$\newcommand{\iixii}[4]{\left(\begin{array}{cc}#1\\#3\end{array}\right)}$

$\newcommand{\matrixa}[1]{\left(\begin{array}{cc}#1\end{array}\right)}$

(i) Given that $\D A= \iixii{4}{-3}{2}{5},$ find the inverse matrix $\D A^{-1}.$ [2]

(ii) Use your answer to part (i) to solve the simultaneous equations

$\D 4x - 3y = -10,$

$\D 2x + 5y = 21. [2]$

2 (CIE 2012, s, paper 22, question 4)

In a competition the contestants search for hidden targets which are classed as difficult, medium

or easy. In the first round, finding a difficult target scores 5 points, a medium target 3 points and

an easy target 1 point. The number of targets found by the two contestants, Claire and Denise, are

shown in the table.

Target Difficult Medium Easy

Contestant

Claire 4 1 7

Denise 2 5 1

In the second round, finding a difficult target scores 8 points, a medium target 4 points and an

easy target 2 points. In the second round Claire finds 2 difficult, 5 medium and 2 easy targets

whilst Denise finds 4 difficult, 3 medium and 6 easy targets.

(i) Write down the sum of two matrix products which, on evaluation, would give the total score

for each contestant. [3]

(ii) Use matrix multiplication and addition to calculate the total score for each contestant. [2]

3 (CIE 2012, w, paper 12, question 2)

(i) Find the inverse of the matrix

$\D \iixii{2}{-1}{-1}{1.5}.$ [2]

(ii) Hence find the matrix A such that $\D \iixii{2}{-1}{-1}{1.5}A=\iixii{1}{6}{-0.5}{4}.$ [3]

4 (CIE 2012, w, paper 21, question 5)

It is given that $\D A=\matrixa{ 4& -2\\

8& -3}, B = \matrixa{2& 0& 4\\

5& -1& 4}$ and $\D C = \matrixa{5\\-2\\3}.$

(i) Calculate $\D ABC.$ [4]

(ii) Calculate $\D A^{-1} B.$ [4]

5 (CIE 2012, w, paper 23, question 2)

(i) Given that $\D A = \iixii{7}{8}{4}{6}$, find the inverse matrix, $\D A^{-1}.$ [2]

(ii) Use your answer to part (i) to solve the simultaneous equations

$\D 7x + 8y = 39,$

$\D 4x + 6y = 23.$ [2]

6 (CIE 2013, s, paper 11, question 6)

(i) Given that $\D A=\iixii{2}{-1}{3}{5}$, find $\D A^{-1}.$ [2]

(ii) Using your answer from part (i), or otherwise, find the values of $\D a, b, c$ and $\D d$ such that

\[A\iixii{a}{b}{c}{-1}=\iixii{7}{5}{17}{d}.\] [5]

7 (CIE 2013, s, paper 12, question 8)

(a) Given that the matrix $\D A =\iixii{4}{2}{3}{-5},$

find

(i) $\D A^2,$ [2]

(ii) $\D 3A + 4I,$ where $\D I$ is the identity matrix. [2]

(b) Find the inverse matrix of $\D \iixii{6}{1}{-9}{3}.$

Hence solve the equations

$\D 6x+y=5$

$\D -9x+3y=\frac{3}{2}.$

8 (CIE 2013, w, paper 11, question 11)

(a) It is given that the matrix

$\D A=\iixii{2}{3}{4}{1}.$

(i) Find $\D A + 2I.$ [1]

(ii) Find $\D A^2.$ [2]

(iii) Using your answer to part (ii) find the matrix $\D B$ such that $\D A^2B = I.$ [2]

(b) Given that the matrix $\D C=\iixii{x}{-1}{x^2-x+1}{x-1},$

show that $\D \det C\not=0.$ [4]

9 (CIE 2013, w, paper 13, question 7)

It is given that $\D A=\iixii{2t}{2}{t^2-t+1}{t}.$

(i) Find the value of $\D t$ for which $\D \det A = 1.$ [3]

(ii) In the case when $\D t = 3,$ find $\D A^{-1}$ and hence solve

$\D 3x + y = 5,$

$\D 7x + 3y = 11.$ [5]

10 (CIE 2014, s, paper 11, question 6)

Matrices $\D A$ and $\D B$ are such that $\D A =\matrixa{-1&4\\7&6\\4&2}.$

and $\D B =\iixii{2}{1}{3}{5}.$

(i) Find $\D AB.$ [2]

(ii) Find $\D B^{-1} .$ [2]

(iii) Using your answer to part (ii), solve the simultaneous equations

$\D 4x+2y=-3$

$\D 6x+10y=-22.$ [3]

11 (CIE 2014, s, paper 12, question 6)

(a) Matrices $\D X, Y$ and $\D Z$ are such that $\D X=\iixii{2}{3}{1}{2}, Y=\matrixa{1&3\\4&5\\6&7}$ and $\D Z=\matrixa{1&2&3}.$

Write down all the matrix products which are possible using any two of these matrices. Do not

evaluate these products. [2]

(b) Matrices $\D A$ and $\D B$ are such that $\D A=\iixii{5}{-2}{-4}{1}$ and $\D AB= \iixii{3}{9}{-6}{-3}.$ Find the matrix $\D B.$ [5]

12 (CIE 2014, s, paper 23, question 3)

In a motor racing competition, the winning driver in each race scores 5 points, the second and third

placed drivers score 3 and 1 points respectively. Each team has two members. The results of the drivers in one team, over a number of races, are shown in the table below.

Driver 1st place 2nd place 3rd place

Alan 3 1 4

Brian 1 4 0

(i) Write down two matrices whose product under matrix multiplication will give the number of

points scored by each of the drivers. Hence calculate the number of points scored by Alan and by

Brian. [3]

(ii) The points scored by Alan and by Brian are added to give the number of points scored by the team. Using your answer to part (i), write down two matrices whose product would give the number of points scored by the team. [1]

13 (CIE 2014, w, paper 11, question 7)

Matrices $\D A$ and $\D B$ are such that $\D \iixii{3a}{2b}{-a}{b}$ and $\D B=\iixii{-a}{b}{2a}{2b},$ where $\D a$ and $\D b$ are non-zero constants.

(i) Find $\D A^{-1}.$ [2]

(ii) Using your answer to part (i), find the matrix $\D X$ such that $\D XA = B.$ [4]

14 (CIE 2014, w, paper 13, question 5)

(a) A drinks machine sells coffee, tea and cola. Coffee costs \$0.50, tea costs \$0.40 and cola costs

\$0.45. The table below shows the numbers of drinks sold over a 4-day period.

Coffee Tea Cola

Tuesday 12 2 1

Wednesday 9 3 0

Thursday 8 5 1

Friday 11 2 0

(i) Write down 2 matrices whose product will give the amount of money the drinks machine

took each day and evaluate this product. [4]

(ii) Hence write down the total amount of money taken by the machine for this 4-day period. [1]

(b) Matrices $\D X$ and $\D Y$ are such that $\D X = \iixii{2}{4}{-5}{1}$

and $\D XY = I,$ where $\D I$ is the identity matrix. Find the matrix $\D Y.$ [3]

### Answers

$\newcommand{\dfrac}[2]{\displaystyle\frac{#1}{#2}}$1.(i) $\D \frac{1}{26}\iixii{5}{3}{-2}{4}$

(ii) $\D x=0.5,y=4$

2(i) $\D \matrixa{4&1&7\\2&5&1} \matrixa{5\\3\\1} +\matrixa{2&5&2\\4&3&6}\matrixa{8\\4\\1}$

(ii) Claire=70, Denise=82

3(i) $\D \frac{1}{2}\matrixa{1.5&1\\1&2}$

(ii) $\D \matrixa{.5&6.5\\0&7}$

4(i) $\D \matrixa{10\\59}$

(ii) $\D \matrixa{1&-0.5&-1\\1&-1&-4}$

5(i) $\D \frac{1}{10}\matrixa{6&-8\\-4&7}$

(ii) $\D x=5,y=0.5$

6(i) $\D A^{-1}\dfrac{1}{13}\matrixa{5&1\\-3&2}$

(ii) $ \D a=4,b=2,c=-1,d=1 $

7 (a)(i) $\D \matrixa{22&-2\\-3&31}$

(ii) $\D \matrixa{16&6\\9&-11}$

(b)(i) $\D \dfrac{1}{27}\matrixa{3&-1\\9&6}$

(ii) $\D x=0.5,y=2$

8(a)(i)$\D \matrixa{4&3\\4&3}$

(ii) $\D A^2=\matrixa{16&9\\12&13}$

(iii) $\D \dfrac{1}{100}\matrixa{13&-9\\-12&16}$

9(i) $\D t=\dfrac{3}{2}$

(ii) $\D A=\matrixa{6&2\\7&3}, A^{-1}=\dfrac{1}{4}\matrixa{3&-2\\-7&6}$

$\D x=2,y=-1$

10 (i) $\D \matrixa{10&19\\32&37\\14&14}$

(ii) $\D B^{-1}=\dfrac{1}{7}\matrixa{5&-1\\-3&2}$

$\D x=0.5,y=-2.5$

11(a) $\D YX,ZY$

(b) $\D \matrixa{3&-1\\6&-7}$

12(i) $\D \matrixa{3&1&4\\1&3&0},\matrixa{5\\3\\1}, \matrixa{22\\17}$

(ii) $\D (1,1),\matrixa{22\\17}$

13(i) $\D A^{-1} =\dfrac{1}{5ab}\matrixa{b&-2b\\ a&3a}$

(ii) $\D X=\matrixa{0&1\\4/5&2/5}$

14(a)(i) $\D \matrixa{0.5&0.4&0.45} \matrixa{12&9&8&11\\ 2&3&5&2\\ 1&0&1&0} =\matrixa{7.25&5.70&6.45&6.30}$

(ii) 25.70

(b) $\D \dfrac{1}{22}\matrixa{1&-4\\5&2}$

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