EXM.1.AHL.TZ0.10

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

Let A = $\left( {\begin{array}{*{20}{c}} 3&2 \\ k&4 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&2 \\ 1&3 \end{array}} \right)$ . Find, in terms of $k$ ,

2A − B.

[3]

det (2A − B).

[3]

Markscheme / solution

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

2A = $\left( {\begin{array}{*{20}{c}} 6&4 \\ {2k}&8 \end{array}} \right)$ (A1)

2A − B = $\left( {\begin{array}{*{20}{c}} 4&2 \\ {2k - 1}&5 \end{array}} \right)$ A2 N3

[3 marks]

Evidence of using the definition of determinant (M1)

Correct substitution (A1)

eg 4(5) − 2(2 $k$ − 1), 20 − 2(2 $k$ − 1), 20 − 4 $k$ + 2

det (2A − B) = 22 − 4 $k$ A1 N3

[3 marks]

Examiners’ report

[N/A]