EXM.1.AHL.TZ0.30

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

Matrices A, B and C are defined as

A = $\left( {\begin{array}{*{20}{c}} 1&5&1 \\ 3&{ - 1}&3 \\ { - 9}&3&7 \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)$ , C = $\left( {\begin{array}{*{20}{c}} 8 \\ 0 \\ { - 4} \end{array}} \right)$ .

Given that AB = $\left( {\begin{array}{*{20}{c}} a&0&0 \\ 0&a&0 \\ 0&0&a \end{array}} \right)$ , find $a$ .

[1]

Hence, or otherwise, find A^–1.

[2]

Find the matrix X, such that AX = C.

[2]

Markscheme / solution

$a = 16$ A1

[1 mark]

A^–1 = $\frac{1}{{16}}\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)$ (M1)A1

[2 marks]

AX = C ⇒ X = A^–1C (M1)

$= \frac{1}{{16}}\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 8 \\ 0 \\ { - 4} \end{array}} \right)$

$= \frac{1}{{16}}\left( {\begin{array}{*{20}{c}} {12} \\ {24} \\ { - 4} \end{array}} \right)\,\,\,\left( { = \left( {\begin{array}{*{20}{c}} {0.75} \\ {1.5} \\ { - 0.25} \end{array}} \right)} \right)$ A1

[2 marks]

Examiners’ report

[N/A]