EXM.1.AHL.TZ0.35

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

Write down the inverse of the matrix

A = $\left( {\begin{array}{*{20}{c}} 1&{ - 3}&1 \\ 2&2&{ - 1} \\ 1&{ - 5}&3 \end{array}} \right)$

[2]

Hence, find the point of intersection of the three planes.

$\begin{array}{*{20}{c}} {x - 3y + z = 1} \\ {2x + 2y - z = 2} \\ {x - 5y + 3z = 3} \end{array}$

[3]

A fourth plane with equation $x + y + z = d$ passes through the point of intersection. Find the value of $d$ .

[1]

Markscheme / solution

A^–1 = $\left( {\begin{array}{*{20}{c}} {0.1}&{0.4}&{0.1} \\ { - 0.7}&{0.2}&{0.3} \\ { - 1.2}&{0.2}&{0.8} \end{array}} \right)$ A2 N2

[2 marks]

For attempting to calculate $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ = A⁻¹ $\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right)$ M1

$x = 1.2{\text{,}}\,\,y = 0.6{\text{,}}\,\,z = 1.6$ (so the point is (1.2, 0.6, 1.6)) A2 N2

[3 marks]

(1.2, 0.6, 1.6) lies on $x + y + z = d$

$\therefore d = 3.4$ A1 N1

[1 mark]

Examiners’ report

[N/A]