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Date May Example question Marks available 3 Reference code EXM.2.AHL.TZ0.22
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number 22 Adapted from N/A

Question

Let M2 = M where M(abcd),bc0(abcd),bc0

Show that a+d=1.

[3]
a.i.

Find an expression for bc in terms of a.

[2]
a.ii.

Hence show that M is a singular matrix.

[3]
b.

If all of the elements of M are positive, find the range of possible values for a.

[3]
c.

Show that (IM)2 = IM where I is the identity matrix.

[3]
d.

Markscheme

Attempting to find M2           M1

M2 = (a2+bcab+bdac+cdbc+d2)           A1

b(a+d)=b or c(a+d)=c           A1

Hence a+d=1   (as b0 or c0)      AG  N0

[3 marks]

a.i.

a2+bc=a        M1

bc=aa2         A1  N1

[2 marks]

a.ii.

METHOD 1

Using det M = adbc        M1

det M = ada(1a) or det M = a(1a)a(1a)

(or equivalent)         A1

      =0 using a+d=1 or d=1a to simplify their expression         R1

Hence M is a singular matrix         AG  N0

 

METHOD 2

Using bc=a(1a) and a+d=1 to obtain bc=ad        M1A1

det M = adbc and adbc=0 as bc=ad         R1

Hence M is a singular matrix         AG  N0

 

[3 marks]

b.

a(1a)>0       (M1)

0 < a < 1        A1A1    N3

Note: Award A1 for correct endpoints and A1 for correct inequality signs.

[3 marks]

c.

METHOD 1

Attempting to expand (I − M)2      M1

(I − M)2 = I − 2M + M2      A1    

              = I − 2M + M          A1    

              = I − M         AG   N0

 

METHOD 2

Attempting to expand (I − M)2 = (1abc1d)2  (or equivalent)      M1

(I − M)2 = ((1a)2+bcb(1a)b(1d)c(1a)c(1d)bc+(1d)2)

(or equivalent)          A1  

Use of a+d=1 and bc=aa2 to show desired result.      M1

Hence (I − M)2 = (1abc1d)      AG   N0

 

[3 marks]

d.

Examiners report

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

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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