EXM.1.AHL.TZ0.50

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

Given that A = $\left( {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 2} \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&{ - 3} \end{array}} \right)$ , find X if BX = A – AB.

Markscheme / solution

METHOD 1

A – AB = $\left( {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 2} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&{ - 9} \\ 2&6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (M1)(A1)

X = B^–1(A – AB) = B^–1 $\left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (M1)

$= - \frac{1}{6}\left( {\begin{array}{*{20}{c}} { - 3}&0 \\ 0&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (A1)

$= \left( {\begin{array}{*{20}{c}} { - 1}&6 \\ {\frac{1}{3}}&{\frac{8}{3}} \end{array}} \right)$ (A2) (C6)

METHOD 2

Attempting to set up a matrix equation (M2)

X = B^–1(A – AB) (A2)

$= \left( {\begin{array}{*{20}{c}} { - 1}&6 \\ {\frac{1}{3}}&{\frac{8}{3}} \end{array}} \right)$ (from GDC) (A2) (C6)

[6 marks]

Examiners’ report

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