EXM.1.AHL.TZ0.8

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

Let A = $\left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&1&2 \\ 2&0&1 \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} {18} \\ {23} \\ {13} \end{array}} \right)$ , and X = $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ .

Consider the equation AX = B.

Write down the inverse matrix A⁻¹.

[2]

Express X in terms of A⁻¹ and B.

[1]

b.i.

Hence, solve for X.

[3]

b.ii.

Markscheme / solution

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

A⁻¹ = $\left( {\begin{array}{*{20}{c}} { - \frac{1}{3}}&{\frac{2}{3}}&{ - \frac{1}{3}} \\ { - \frac{1}{3}}&{\frac{5}{3}}&{ - \frac{7}{3}} \\ {\frac{2}{3}}&{ - \frac{4}{3}}&{\frac{5}{3}} \end{array}} \right)$ or $\left( {\begin{array}{*{20}{c}} { - 0.333}&{0.667}&{ - 0.333} \\ { - 0.333}&{1.67}&{ - 2.33} \\ {0.667}&{ - 1.33}&{1.67} \end{array}} \right)$ A2 N2

[2 marks]

X = A⁻¹B A1 N1

[1 mark]

b.i.

X = $\left( {\begin{array}{*{20}{c}} 5 \\ 2 \\ 3 \end{array}} \right)$ A3 N3

[3 marks]

b.ii.

Examiners’ report

[N/A]

b.i.

[N/A]

b.ii.