IBDP Maths AI: Topic : AHL 1.15: Eigenvalues and eigenvectors: IB style Questions HL Paper 1

(i)     let \(\boldsymbol{M} = \left( {\begin{array}{*{20}{c}}
a&b\\
b&c
\end{array}} \right)\)     (M1)

the eigenvalues satisfy

\(\det (\boldsymbol{M} – \lambda \boldsymbol{I}) = 0\)     (M1)

\((a – \lambda )(c – \lambda ) – {b^2} = 0\)     (A1)

\({\lambda ^2} – \lambda (a + c) + ac – {b^2} = 0\)     A1

discriminant \( = {(a + c)^2} – 4(ac – {b^2})\)     M1

\( = {(a – c)^2} + 4{b^2} \ge 0\)     A1

this shows that the eigenvalues are real     AG

(ii)     let the distinct eigenvalues be \({\lambda _1},{\lambda _2}\)  , with eigenvectors \({{\boldsymbol{X}}_1}\), \({{\boldsymbol{X}}_2}\)

then

\({\lambda _1}{{\boldsymbol{X}}_1} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\) and \({\lambda _2}{{\boldsymbol{X}}_2} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\)     M1

transpose the first equation and postmultiply by \({{\boldsymbol{X}}_2}\) to give

\({\lambda _1}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\)     A1

premultiply the second equation by \({\boldsymbol{X}}_1^T\)

\({\lambda _2}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\)     A1

it follows that

\((\lambda 1 – {\lambda _2}){\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\)     A1

since \(\lambda 1 \ne {\lambda _2}\) , it follows that \({\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\) so that the eigenvectors are orthogonal     R1

[11 marks]

the eigenvalues satisfy \(\left| \begin{array}{l}
11 – \lambda \\
\sqrt 3
\end{array} \right.\left. \begin{array}{l}
\sqrt 3 \\
9 – \lambda
\end{array} \right| = 0\)     M1A1

\({\lambda ^2} – 20\lambda  + 96 = 0\)    A1

\(\lambda  = 8,12\)     A1

first eigenvector satisfies

\(\left( {\begin{array}{*{20}{c}}
3&{\sqrt 3 }\\
{\sqrt 3 }&1
\end{array}} \right)\left( \begin{array}{l}
x\\
y
\end{array} \right) = \left( \begin{array}{l}
0\\
0
\end{array} \right)\)     M1

\(\left( \begin{array}{l}
x\\
y
\end{array} \right) = \) (any multiple of) \(\left( {\begin{array}{*{20}{c}}
1\\
{ – \sqrt 3 }
\end{array}} \right)\)     A1

second eigenvector satisfies

\(\left( {\begin{array}{*{20}{c}}
{ – 1}&{\sqrt 3 }\\
{\sqrt 3 }&{ – 3}
\end{array}} \right)\left( \begin{array}{l}
x\\
y
\end{array} \right) = \left( \begin{array}{l}
0\\
0
\end{array} \right)\)

\(\left( \begin{array}{l}
x\\
y
\end{array} \right) = \) (any multiple of ) \(\left( {\begin{array}{*{20}{c}}
{\sqrt 3 }\\
1
\end{array}} \right)\)     A1

[7 marks]

(i)     consider the rotation in which \((x,y)\) is transformed onto \((x’,y’)\) defined by

\(\left( \begin{array}{l}
{x’}\\
{y’}
\end{array} \right) = \left( {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{ – \frac{{\sqrt 3 }}{2}}\\
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
\end{array}} \right)\left( \begin{array}{l}
x\\
y
\end{array} \right)\) so that \(\left( \begin{array}{l}
x\\
y
\end{array} \right) = \left( {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\
{ – \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
\end{array}} \right)\left( \begin{array}{l}
{x’}\\
{y’}
\end{array} \right)\)     M1A1

the ellipse \(E\) becomes

\(\left( {\begin{array}{*{20}{c}}
{x’}&{y’}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{ – \frac{{\sqrt 3 }}{2}}\\
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{11}&{\sqrt 3 }\\
{\sqrt 3 }&9
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\
{ – \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
\end{array}} \right)\left( \begin{array}{l}
{x’}\\
{y’}
\end{array} \right) = 24\)     M1A1

\(\left( {\begin{array}{*{20}{c}}
{x’}&{y’}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
8&0\\
0&{12}
\end{array}} \right)\left( \begin{array}{l}
{x’}\\
{y’}
\end{array} \right) = 24\)     A1

\(2{(x’)^2} + 3{(y’)^2} = 6\)    AG

(ii)     the angle of rotation is given by \(\cos \theta  = \frac{1}{2},\sin \theta  = \frac{{\sqrt 3 }}{2}\)     M1

since a rotational matrix has the form \(\left( {\begin{array}{*{20}{c}}
{\cos \theta }&{ – \sin \theta }\\
{\sin \theta }&{\cos \theta }
\end{array}} \right)\)

so \(\theta  = {60^ \circ }\) (anticlockwise)     A1

[7 marks]