EXM.1.AHL.TZ0.46

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

If A = $\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right)$ , find 2 values of $x$ and $y$ , given that AB = BA.

Markscheme / solution

AB = $\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2x + 32}&{xy + 16} \\ {24}&{4y + 8} \end{array}} \right)$ (A1)

BA = $\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2x + 4y}&{2y + 8} \\ {8x + 16}&{40} \end{array}} \right)$ (A1)

AB = BA ⇒ 8 $x$ + 16 = 24 and 4 $y$ + 8 = 40

This gives $x = 1$ and $y = 8$ . (A1) (C3)

[3 marks]

Examiners’ report

[N/A]