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Date May Example question Marks available 8 Reference code EXM.3.AHL.TZ0.5
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 5 Adapted from N/A

Question

This question will diagonalize a matrix and apply this to the transformation of a curve.

Let the matrix  M = ( 5 2 1 2 1 2 5 2 ) .

Let  ( 1 2 1 2 1 2 1 2 ) = R 1 .

Let  R ( x y ) = ( X Y ) .

Let  ( 1 3 0 0 1 2 ) ( X Y ) = ( u v ) .

Hence state the geometrical shape represented by

Find the eigenvalues for M . For each eigenvalue find the set of associated eigenvectors.

[8]
a.

Show that the matrix equation  ( x y ) M ( x y ) = ( 6 )  is equivalent to the Cartesian equation  5 2 x 2 + x y + 5 2 y 2 = 6 .

[2]
b.

Show that  ( 1 2 1 2 ) and  ( 1 2 1 2 )  are unit eigenvectors and that they correspond to different eigenvalues.

[2]
c.i.

Hence, show that M ( 1 2 1 2 1 2 1 2 ) = ( 1 2 1 2 1 2 1 2 ) ( 2 0 0 3 ) .

[1]
c.ii.

Find matrix R.

[2]
d.i.

Show that  M = R 1 ( 2 0 0 3 ) R .

[1]
d.ii.

Verify that  ( X Y ) = ( x y ) R 1 .

[3]
e.i.

Hence, find the Cartesian equation satisfied by X and  Y .

[2]
e.ii.

Find the Cartesian equation satisfied by u and  v and state the geometric shape that this curve represents.

[2]
f.

State geometrically what transformation the matrix R represents.

[2]
g.

the curve in X and  Y in part (e) (ii), giving a reason.

[2]
h.i.

the curve in x and  y in part (b).

[1]
h.ii.

Write down the equations of two lines of symmetry for the curve in x and  y in part (b).

[2]
i.

Markscheme

| 5 2 λ 1 2 1 2 5 2 λ | = 0 ( 5 2 λ ) 2 ( 1 2 ) 2 = 0 5 2 λ = ± 1 2 λ = 2 or 3        M1M1A1A1

λ = 2       ( 1 2 1 2 1 2 1 2 ) ( p q ) = ( 0 0 ) q = p       eigenvalues are of the form  t ( 1 1 )        M1A1

λ = 3       ( 1 2 1 2 1 2 1 2 ) ( p q ) = ( 0 0 ) q = p      eigenvalues are of the form  t ( 1 1 )        M1A1

[8 marks]

a.

( x y ) ( 5 2 1 2 1 2 5 2 ) ( x y ) = ( 6 ) ( 5 2 x + 1 2 y 1 2 x + 5 2 y ) ( x y ) = ( 6 )       M1A1

( 5 2 x 2 + 1 2 x y + 1 2 x y + 5 2 y 2 ) = ( 6 ) 5 2 x 2 + x y + 5 2 y 2 = 6.        AG

[2 marks]

b.

( 1 2 1 2 ) = 1 2 ( 1 1 )  corresponding to λ = 2 ,      ( 1 2 1 2 ) = 1 2 ( 1 1 )  corresponding to  λ = 3      R1R1

[2 marks]

c.i.

M ( 1 2 1 2 ) = 2 ( 1 2 1 2 ) and M ( 1 2 1 2 ) = 3 ( 1 2 1 2 ) M ( 1 2 1 2 1 2 1 2 ) = ( 1 2 1 2 1 2 1 2 ) ( 2 0 0 3 )       A1AG

[1 mark]

c.ii.

Determinant is 1.    R = ( 1 2 1 2 1 2 1 2 )         M1A1

[2 marks]

d.i.

M R 1 = R 1 ( 2 0 0 3 )  so post multiplying by R gives  M = R 1 ( 2 0 0 3 ) R        M1AG

[1 mark]

d.ii.

( 1 2 1 2 1 2 1 2 ) ( x y ) = ( X Y ) ( 1 2 x 1 2 y 1 2 x + 1 2 y ) = ( X Y ) ( X Y ) = ( 1 2 x 1 2 y 1 2 x + 1 2 y )          M1A1

and  ( x y ) ( 1 2 1 2 1 2 1 2 ) = ( 1 2 x 1 2 y 1 2 x + 1 2 y ) completing the proof       A1AG

[3 marks]

e.i.

( x y ) M ( x y ) = ( 6 ) ( x y ) R 1 ( 2 0 0 3 ) R ( x y ) = ( 6 ) ( X Y ) ( 2 0 0 3 ) ( X Y ) = ( 6 )

( 2 X 2 + 3 Y 2 ) = ( 6 ) X 2 3 + Y 2 2 = 1        M1A1

[2 marks]

e.ii.

X 3 = u , Y 2 = v u 2 + v 2 = 1 , a circle (centre at the origin radius of 1)     A1A1

[2 marks]

f.

A rotation about the origin through an angle of 45° anticlockwise.    A1A1

[2 marks]

g.

an ellipse, since the matrix represents a vertical and a horizontal stretch    R1A1

[2 marks]

h.i.

an ellipse      A1

[1 mark]

h.ii.

y = x y = x       A1A1

[2 marks]

i.

Examiners report

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a.
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b.
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c.i.
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c.ii.
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d.i.
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d.ii.
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e.i.
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e.ii.
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f.
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g.
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h.i.
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h.ii.
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i.

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Eigenvalues and eigenvectors
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Topic 1—Number and algebra

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