EXM.1.AHL.TZ0.26

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

Consider the matrices

A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} 1&3 \\ 2&{ - 2} \end{array}} \right)$ .

Find BA.

[2]

Calculate det (BA).

[2]

Find A(A^–1B + 2A^–1)A.

[3]

Markscheme / solution

BA = $\left( {\left( {\begin{array}{*{20}{c}} 1&3 \\ 2&{ - 2} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} {18}&{ - 14} \\ { - 4}&4 \end{array}} \right)$ A2

Note: Award A1 for one error, A0 for two or more errors.

[2 marks]

det(BA) = (72 – 56) = 16 (M1)A1

[2 marks]

EITHER

A(A^–1B + 2A^–1)A = BA + 2A (M1)A1

$= \left( {\begin{array}{*{20}{c}} {24}&{ - 18} \\ 6&{ - 4} \end{array}} \right)$ A1

A^–1 $= - \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 4}&2 \\ { - 5}&3 \end{array}} \right)$ (A1)

an attempt to evaluate (M1)

A^–1B + 2A^–1 $= - \frac{1}{2}\left( {\begin{array}{*{20}{c}} 0&{ - 16} \\ 1&{ - 21} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 4}&2 \\ { - 5}&3 \end{array}} \right)$

A(A^–1B + 2A^–1)A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 4&6 \\ {4.5}&{7.5} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)$

$= \left( {\begin{array}{*{20}{c}} 3&3 \\ 2&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {24}&{ - 18} \\ 6&{ - 4} \end{array}} \right)$ A1

[3 marks]

Examiners’ report

[N/A]