EXM.2.AHL.TZ0.3

pestleMathematicsAIHLPaper 2EXM· ahl-1-14-introduction-to-matricessource ↗

Let $M = \left( {\begin{array}{*{20}{c}} a&2 \\ 2&{ - 1} \end{array}} \right)$ , where $a \in \mathbb{Z}$ .

Find ${M^2}$ in terms of $a$ .

[4]

If ${M^2}$ is equal to $\left( {\begin{array}{*{20}{c}} 5&{ - 4} \\ { - 4}&5 \end{array}} \right)$ , find the value of $a$ .

[2]

Using this value of $a$ , find ${M^{ - 1}}$ and hence solve the system of equations:

$- x + 2y = - 3$

$2x - y = 3$

[6]

Markscheme / solution

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

${M^2} = \left( {\begin{array}{*{20}{c}} a&2 \\ 2&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&2 \\ 2&{ - 1} \end{array}} \right)$

$= \left( {\begin{array}{*{20}{c}} {{a^2} + 4}&{2a - 2} \\ {2a - 2}&5 \end{array}} \right)$ (A1)(A1)(A1)(A1)

[4 marks]

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

$2a - 2 = - 4$

$\Rightarrow a = - 1$ (A1)

Substituting: ${a^2} + 4 = {\left( { - 1} \right)^2} + 4 = 5$ (A1)

Note: Candidates may solve ${a^2} + 4 = 5$ to give $a = \pm 1$ , and then show that only $a = - 1$ satisfies $2a - 2 = - 4$ .

[2 marks]

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

$M = \left( {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&{ - 1} \end{array}} \right)$

${M^{ - 1}} = - \frac{1}{3}\left( {\begin{array}{*{20}{c}} { - 1}&{ - 2} \\ { - 2}&{ - 1} \end{array}} \right)$ (M1)

$= \frac{1}{3}\left( {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right)$ or $\left( {\begin{array}{*{20}{c}} {\frac{1}{3}}&{\frac{2}{3}} \\ {\frac{2}{3}}&{\frac{1}{3}} \end{array}} \right)$ (A1)

$- x + 2y = - 3$

$2x - y = 3$

$\Rightarrow \left( {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 3} \\ 3 \end{array}} \right)$ (M1)(M1)

$\Rightarrow \left( {\begin{array}{*{20}{c}} {\frac{1}{3}}&{\frac{2}{3}} \\ {\frac{2}{3}}&{\frac{1}{3}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{3}}&{\frac{2}{3}} \\ {\frac{2}{3}}&{\frac{1}{3}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} { - 3} \\ 3 \end{array}} \right)$ (A1)

$\Rightarrow \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ (A1)

ie $x = 1$

$y = - 1$

Note: The solution must use matrices. Award no marks for solutions using other methods.

[6 marks]

Examiners’ report

[N/A]