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Date May Example question Marks available 3 Reference code EXM.2.AHL.TZ0.21
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Justify Question number 21 Adapted from N/A

Question

Let A ( 1 1 1 0 1 1 0 0 1 ) and B = ( 1 0 0 1 1 0 1 1 1 ) .

Given that X = B A–1 and Y = B–1 – A,

You are told that A n = ( 1 n n ( n + 1 ) 2 0 1 n 0 0 1 ) , for  n Z + .

Given that  ( A n ) 1 = ( 1 x y 0 1 x 0 0 1 ) , for  n Z + ,

find X and Y.

[2]
a.i.

does X–1 + Y–1 have an inverse? Justify your conclusion.

[3]
a.ii.

find x and y in terms of n .

[5]
b.i.

and hence find an expression for  A n + ( A n ) 1 .

[1]
b.ii.

Markscheme

X = BA–1 =  ( 1 0 0 1 1 0 1 1 1 ) ( 1 1 0 0 1 1 0 0 1 ) = ( 0 1 0 1 0 1 1 1 0 )         A1

Y = B–1  A ( 1 0 0 1 1 0 1 1 1 ) ( 1 1 1 0 1 1 0 0 1 ) = ( 0 1 1 1 0 1 0 1 0 )         A1

 

[2 marks]

a.i.

X–1 + Y–1 =  ( 0 1 0 1 0 1 0 1 0 )          (A1)

X–1 + Y–1 has no inverse           A1

as det(X–1 + Y–1) = 0        R1

[3 marks]

a.ii.

A n ( A n ) 1 = I ( 1 n n ( n + 1 ) 2 0 1 n 0 0 1 ) ( 1 x y 0 1 x 0 0 1 ) = ( 1 0 0 0 1 0 0 0 1 )        M1

( 1 x + n y + n x + n ( n + 1 ) 2 0 1 x + n 0 0 1 ) = ( 1 0 0 0 1 0 0 0 1 )            A1

solve simultaneous equations to obtain

x + n = 0 and  y + n x + n ( n + 1 ) 2 = 0         M1

x = n and  y = n ( n 1 ) 2           A1A1N2

[5 marks]

b.i.

A n + ( A n ) 1 = ( 1 n n ( n + 1 ) 2 0 1 n 0 0 1 ) + ( 1 n n ( n 1 ) 2 0 1 n 0 0 1 ) = ( 2 0 n 2 0 2 0 0 0 2 )           A1

 

[1 mark]

b.ii.

Examiners report

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

Syllabus sections

Topic 1—Number and algebra » AHL 1.14—Introduction to matrices
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Topic 1—Number and algebra

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