-\frac 1 {\sqrt 2}\right)} \begin{pmatrix}-\frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2}\end{pmatrix} \\ & = -\begin{pmatrix}-\frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2}\end{pmatrix} \\ & = \begin{pmatrix}\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2} \\ -\frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2}\end{pmatrix} \\ & = R \end{align*}

(or note that R^2 = I [notice that this means that \det R = \pm 1 so R is invertible], then R^2 R^{-1} = R^{-1}, giving R = R^{-1})

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