EXM.1.AHL.TZ0.29

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

Let M = $\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right)$ where $a$ and $b$ are non-zero real numbers.

Show that M is non-singular.

[2]

Calculate M².

[2]

Show that det(M²) is positive.

[2]

Markscheme / solution

finding det M $= {a^2} + {b^2}$ A1

${a^2} + {b^2} > 0$ , therefore M is non-singular or equivalent statement R1

[2 marks]

M² = $\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a^2} - {b^2}}&{2ab} \\ { - 2ab}&{{a^2} - {b^2}} \end{array}} \right)$ M1A1

[2 marks]

EITHER

det(M²) $= \left( {{a^2} - {b^2}} \right)\left( {{a^2} - {b^2}} \right) + \left( {2ab} \right)\left( {2ab} \right)$ A1

det(M²) $= {\left( {{a^2} - {b^2}} \right)^2} + {\left( {2ab} \right)^2}$ $\left( { = {{\left( {{a^2} + {b^2}} \right)}^2}} \right)$

since the first term is non-negative and the second is positive R1

therefore det(M2) > 0

Note: Do not penalise first term stated as positive.

det(M²) = (det M)² A1

since det M is positive so too is det (M²) R1

[2 marks]

Examiners’ report

[N/A]