Question
Let A2 = 2A + I where A is a 2 × 2 matrix.
Show that A4 = 12A + 5I.
Let B = \(\left[ {\begin{array}{*{20}{c}}
4&2 \\
1&{ – 3}
\end{array}} \right]\).
Given that B2 – B – 4I = \(\left[ {\begin{array}{*{20}{c}}
k&0 \\
0&k
\end{array}} \right]\), find the value of \(k\).
Answer/Explanation
Markscheme
METHOD 1
A4 = 4A2 + 4AI + I2 or equivalent M1A1
= 4(2A + I) + 4A + I A1
= 8A + 4I + 4A + I
= 12A + 5I AG
[3 marks]
METHOD 2
A3 = A(2A + I) = 2A2 + AI = 2(2A + I) + A(= 5A + 2I) M1A1
A4 = A(5A + 2I) A1
= 5A2 + 2A = 5(2A + I) + 2A
= 12A + 5I AG
[3 marks]
B2 = \(\left[ {\begin{array}{*{20}{c}}
{18}&2 \\
1&{11}
\end{array}} \right]\) (A1)
\(\left[ {\begin{array}{*{20}{c}}
{18}&2 \\
1&{11}
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
4&2 \\
1&{ – 3}
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
4&0 \\
0&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{10}&0 \\
0&{10}
\end{array}} \right]\) (A1)
\( \Rightarrow k = 10\) A1
[3 marks]
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 \(y = \left( {{\text{tan}}\,\theta } \right)x\).
determine the 2 × 2 matrix Q which represents an anticlockwise rotation of θ about the origin.
Describe the transformation represented by the matrix PQ.
A matrix M is said to be orthogonal if M TM = I where I is the identity. Show that Q is orthogonal.
Answer/Explanation
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]
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]
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]
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]