EXM.1.AHL.TZ0.3

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

$A$ and $B$ are 2 × 2 matrices, where $A = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 2&0 \end{array}} \right]$ and $BA = \left[ {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right]$ . Find $B$

Markscheme / solution

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

$B = \left( {BA} \right){A^{ - 1}}$ (M1)

$= - \frac{1}{4}\left( {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&{ - 2} \\ { - 2}&5 \end{array}} \right)$ (M1)

$= - \frac{1}{4}\left( {\begin{array}{*{20}{c}} { - 4}&{ - 12} \\ { - 16}&{ - 48} \end{array}} \right)$ (A1)

$= \left( {\begin{array}{*{20}{c}} 1&3 \\ 4&{12} \end{array}} \right)$ (A1)

$\left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5&2 \\ 2&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right)$ (M1)

$\left. { \Rightarrow \begin{array}{*{20}{c}} {5a + 2b = 11} \\ {2a\,\,\,\,\,\,\,\,\,\,\,\,\, = 2} \end{array}} \right\}$

$\Rightarrow a = 1$ , $b = 3$ (A1)

$\left. {\begin{array}{*{20}{c}} {5c + 2d = 44} \\ {2c{\text{ }} = 8} \end{array}} \right\}$

$\Rightarrow c = 4$ , $d = 12$ (A1)

$B = \left( {\begin{array}{*{20}{c}} 1&3 \\ 4&{12} \end{array}} \right)$ (A1) (C4)

Note: Correct solution with inversion (ie AB instead of BA) earns FT marks, (maximum [3 marks]).

[4 marks]

Examiners’ report

[N/A]