EXM.1.AHL.TZ0.6

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

Let C = $\left( {\begin{array}{*{20}{c}} { - 2}&4 \\ 1&7 \end{array}} \right)$ and D = $\left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$ .

The 2 × 2 matrix Q is such that 3Q = 2C – D

Find Q.

[3]

Find CD.

[4]

Find D^–1.

[2]

Markscheme / solution

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

3Q = $\left( {\begin{array}{*{20}{c}} { - 4}&8 \\ 2&{14} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$ (A1)

3Q = $\left( {\begin{array}{*{20}{c}} { - 9}&6 \\ 3&{14 - a} \end{array}} \right)$ (A1)

Q = $\left( {\begin{array}{*{20}{c}} { - 3}&2 \\ 1&{\frac{{14 - a}}{3}} \end{array}} \right)$ (A1) (N3)

[3 marks]

CD = $\left( {\begin{array}{*{20}{c}} { - 2}&4 \\ 1&7 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$

$= \left( {\begin{array}{*{20}{c}} { - 14}&{ - 4 + 4a} \\ { - 2}&{2 + 7a} \end{array}} \right)$ (A1)(A1)(A1)(A1) (N4)

[4 marks]

det D = 5 $a$ + 2 (may be implied) (A1)

D^–1 = $\frac{1}{{5a + 2}}\left( {\begin{array}{*{20}{c}} a&{ - 2} \\ 1&5 \end{array}} \right)$ (A1) (N2)

[2 marks]

Examiners’ report

[N/A]