EXM.1.AHL.TZ0.33

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

Given that M = $\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right)$ and that M² – 6M + kI = 0 find k.

Markscheme / solution

M² = $\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7&{ - 6} \\ { - 18}&{19} \end{array}} \right)$ M1A1

$\Rightarrow \left( {\begin{array}{*{20}{c}} 7&{ - 6} \\ { - 18}&{19} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {12}&{ - 6} \\ { - 18}&{24} \end{array}} \right)$ + kI = 0 (M1)

$\Rightarrow \left( {\begin{array}{*{20}{c}} { - 5}&0 \\ 0&{ - 5} \end{array}} \right)$ + kI = 0 (A1)

⇒ k = 5 A1

[5 marks]

Examiners’ report

[N/A]