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Date May 2017 Marks available 7 Reference code 17M.2.hl.TZ0.4
Level HL only Paper 2 Time zone TZ0
Command term Prove that Question number 4 Adapted from N/A

Question

Consider the matrix

A =[λ3224λ373].

Suppose now that λ=1 so consider the matrix

B =[132241373].

Find an expression for det(A) in terms of λ, simplifying your answer.

[3]
a.i.

Hence show that A is singular when λ=1 and find the other value of λ for which A is singular.

[2]
a.ii.

Explain how it can be seen immediately that B is singular without calculating its determinant.

[1]
b.i.

Determine the null space of B.

[4]
b.ii.

Explain briefly how your results verify the rank-nullity theorem.

[[N/A]]
b.iii.

Prove, using mathematical induction, that

Bn=8n2B2 for nZ+, n.

[7]
c.

Markscheme

det(A) = \lambda (12 - 7\lambda ) + 3(3\lambda  - 6) + 2(14 - 12)     M1A1

= 12\lambda  - 7{\lambda ^2} + 9\lambda  - 18 + 4

=  - 7{\lambda ^2} + 21\lambda  - 14     A1

[??? marks]

a.i.

A is singular when \lambda  = 1 because the determinant is zero     R1

 

Note:     Do not award the R1 if the determinant has not been obtained.

 

the other value is 2     A1

[??? marks]

a.ii.

the third row is the sum of the first two rows     A1

[??? marks]

b.i.

the null space satisfies

\left[ {\begin{array}{*{20}{c}} 1&3&2 \\ 2&4&1 \\ 3&7&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right]     M1

x + 3y + 2z = 0

2x + 4y + z = 0     (A1)

3x + 7y + 3z = 0

the solution is (by GDC or otherwise)

\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 3} \\ 2 \end{array}} \right]\alpha where \alpha  \in \mathbb{R}     M1A1

[??? marks]

b.ii.

the rank-nullity theorem for square matrices states that

rank of matrix + dimension of null space = number of columns     A1

here, rank = 2, dimension of null space = 1 and number of columns = 3     A1

[??? marks]

b.iii.

first show that the result is true for n = 3

B^2 = \left[ {\begin{array}{*{20}{c}} {13}&{29}&{11} \\ {13}&{29}&{11} \\ {26}&{58}&{22} \end{array}} \right]     A1

B^3 = \left[ {\begin{array}{*{20}{c}} {104}&{232}&{88} \\ {104}&{232}&{88} \\ {208}&{464}&{176} \end{array}} \right]     A1

therefore B^3 = 8B^2 so true for n = 3     R1

assume the result is true for n = k, that is B^k = {8^{k - 2}}B^2     M1

consider B^{k + 1} = {8^{k - 2}}B^2     M1

= {8^{k - 2}}8B^2

= {8^{k - 1}}B^2     A1

therefore, true for n = k \Rightarrow true for n = k + 1 and since the result is true for n = 3, it is true for n \geqslant 3     R1

[7 marks]

c.

Examiners report

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

Syllabus sections

Topic 1 - Linear Algebra » 1.3

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