EXM.2.AHL.TZ0.11

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

Let M = $\left( {\begin{array}{*{20}{c}} 2&1 \\ 2&{ - 1} \end{array}} \right)$ .

Write down the determinant of M.

[1]

Write down M⁻¹.

[2]

Hence solve M $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ 8 \end{array}} \right)$ .

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

det M = −4 A1 N1

[1 mark]

M⁻¹ = $- \frac{1}{4}\left( {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\ { - 2}&2 \end{array}} \right)\,\,\,\left( { = \left( {\begin{array}{*{20}{c}} {\frac{1}{4}}&{\frac{1}{4}} \\ {\frac{1}{2}}&{ - \frac{1}{2}} \end{array}} \right)} \right)$ A1A1 N2

Note: Award A1 for $- \frac{1}{4}$ and A1 for the correct matrix.

[2 marks]

X = M⁻¹ $\left( {\begin{array}{*{20}{c}} 4 \\ 8 \end{array}} \right)$ $\left( {{\text{X}} = - \frac{1}{4}\left( {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\ { - 2}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 4 \\ 8 \end{array}} \right)} \right)$ M1

X = $\left( {\begin{array}{*{20}{c}} 3 \\ { - 2} \end{array}} \right)$ ( $x = 3$ , $y = - 2$ ) A1A1 N0

Note: Award no marks for an algebraic solution of the system $2x + y = 4$ , $2x - y = 8$ .

[3 marks]

Examiners’ report

[N/A]