EXM.1.AHL.TZ0.43

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

Find the values of $a$ and $b$ given that the matrix $A = \left( {\begin{array}{*{20}{c}} a&{ - 4}&{ - 6} \\ { - 8}&5&7 \\ { - 5}&3&4 \end{array}} \right)$ is the inverse of the matrix $B = \left( {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ 3&b&1 \\ { - 1}&1&{ - 3} \end{array}} \right)$ .

[2]

For the values of $a$ and $b$ found in part (a), solve the system of linear equations

$\begin{array}{*{20}{c}} {x + 2y - 2z = 5} \\ {3x + by + z = 0} \\ { - x + y - 3z = a - 1.} \end{array}$

[2]

Markscheme / solution

AB = I

(AB)₁₁ = 1 ⇒ $a$ – 12 + 6 = 1, giving $a$ = 7 (A1) (C1)

(AB)₂₂ = 1 ⇒ –16 + 5 $b$ + 7 = 1, giving $b$ = 2 (A1) (C1)

[2 marks]

the system is $BX = \left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right)$ where $X = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ .

Then, $X = A\left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7&{ - 4}&{ - 6} \\ { - 8}&5&7 \\ { - 5}&3&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right).$ (M1)

Thus $x = - 1$ , $y = 2$ , $z = - 1$ (A1) (C2)

[2 marks]

Examiners’ report

[N/A]