EXM.2.AHL.TZ0.9

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

Write down the inverse of the matrix A = $\left( {\begin{array}{*{20}{c}} 1&{ - 3}&0 \\ 2&0&1 \\ 4&1&3 \end{array}} \right)$ .

[2]

Hence or otherwise solve

$x - 3y = 1$

$2x + z = 2$

$4x + y + 3z = - 1$

[4]

Markscheme / solution

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

A⁻¹ = $\left( {\begin{array}{*{20}{c}} { - 0.2}&{1.8}&{ - 0.6} \\ { - 0.4}&{0.6}&{ - 0.2} \\ {0.4}&{ - 2.6}&{1.2} \end{array}} \right)$ A2 N2

[2 marks]

For recognizing that the equations may be written as A $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)$ (M1)

For attempting to calculate $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = {{\text{A}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)$ (M1)

$x$ = 4, $y$ = 1, $z$ = −6 A2 N4

[4 marks]

Examiners’ report

[N/A]