EXM.2.AHL.TZ0.21

pestleMathematicsAIHLPaper 2EXM· ahl-1-14-introduction-to-matricessource ↗

Let A = $\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 1&1&0 \\ 1&1&1 \end{array}} \right)$ .

Given that X = B – A^–1 and Y = B^–1 – A,

You are told that ${A^n} = \left( {\begin{array}{*{20}{c}} 1&n&{\frac{{n\left( {n + 1} \right)}}{2}} \\ 0&1&n \\ 0&0&1 \end{array}} \right)$ , for $n \in {\mathbb{Z}^ + }$ .

Given that ${\left( {{A^n}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}} 1&x&y \\ 0&1&x \\ 0&0&1 \end{array}} \right)$ , for $n \in {\mathbb{Z}^ + }$ ,

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 ${A^n} + {\left( {{A^n}} \right)^{ - 1}}$ .

[1]

b.ii.

Markscheme / solution

X = B – A^–1 = $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 1&1&0 \\ 1&1&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\ 0&1&{ - 1} \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&1&0 \\ 1&0&1 \\ 1&1&0 \end{array}} \right)$ A1

Y = B^–1 – A = $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ { - 1}&1&0 \\ 1&{ - 1}&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{ - 1}&{ - 1} \\ { - 1}&0&{ - 1} \\ 0&{ - 1}&0 \end{array}} \right)$ A1

[2 marks]

a.i.

X^–1 + Y^–1 = $\left( {\begin{array}{*{20}{c}} 0&{ - 1}&0 \\ 1&0&{ - 1} \\ 0&1&0 \end{array}} \right)$ (A1)

X^–1 + Y^–1 has no inverse A1

as det(X^–1 + Y^–1) = 0 R1

[3 marks]

a.ii.

${A^n}{\left( {{A^n}} \right)^{ - 1}} = I \Rightarrow \left( {\begin{array}{*{20}{c}} 1&n&{\frac{{n\left( {n + 1} \right)}}{2}} \\ 0&1&n \\ 0&0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&x&y \\ 0&1&x \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$ M1

$\Rightarrow \left( {\begin{array}{*{20}{c}} 1&{x + n}&{y + nx + \frac{{n\left( {n + 1} \right)}}{2}} \\ 0&1&{x + n} \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$ A1

solve simultaneous equations to obtain

$x + n = 0$ and $y + nx + \frac{{n\left( {n + 1} \right)}}{2} = 0$ M1

$x = - n$ and $y = \frac{{n\left( {n - 1} \right)}}{2}$ A1A1N2

[5 marks]

b.i.

${A^n} + {\left( {{A^n}} \right)^{ - 1}} = \, \Rightarrow \left( {\begin{array}{*{20}{c}} 1&n&{\frac{{n\left( {n + 1} \right)}}{2}} \\ 0&1&n \\ 0&0&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&{ - n}&{\frac{{n\left( {n - 1} \right)}}{2}} \\ 0&1&{ - n} \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2&0&{{n^2}} \\ 0&2&0 \\ 0&0&2 \end{array}} \right)$ A1

[1 mark]

b.ii.

Examiners’ report

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