EXM.2.AHL.TZ0.4

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The matrices A, B, X are given by

A = $\left( {\begin{array}{*{20}{c}} 3&1 \\ { - 5}&6 \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$ , X = $\left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)$ , ${\text{where}}$ $a$ , $b$ , $c$ , $d \in \mathbb{Q}$ .

Given that AX + X = Β, find the exact values of $a$ , $b$ , $c$ and $d$ .

Markscheme / solution

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

$\left( {\begin{array}{*{20}{c}} 3&1 \\ { - 5}&6 \end{array}} \right)$ X + $\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)$ X = $\left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} 4&1 \\ { - 5}&7 \end{array}} \right)$ X = $\left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$ (M1)

Pre-multiply by inverse of $\left( {\begin{array}{*{20}{c}} 4&1 \\ { - 5}&7 \end{array}} \right)$ (M1)

X = $\frac{1}{{33}}\left( {\begin{array}{*{20}{c}} 7&{ - 1} \\ 5&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$ (A1)(A1)

Note: Award (A1) for determinant, (A1) for matrix $\left( {\begin{array}{*{20}{c}} 7&{ - 1} \\ 5&4 \end{array}} \right)$ .

$= \frac{1}{{33}}\left( {\begin{array}{*{20}{c}} {28}&{59} \\ {20}&{28} \end{array}} \right)$ (A1)(A1)(A1)(A1)

$\left( { \Rightarrow a = \frac{{28}}{{33}}{\text{,}}\,\,b = \frac{{59}}{{33}}{\text{,}}\,\,c = \frac{{20}}{{33}}{\text{,}}\,\,d = \frac{{28}}{{33}}} \right)$

$\left( {\begin{array}{*{20}{c}} 3&1 \\ { - 5}&6 \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right) + \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$ (A1)

$\left( {\begin{array}{*{20}{c}} {3a + c}&{3b + d} \\ { - 5a + 6c}&{ - 5b + 6d} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4&8 \\ 0&{ - 3} \end{array}} \right)$ (A1)

$4a + c = 4$

$- 5a + 7c = 0$ (A1)

$4b + d = 8$

$- 5b + 7d = - 3$ (A1)

Notes: Award (A1) for each pair of equations.
Allow ft from their equations.

${a = \frac{{28}}{{33}}{\text{,}}\,\,b = \frac{{59}}{{33}}{\text{,}}\,\,c = \frac{{20}}{{33}}{\text{,}}\,\,d = \frac{{28}}{{33}}}$ (A1)(A1)(A1)(A1)

Note: Award (A0)(A0)(A1)(A1) if the final answers are given as decimals ie 0.848, 1.79, 0.606, 0.848.

[8 marks]

Examiners’ report

[N/A]