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Date May 2018 Marks available 5 Reference code 18M.1.hl.TZ0.10
Level HL only Paper 1 Time zone TZ0
Command term Describe Question number 10 Adapted from N/A

Question

By considering the images of the points (1, 0) and (0, 1),

determine the 2 × 2 matrix P which represents a reflection in the line \(y = \left( {{\text{tan}}\,\theta } \right)x\).

[3]
a.i.

determine the 2 × 2 matrix Q which represents an anticlockwise rotation of θ about the origin.

[2]
a.ii.

Describe the transformation represented by the matrix PQ.

[5]
b.

A matrix M is said to be orthogonal if M TM = I where I is the identity. Show that Q is orthogonal.

[2]
c.

Markscheme

      (M1)

using the transformation of the unit square:

\(\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,2\theta } \\
{{\text{sin}}\,2\theta }
\end{array}} \right)\) and \(\left( \begin{gathered}
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{sin}}\,2\theta } \\
{ - {\text{cos}}\,2\theta }
\end{array}} \right)\)       (M1)

hence the matrix P is \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,2\theta } \\
{{\text{sin}}\,2\theta }
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,2\theta } \\
{ - {\text{cos}}\,2\theta }
\end{array}} \right)\)       A1

[3 marks]

a.i.

using the transformation of the unit square:

\(\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}} \right)\) and \(\left( \begin{gathered}
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{ - {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\)      (M1)

hence the matrix Q is \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ - {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\)      A1

[2 marks]

a.ii.

PQ = \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta \,{\text{cos}}\,2\theta + {\text{sin}}\,\theta \,{\text{sin}}\,2\theta } \\
{ - {\text{cos}}\,2\theta \,{\text{sin}}\,\theta + \,{\text{sin}}\,2\theta \,{\text{cos}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta \,{\text{sin}}\,2\theta \, - {\text{sin}}\,\theta \,{\text{cos}}\,2\theta } \\
{ - {\text{sin}}\,\theta \,{\text{sin}}\,2\theta - {\text{cos}}\,\theta \,{\text{cos}}\,2\theta \,}
\end{array}} \right)\)      M1A1

\( = \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\left( {2\theta - \theta } \right)} \\
{{\text{sin}}\,\left( {2\theta - \theta } \right)}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\left( {2\theta - \theta } \right)} \\
{ - {\text{cos}}\,\left( {2\theta - \theta } \right)}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\theta } \\
{ - {\text{cos}}\,\theta }
\end{array}} \right)\)     M1A1

this is a reflection in the line \(y = \left( {{\text{tan}}\,\frac{1}{2}\theta } \right)x\)      A1

[5 marks]

b.

Q TQ = \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{ - {\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ - {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}
{{\text{co}}{{\text{s}}^2}\,\theta + {\text{si}}{{\text{n}}^2}\,\theta } \\
{ - {\text{sin}}\,\theta \,{\text{cos}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ - {\text{cos}}\,\theta \,{\text{sin}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta } \\
{{\text{si}}{{\text{n}}^2}\,\theta + {\text{co}}{{\text{s}}^2}\,\theta }
\end{array}} \right)\)      M1A1

\( = \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)\)      AG

[2 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 1 - Linear Algebra » 1.6 » Linear transformations: \(T(u + v) = T(u) + T(v)\) , \(T(ku) = kT(u)\) .

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