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Date May Example question Marks available 3 Reference code EXM.2.AHL.TZ0.23
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Explain Question number 23 Adapted from N/A

Question

Let = ( 3 1 4 3 ) .

Let A2 + m + n I = O where m , n Z and = ( 0 0 0 0 ) .

Find the values of λ for which the matrix (A λ I) is singular.

[5]
a.

Find the value of m  and of n .

[5]
b.i.

Hence show that I = 1 5 A (6IA).

[4]
b.ii.

Use the result from part (b) (ii) to explain why A is non-singular.

[3]
b.iii.

Use the values from part (b) (i) to express A4 in the form p A+ q I where p , q Z .

[5]
c.

Markscheme

A −  λ I = ( 3 λ 1 4 3 λ )            A1

If A −  λ I is singular then det (A −  λ I) = 0           (R1)

det (A −  λ I = ( 3 λ ) 2 4 ( = λ 2 6 λ + 5 )            (A1)

Attempting to solve  ( 3 λ ) 2 4 = 0  or equivalent for  λ          M1

λ = 1, 5      A1  N2

Note: Candidates need both values of λ for the final A1.

[5 marks]

a.

( 3 1 4 3 ) 2 + m ( 3 1 4 3 ) + n ( 1 0 0 1 ) = ( 0 0 0 0 )            A1

( 3 1 4 3 ) 2 = ( 13 6 24 13 )            (A1)

Forming any two independent equations           M1

(eg  6 + m = 0 13 + 3 m + n = 0 or equivalent)

Note: Accept equations in matrix form.

Solving these two equations      (M1)

m = 6 and  n = 5       A1  N2

[5 marks]

b.i.

A2 − 6A + 5I = O        (M1)

5I = 6A − A2         A1

= A(6I − A)          A1A1

Note: Award A1 for A and A1 for (6I − A).

I = 1 5 A(6I − A)     AG  N0

Special Case: Award M1A0A0A0 only for candidates following alternative methods.

[5 marks]

b.ii.

METHOD 1

I = 1 5 A(6I − A) = A ×  1 5 (6I − A)         M1

Hence by definition 1 5 (6I − A) is the inverse of A.     R1

Hence A−1 exists and so A is non-singular       R1   N0

 

METHOD 2

As det I = 1 (≠ 0), then          R1

det  1 5 A(6I − A) = 1 5 det A × det (6I − A) (≠ 0)      M1

⇒ det A ≠ 0 and so A is non-singular.         R1   N0

 

[3 marks]

b.iii.

METHOD 1

A2 = 6A − 5I              (A1)

A4 = (6A − 5I)2              M1

     = 36A2 − 60AI + 25I2              A1

     = 36(6A − 5I) − 60A + 25I              M1

     = 156A − 155I ( p  = 156, q  = −155)              A1  N0

 

METHOD 2

A2 = 6A − 5I              (A1)

A3 = 6A2 − 5A where A2 = 6A − 5I              M1

     = 31A − 30I              A1

A4 = 31A2 − 30A where A2 = 6A − 5I              M1

     = 156A − 155I ( p  = 156, q  = −155)              A1  N0

 

Note: Do not accept methods that evaluate A4 directly from A.

 

[5 marks]

c.

Examiners report

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

Syllabus sections

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