Question 1
[2 marks]A plotter uses x'=2x+y+4, y'=3y-1.
Find the image of (3,-2).
[2]
Show solutionHide solution
x'=6-2+4 and y'=-6-1.
M1
The image is (8,-7).
A1
Final answer
(8,\,-7).
Free worked solutions · 2026 Edition
IB Maths AI HL Chapter 11: Matrix transformations worked solutions
Worked solutions to the practice questions in Chapter 11 of Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.
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Basics
One idea at a time, in the order the chapter teaches them.
Question 2
[2 marks]A coordinate rule is (x,y)\mapsto(2x-y,4x+3y).
Write its matrix and calculate the image of (2,-1).
[2]
Show solutionHide solution
The matrix is \begin{pmatrix}2&-1\\4&3\end{pmatrix}.
A1
The image is (5,5).
A1
Final answer
\begin{pmatrix}2&-1\\4&3\end{pmatrix}; image (5,\,5).
Question 3
[2 marks]A badge is enlarged about the origin by scale factor −2.
Write its matrix and find the image of (5,-3).
[2]
Show solutionHide solution
The matrix is \begin{pmatrix}-2&0\\0&-2\end{pmatrix}.
A1
The image is (-10,6).
A1
Final answer
\begin{pmatrix}-2&0\\0&-2\end{pmatrix}; image (-10,\,6).
Question 4
[3 marks]A point P is rotated 90^\circ anticlockwise about the origin. Its image is (-9,2).
Find P and its image after a further 90^\circ anticlockwise rotation.
[3]
Show solutionHide solution
(-y,x)=(-9,2) gives P=(2,9). M1A1
The further image is (-2,-9).
A1
Final answer
P(2,\,9); (-2,\,-9).
Question 5
[4 marks]A rotation about the origin sends (4,0) to (2\sqrt3,-2). The rotation angle is between -180^\circ and 180^\circ, with anticlockwise positive.
Determine the signed angle and use it to find the image of (0,7).
[4]
Show solutionHide solution
\cos\theta=\sqrt3/2 and \sin\theta=-1/2, so \theta=-30^\circ. M1A1
The image is (7/2,7\sqrt3/2). M1A1
Final answer
-30^\circ (30^\circ clockwise); image \left(\frac72,\,\frac{7\sqrt3}{2}\right)=(3.5,\,6.06).
Question 6
[3 marks]A map has matrix D=\begin{pmatrix}4&0\\0&1\end{pmatrix}.
Which of (0,6), (2,0) and (-1,3) are unchanged by the map? Show the images.
[3]
Show solutionHide solution
The images are (0,6), (8,0) and (-4,3). M1A1
Only (0,6) is unchanged.
R1
Final answer
Images (0,\,6), (8,\,0), (-4,\,3); only (0,\,6) is unchanged.
Question 7
[3 marks]A map first translates by (2,-5), then rotates 90^\circ anticlockwise about the origin.
Express the result as \mathbf x\prime=M\mathbf x+\mathbf t.
[3]
Show solutionHide solution
R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}; R(\begin{pmatrix}2\\-5\end{pmatrix})=\begin{pmatrix}5\\2\end{pmatrix}. M1A1
\mathbf x\,\prime=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\mathbf x+\begin{pmatrix}5\\2\end{pmatrix}.
A1
Final answer
\mathbf x'=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\mathbf x+\begin{pmatrix}5\\2\end{pmatrix}.
Question 8
[4 marks]A screen map is x'=3x+y+2, y'=2y-6. A displayed point is (19,4).
Recover the original point.
[4]
Show solutionHide solution
2y-6=4 gives y=5. M1A1
3x+5+2=19 gives x=4; original point (4,5). M1A1
Final answer
(4,\,5).
Question 9
[3 marks]A polygon of area 18\text{ cm}^2 undergoes M=\begin{pmatrix}2&-1\\1&3\end{pmatrix}.
Calculate the image area.
[3]
Show solutionHide solution
\det M=6+1=7.
M1
Area multiplier |7| gives 126\text{ cm}^2. M1A1
Final answer
126\ \text{cm}^2.
Question 10
[3 marks]A cut-out of area 710\text{ mm}^2 is repeatedly transformed by M=\begin{pmatrix}0.7&0\\0&0.6\end{pmatrix}.
Find the least number of applications for the area to become less than 17\text{ mm}^2.
[3]
Show solutionHide solution
After n applications, area is 710(0.42)^n.
M1
710(0.42)^4=22.0930416 and 710(0.42)^5=9.279077472.
M1
The least number is 5.
A1
Final answer
5 applications.
Tutorial
Exam-style questions on each question type, with the marks the IB would give.
Question 11
[6 marks]A floor-plan scanner converts a position \mathbf{r} on a photographed plan into a position on the real floor using an affine map \mathbf{r}' = M\mathbf{r} + \mathbf{t}. A technician finds that the map sends the origin O to (3, -2), the point (2, 0) to (7, 0) and the point (0, 3) to (0, 1).
(a) Write down the vector \mathbf{t}.
[1]
(b) Find the matrix M.
[3]
(c) Find the image of the point (6, -2).
[2]
Show solutionHide solution
(a) The image of the origin is M\mathbf{0} + \mathbf{t} = \mathbf{t}, so \mathbf{t} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}.
[1]
(b) Image of (2,0) is 2\times\text{column 1} + \mathbf{t}, so 2\times\text{column 1} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} and column 1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}.
[2]
Image of (0,3) is 3\times\text{column 2} + \mathbf{t}, so 3\times\text{column 2} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \end{pmatrix}, column 2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix} and M = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}.
[1]
(c) M\begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 14 \\ 4 \end{pmatrix}, then + \begin{pmatrix} 3 \\ -2 \end{pmatrix} gives \begin{pmatrix} 17 \\ 2 \end{pmatrix}.
[1]
The image is (17,\,2).
[1]
Final answer
- \mathbf t=\begin{pmatrix}3\\-2\end{pmatrix}; M=\begin{pmatrix}2&-1\\1&1\end{pmatrix}
- image (17,\,2).
Question 12
[6 marks]In Sakura's workshop a robotic arm has its base at the origin O. The gripper is at G(3.5, -2) (metres). The arm turns through \dfrac{5\pi}{12} radians clockwise about O.
(a) Write down the rotation matrix for this turn, with each entry to 3 significant figures.
[2]
(b) Find the new position of the gripper.
[2]
(c) Sakura's GDC was left in degree mode and she entered the angle as the number \dfrac{5\pi}{12} \approx 1.309. Find the position she would obtain, and comment.
[2]
Show solutionHide solution
(a) Radian mode. Clockwise means \theta = -\dfrac{5\pi}{12}: \cos\theta = 0.259, \sin\theta = -0.966, so R = \begin{pmatrix} 0.259 & 0.966 \\ -0.966 & 0.259 \end{pmatrix}.
[2]
(b) R\begin{pmatrix} 3.500 \\ -2 \end{pmatrix} = \begin{pmatrix} -1.0260 \\ -3.8984 \end{pmatrix}, so the gripper moves to (-1.03,\,-3.90).
[2]
(c) In degree mode the angle is -1.309^\circ, so R = \begin{pmatrix} 0.99974 & 0.022844 \\ -0.022844 & 0.99974 \end{pmatrix} and the position is (3.45,\,-2.08), which is almost unchanged from G.
[1]
The mode must match the angle: 5\pi/12 rad = 75^\circ, not 1.309^\circ, so the arm would appear hardly to have turned.
[1]
Final answer
(a) \begin{pmatrix}0.259&0.966\\-0.966&0.259\end{pmatrix}. (b) (-1.03,\,-3.90). (c) (3.45,\,-2.08): the arm has hardly turned.
Question 13
[7 marks]A drawing program stores a mirror tool as the matrix B = \begin{pmatrix} 0.28 & 0.96 \\ 0.96 & -0.28 \end{pmatrix}, which acts on points relative to the origin O of the canvas.
(a) Find \det B and hence explain why B represents a reflection and not a rotation.
[2]
(b) The mirror line is y = ax where a > 0. Find the value of a.
[3]
(c) Show that the point (4, 3) is mapped to itself, and explain why this is expected.
[2]
Show solutionHide solution
(a) \det B = (0.28)(-0.28) - (0.96)(0.96) = -0.0784 - 0.9216 = -1.
[1]
A rotation has determinant +1; −1 means orientation is reversed, so B is a reflection.
[1]
(b) Compare with \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}: \cos 2\alpha = 0.28 and \sin 2\alpha = 0.96.
[1]
2\alpha = \cos^{-1}0.28 = 73.740^\circ, so \alpha = 36.870^\circ.
[1]
a = \tan\alpha = 0.7500 = 0.75.
[1]
(c) B\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}, so (4, 3) is unchanged.
[1]
(4, 3) has gradient \frac34 = a, so it lies on the mirror line, and points on a mirror do not move.
[1]
Final answer
(a) \det B=-1, a reflection. (b) a=0.75. (c) (4,\,3) lies on the mirror line.
Question 14
[8 marks]A planner draws a solar-panel array as the rectangle PQRS with P(2, 1), Q(10, 1), R(10, 5) and S(2, 5) (units of 10 m) on a site map. To match a satellite photograph she applies the map \mathbf{r}' = \begin{pmatrix} 1.2 & 0 \\ 0 & 0.8 \end{pmatrix}\mathbf{r} + \begin{pmatrix} -3 \\ 10 \end{pmatrix}.
(a) Find the coordinates of P', Q', R' and S'.
[3]
(b) Find the lengths P'Q' and Q'R' and hence the area of the image rectangle.
[3]
(c) Use the determinant of the matrix to confirm your area, and explain why the translation does not affect it.
[2]
Show solutionHide solution
(a) Stretch first, then translate: P' = \begin{pmatrix} 1.2(2) \\ 0.8(1) \end{pmatrix} + \begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -0.600 \\ 10.8 \end{pmatrix}.
[2]
Q' = (9,\,10.8), R' = (9,\,14), S' = (-0.600,\,14).
[1]
(Check: P' = (-0.6, 10.8), Q' = (9, 10.8), R' = (9, 14), S' = (-0.6, 14).)
(b) P'Q' = 9 - (-0.6) = 9.6 and Q'R' = 14 - 10.8 = 3.2 (units of 10 m).
[2]
Area = 9.6 \times 3.2 = 30.72 (units of 100\ \text{m}^2), i.e. 3072\ \text{m}^2.
[1]
(c) \det = 1.2 \times 0.8 = 0.96 and the original area is 8 \times 4 = 32, so the image area is 0.96 \times 32 = 30.72, which agrees.
[1]
A translation slides every point the same way, so it cannot change any length or area; only the matrix does.
[1]
Final answer
(a) P'(-0.6,\,10.8), Q'(9,\,10.8), R'(9,\,14), S'(-0.6,\,14). (b) 9.6\times3.2=30.72 (3072 m^2). (c) \det=0.96, 0.96\times32=30.72.
Question 15
[7 marks]A revolving stage rotates through 120^\circ anticlockwise about its pivot C(4, -2) on a plan (metres), so C does not move.
(a) Write the rotation as \mathbf{r}' = R\mathbf{r} + \mathbf{t}, giving \mathbf{t} to 3 significant figures.
[3]
(b) A performer stands at P(10, -2). Find where the performer ends up.
[2]
(c) Show that the performer is the same distance from the pivot before and after the turn.
[2]
Show solutionHide solution
(a) \mathbf{r}' = R(\mathbf{r} - \mathbf{c}) + \mathbf{c} gives \mathbf{t} = \mathbf{c} - R\mathbf{c} with R = \begin{pmatrix} - \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} \end{pmatrix}.
[1]
R\mathbf{c} = \begin{pmatrix} -0.26795 \\ 4.4641 \end{pmatrix}.
[1]
\mathbf{t} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} - R\mathbf{c} = \begin{pmatrix} 4.27 \\ -6.46 \end{pmatrix}.
[1]
(b) R\begin{pmatrix} 10 \\ -2 \end{pmatrix} + \mathbf{t} = \begin{pmatrix} 1 \\ 3.20 \end{pmatrix}; the performer ends at (1,\,3.20).
[2]
(Check: \mathbf{p} - \mathbf{c} = (6, 0) turns to (-3, 3\sqrt3), and adding \mathbf{c} gives (1, 3.196).)
(c) Before: |CP| = \sqrt{6^2 + 0^2} = 6 m.
[1]
After: |CP'| = \sqrt{(1-4)^2 + (3.196+2)^2} = \sqrt{9 + 27} = 6 m, the same (a rotation about C preserves distances from C).
[1]
Final answer
(a) \mathbf t=\begin{pmatrix}4.27\\-6.46\end{pmatrix}. (b) (1,\,3.20). (c) both distances are 6 m.
Question 16
[8 marks]A laser beam travels through a lighting rig. The first mirror lies along the line y = x\tan 30^\circ and the second along the x-axis; both pass through the origin O. The beam's position is reflected first in mirror 1 and then in mirror 2.
(a) Write down the matrices M_1 and M_2 of the two reflections, to 3 significant figures.
[2]
(b) Find the single matrix for the two reflections in this order.
[2]
(c) Show that the combined transformation is a rotation and find its angle and direction.
[3]
(d) Write down how the rotation angle is related to the angle between the two mirrors.
[1]
Show solutionHide solution
(a) M_1 = \begin{pmatrix} 0.500 & 0.866 \\ 0.866 & -0.500 \end{pmatrix} (using 2\alpha = 60^\circ) [1] and M_2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
[1]
(b) Mirror 1 acts first, so the combined matrix is M_2M_1 = \begin{pmatrix} 0.500 & 0.866 \\ -0.866 & 0.500 \end{pmatrix}.
[2]
(c) \det(M_2M_1) = (-1)(-1) = 1 (also 0.5^2 + 0.866^2 = 1), so it preserves orientation and has the form of a rotation.
[1]
\cos\theta = 0.5 and \sin\theta = -0.866 give \theta = -60^\circ, so it is a rotation through 60^\circ clockwise about O.
[2]
(d) The rotation angle is twice the angle between the mirrors (2 \times 30^\circ = 60^\circ). Turning from mirror 1 towards mirror 2 here is a clockwise turn, so the rotation is clockwise.
[1]
Final answer
(a) M_1=\begin{pmatrix}0.5&0.866\\0.866&-0.5\end{pmatrix}, M_2=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. (b) \begin{pmatrix}0.5&0.866\\-0.866&0.5\end{pmatrix}. (c) rotation of 60^\circ clockwise about O. (d) twice the angle between the mirrors.
Question 17
[7 marks]A drone camera converts a ground position \mathbf{r} (metres) into a pixel position \mathbf{r}' using \mathbf{r}' = M\mathbf{r} + \mathbf{t}, where M = \begin{pmatrix} 3 & -1 \\ 2 & 2 \end{pmatrix}, \qquad \mathbf{t} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}.
(a) Find M^{-1}.
[2]
(b) A target appears at the pixel position (23, 5). Find its ground position.
[3]
(c) A field has area 36\ \text{m}^2 on the ground. Find the area of its image in pixel units.
[2]
Show solutionHide solution
(a) \det M = (3)(2) - (-1)(2) = 8.
[1]
M^{-1} = \dfrac18\begin{pmatrix} 2 & 1 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & \frac{1}{8} \\ - \frac{1}{4} & \frac{3}{8} \end{pmatrix}.
[1]
(b) From \mathbf{r}' = M\mathbf{r} + \mathbf{t}: \mathbf{r} = M^{-1}(\mathbf{r}' - \mathbf{t}).
[1]
\mathbf{r}' - \mathbf{t} = \begin{pmatrix} 18 \\ 8 \end{pmatrix} and M^{-1}\begin{pmatrix} 18 \\ 8 \end{pmatrix} = \begin{pmatrix} \frac{11}{2} \\ - \frac{3}{2} \end{pmatrix}.
[1]
The target is at (\frac{11}{2},\,- \frac{3}{2}), i.e. (5.5,\ -1.5).
[1]
(c) The area scale factor is |\det M| = 8 (the translation has no effect), so the image area is 8 \times 36 = 288 pixel units.
[2]
Final answer
(a) \frac18\begin{pmatrix}2&1\\-2&3\end{pmatrix}. (b) (5.5,\,-1.5). (c) 288.
Question 18
[8 marks]Lena's video editor has a Pulse button that transforms every point of a clip, relative to the screen centre O, by the matrix M = \begin{pmatrix} 1.5 & -2 \\ 2 & 1.5 \end{pmatrix}.
(a) The matrix has the form k\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}. Find k and \theta (in degrees, to 3 significant figures), and describe the transformation fully.
[4]
(b) Show that \det M = k^2.
[1]
(c) The button is pressed 3 times. Find the single matrix for the three presses, and the total enlargement factor and angle of turn.
[3]
Show solutionHide solution
(a) Matching entries: k\cos\theta = 1.5 and k\sin\theta = 2.
[1]
k = \sqrt{1.5^2 + 2^2} = \sqrt{6.25} = 2.5.
[1]
\tan\theta = \dfrac{2}{1.5}, so \theta = 53.130^\circ \approx 53.1^\circ (\cos\theta = 0.6, \sin\theta = 0.8).
[1]
An enlargement of scale factor 2.5 about O together with a rotation of 53.1^\circ anticlockwise about O.
[1]
(b) \det M = (1.5)(1.5) - (-2)(2) = 2.25 + 4 = 6.25 = 2.5^2 = k^2.
[1]
(c) Three presses: M^3 = \begin{pmatrix} -14.62 & -5.500 \\ 5.500 & -14.62 \end{pmatrix} (GDC: M^ 3).
[2]
Enlargement factor 2.5^3 = 15.625 and rotation 3 \times 53.1^\circ = 159.4^\circ anticlockwise.
[1]
Final answer
(a) k=2.5, \theta=53.1^\circ anticlockwise about O. (b) Shown. (c) \begin{pmatrix}-14.6&-5.50\\5.50&-14.6\end{pmatrix}; enlargement factor 15.625, turn 159^\circ.
Question 19
[11 marks]A ripple effect on a phone screen moves a dot according to \mathbf{r}_{n+1} = M\mathbf{r}_n + \mathbf{t} each frame, where M = \begin{pmatrix} 0.6 & -0.3 \\ 0.3 & 0.6 \end{pmatrix}, \qquad \mathbf{t} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} (units: mm), and the dot starts at the origin O.
(a) Show that a fixed point \mathbf{r} of the map satisfies (M - I)\mathbf{r} = -\mathbf{t}.
[2]
(b) Find the fixed point F.
[3]
(c) The matrix M is an enlargement of scale factor k combined with a rotation. Find k and use it to explain why the dot spirals in towards F.
[3]
(d) Find the number of the first frame at which the dot is within 0.1 mm of F.
[3]
Show solutionHide solution
(a) A fixed point satisfies M\mathbf{r} + \mathbf{t} = \mathbf{r}.
[1]
Rearranging, M\mathbf{r} - \mathbf{r} = -\mathbf{t}, so (M - I)\mathbf{r} = -\mathbf{t}.
[1]
(b) M - I = \begin{pmatrix} - \frac{2}{5} & - \frac{3}{10} \\ \frac{3}{10} & - \frac{2}{5} \end{pmatrix}, with \det(M - I) = 0.16 + 0.09 = 0.25 ≠ 0, so the fixed point is unique.
[1]
\mathbf{r} = (M - I)^{-1}(-\mathbf{t}) = \begin{pmatrix} - \frac{8}{5} & \frac{6}{5} \\ - \frac{6}{5} & - \frac{8}{5} \end{pmatrix}\begin{pmatrix} -4 \\ 2 \end{pmatrix} (GDC).
[1]
F = (\frac{44}{5},\,\frac{8}{5}).
[1]
(c) k = \sqrt{0.6^2 + 0.3^2} = \sqrt{0.45} = 0.6708 (with a rotation of 26.57^\circ).
[1]
Writing \mathbf{r}_{n+1} - F = M(\mathbf{r}_n - F), each frame multiplies the distance from F by k < 1 and turns the dot about F.
[1]
So the dot spirals inwards and the distance tends to 0.
[1]
(d) |\mathbf{r}_0 - F| = \sqrt{8.8^2 + 1.6^2} = 8.944 mm, so the distance after n frames is 8.944(0.6708)^n.
[1]
Need 8.944(0.6708)^n < 0.1, so n > \dfrac{\ln(0.1/8.944)}{\ln 0.6708} = 11.25.
[1]
The first such frame is n = 12 (table check: 0.111 mm at frame 11 and 0.0743 mm at frame 12).
[1]
Final answer
(a) Shown. (b) F(8.8,\,1.6). (c) k=0.671; the factor is below 1. (d) Frame 12.
Question 20
[7 marks]A security camera pans about a vertical axis through the origin O of its floor plan. A single pan turns it 25^\circ anticlockwise (matrix R_{25}); a second kind of pan turns it 35^\circ anticlockwise (matrix R_{35}).
(a) Find the single matrix for a 25^\circ pan followed by a 35^\circ pan, with entries to 3 significant figures.
[3]
(b) Describe this single transformation fully.
[1]
(c) The camera repeats the 25^\circ pan again and again. Find the smallest number of pans after which it first points in its starting direction again.
[3]
Show solutionHide solution
(a) The 25^\circ pan is applied first, so the matrix is R_{35}R_{25}.
[1]
R_{25} = \begin{pmatrix} 0.906 & -0.423 \\ 0.423 & 0.906 \end{pmatrix} and R_{35} = \begin{pmatrix} 0.819 & -0.574 \\ 0.574 & 0.819 \end{pmatrix} (degree mode).
[1]
R_{35}R_{25} = \begin{pmatrix} 0.500 & -0.866 \\ 0.866 & 0.500 \end{pmatrix}.
[1]
(b) Entries \cos 60^\circ = 0.5 and \sin 60^\circ = 0.866 show a rotation through 60^\circ anticlockwise about O (angles add: 25 + 35 = 60).
[1]
(c) After n pans the turn is 25n^\circ, which must be a multiple of 360^\circ.
[1]
25n = 360m gives 5n = 72m, and since \gcd(5, 72) = 1 the smallest solution is n = 72 (m = 5).
[2]
(GDC check: R25^ 72 gives the identity matrix and no smaller power does.)
Final answer
(a) \begin{pmatrix}0.5&-0.866\\0.866&0.5\end{pmatrix}. (b) rotation of 60^\circ anticlockwise about O. (c) 72 pans.
Mastery
Full exam questions that mix question types, at Paper 1 and Paper 2 standard.
Question 21
[11 marks]A camera-calibration program undoes distortions by applying transformations in reverse. A distortion is created by reflecting an image in the x-axis (matrix B) and then rotating it 90^{\circ} anticlockwise about the origin (matrix A).
(a) Write down the matrices B (reflection in x-axis) and A (rotation 90^{\circ}).
[2]
(b) Find the single matrix D=AB representing “reflect then rotate”.
[2]
(c) Describe fully the single transformation represented by D.
[2]
(d) (i) Find the inverse matrix D^{-1}.
(ii) The distorted image of a feature is at (-5,\,-1). Use D^{-1} to recover the original position of the feature.
[5]
Show solutionHide solution
(a) B=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.
[2]
(b) D=AB=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
[2]
(c) D=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} swaps coordinates (x,y)\mapsto(y,x), which is a reflection in the line y=x.
[2]
(d)(i) \det D=(0)(0)-(1)(1)=-1, so D^{-1}=\dfrac{1}{-1}\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} (equal to D, since reflection is self-inverse).
(ii) D^{-1}\begin{pmatrix}-5\\-1\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}-5\\-1\end{pmatrix}=\begin{pmatrix}-1\\-5\end{pmatrix}, so the original feature was at (-1,\,-5).
[5]
Final answer
(a) B=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. (b) D=\begin{pmatrix}0&1\\1&0\end{pmatrix}. (c) reflection in y=x. (d) D^{-1}=D; original (-1,\,-5).
Question 22
[12 marks]A logo animator builds a “spin and grow” effect from two matrix transformations applied to a shape about the origin:
first an anticlockwise rotation of 45^{\circ} (matrix A), then an enlargement of scale factor 2, centre the origin (matrix E).
(a) Write down the matrices A (exact entries) and E.
[3]
(b) Find the single matrix C=EA that represents the whole effect, giving entries correct to 3 significant figures.
[2]
(c) A marker on the logo is at (3,\,1). Find its position after the complete effect, correct to 3 significant figures.
[3]
(d) (i) Find \det C.
(ii) A region of the logo has area 9\text{ cm}^2 before the effect. Find its area afterwards.
[4]
Show solutionHide solution
(a) A=\begin{pmatrix} \cos 45^{\circ} & -\sin 45^{\circ} \\ \sin 45^{\circ} & \cos 45^{\circ} \end{pmatrix}=\begin{pmatrix} \tfrac{\sqrt2}{2} & -\tfrac{\sqrt2}{2} \\ \tfrac{\sqrt2}{2} & \tfrac{\sqrt2}{2} \end{pmatrix}, E=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.
[3]
(b) C=EA=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} \tfrac{\sqrt2}{2} & -\tfrac{\sqrt2}{2} \\ \tfrac{\sqrt2}{2} & \tfrac{\sqrt2}{2} \end{pmatrix}=\begin{pmatrix} \sqrt2 & -\sqrt2 \\ \sqrt2 & \sqrt2 \end{pmatrix}\approx\begin{pmatrix} 1.41 & -1.41 \\ 1.41 & 1.41 \end{pmatrix}.
[2]
(c) C\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}\sqrt2(3)-\sqrt2(1)\\ \sqrt2(3)+\sqrt2(1)\end{pmatrix}=\begin{pmatrix}2\sqrt2\\4\sqrt2\end{pmatrix}\approx(2.83,\,5.66).
[3]
(d)(i) \det C=(\sqrt2)(\sqrt2)-(-\sqrt2)(\sqrt2)=2+2=4.
(ii) New area =9\times|\det C|=9\times 4=36\text{ cm}^2.
[4]
Final answer
(a) A=\begin{pmatrix}\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\\\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\end{pmatrix}, E=2I. (b) \begin{pmatrix}1.41&-1.41\\1.41&1.41\end{pmatrix}. (c) (2.83,\,5.66). (d) \det C=4; 36\ \text{cm}^2.
Question 23
[14 marks]A technician tests a worn pen plotter. A position \mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix} in the design file (in mm) is drawn on the paper at the position \mathbf{r}' = M\mathbf{r} + \mathbf{t} (in mm). Three test marks are made: the file point (0, 0) is drawn at (12, -4), the file point (1, 0) is drawn at (13.5, -3.6), and the file point (0, 1) is drawn at (11.2, -2.7).
(a) Find \mathbf{t} and the matrix M.
[3]
(b) Find \det M. A shape in the design file has area 840\ \text{mm}^2. Find the area of the shape as drawn on the paper.
[3]
(c) A mark on the paper is at (45.3,\ 18.5). Find the file position it came from.
[3]
(d) There is one point that the plotter draws exactly where the file says (its drawn position has the same coordinates as its file position). Find it, to 3 significant figures.
[3]
(e) State what the sign of \det M tells you about the drawing, and give one reason why an affine map may not model a real worn plotter accurately over the whole sheet.
[2]
Show solutionHide solution
(a) The image of the origin is \mathbf{t}, so \mathbf{t} = \begin{pmatrix} 12 \\ -4 \end{pmatrix}.
[1]
Column 1 = \begin{pmatrix} 13.50 \\ -3.600 \end{pmatrix} - \begin{pmatrix} 12 \\ -4 \end{pmatrix} = \begin{pmatrix} 1.500 \\ 0.4000 \end{pmatrix} and column 2 = \begin{pmatrix} 11.20 \\ -2.700 \end{pmatrix} - \begin{pmatrix} 12 \\ -4 \end{pmatrix} = \begin{pmatrix} -0.8000 \\ 1.300 \end{pmatrix}.
[1]
M = \begin{pmatrix} 1.50 & -0.800 \\ 0.400 & 1.30 \end{pmatrix}.
[1]
(b) \det M = (1.5)(1.3) - (-0.8)(0.4) = 1.95 + 0.32 = 2.27.
[1]
Every area is multiplied by 2.27, so the drawn shape has area 2.27 \times 840 = 1906.8\ \text{mm}^2.
[2]
(c) \mathbf{r} = M^{-1}(\mathbf{r}' - \mathbf{t}) with \mathbf{r}' - \mathbf{t} = \begin{pmatrix} 33.3 \\ 22.5 \end{pmatrix}.
[1]
M^{-1} = \dfrac{1}{2.27}\begin{pmatrix} 1.3 & 0.8 \\ -0.4 & 1.5 \end{pmatrix} (GDC), so \mathbf{r} = \begin{pmatrix} 27 \\ 9 \end{pmatrix}.
[1]
The mark came from the file position (27,\,9).
[1]
(d) Solve M\mathbf{f} + \mathbf{t} = \mathbf{f}, i.e. (M - I)\mathbf{f} = -\mathbf{t} with M - I = \begin{pmatrix} 0.500 & -0.800 \\ 0.400 & 0.300 \end{pmatrix}.
[1]
\det(M - I) = 0.15 + 0.32 = 0.47, so \mathbf{f} = \begin{pmatrix} 0.6383 & 1.702 \\ -0.8511 & 1.064 \end{pmatrix}\begin{pmatrix} -12 \\ 4 \end{pmatrix} (GDC).
[1]
\mathbf{f} = \begin{pmatrix} -0.8511 \\ 14.47 \end{pmatrix}, so the point is (-0.851,\,14.5).
[1]
(e) \det M > 0, so orientation is preserved: the drawing is not a mirror image of the design.
[1]
An affine map scales every region by the same factor, but the wear in a real plotter (a pen arm that swings in an arc, a stretching belt) distorts different parts of the sheet by different amounts, which no single affine map can model.
[1]
Final answer
(a) \mathbf t=\begin{pmatrix}12\\-4\end{pmatrix}, M=\begin{pmatrix}1.5&-0.8\\0.4&1.3\end{pmatrix}. (b) \det M=2.27; 1906.8\ \text{mm}^2. (c) (27,\,9). (d) (-0.851,\,14.5). (e) positive: orientation kept.
Question 24
[8 marks]R(\theta) is the matrix of a rotation through \theta anticlockwise about the origin. You may use \sin2\alpha=2\sin\alpha\cos\alpha and \cos2\alpha=1-2\sin^2\alpha.
(a) Write down R(2\alpha) with every entry in terms of \sin\alpha and \cos\alpha.
[2]
(b) Explain why R(2\alpha)R(\alpha)=R(3\alpha).
[1]
(c) By comparing the bottom-left entries of R(2\alpha)R(\alpha) and R(3\alpha), show that \sin3\alpha=3\sin\alpha-4\sin^3\alpha.
[3]
(d) Verify the identity in part (c) when \alpha=30^\circ.
[2]
Show solutionHide solution
(a) R(2\alpha)=\begin{pmatrix}1-2\sin^2\alpha&-2\sin\alpha\cos\alpha\\2\sin\alpha\cos\alpha&1-2\sin^2\alpha\end{pmatrix}. A1A1
(b) A turn through \alpha followed by a turn through 2\alpha about O is one turn through 3\alpha.
R1
(c) Bottom-left of the product: 2\sin\alpha\cos\alpha\cdot\cos\alpha+(1-2\sin^2\alpha)\sin\alpha.
M1
=2\sin\alpha(1-\sin^2\alpha)+\sin\alpha-2\sin^3\alpha.
M1
=3\sin\alpha-4\sin^3\alpha.
A1
The bottom-left entry of R(3\alpha) is \sin3\alpha, so \sin3\alpha=3\sin\alpha-4\sin^3\alpha.
AG
(d) \sin90^\circ=1.
A1
3\left(\tfrac12\right)-4\left(\tfrac18\right)=1.
A1
Final answer
(a) \begin{pmatrix}1-2\sin^2\alpha&-2\sin\alpha\cos\alpha\\2\sin\alpha\cos\alpha&1-2\sin^2\alpha\end{pmatrix}. (b) Two turns about O make one turn. (c) Shown. (d) both sides equal 1.
Question 25
[17 marks]A drone light show begins with four drones at the corners of a square: A(0, 0), B(10, 0), C(10, 10) and D(0, 10) (metres on a ground plan). The drones carry out three moves in this order:
Move 1: a rotation of 30^\circ anticlockwise about the origin O;
Move 2: an enlargement of the whole formation, scale factor 1.5, centre O;
Move 3: a translation by \begin{pmatrix} 20 \\ -5 \end{pmatrix}.
(a) Find the single matrix M for moves 1 and 2 together, to 3 significant figures.
[3]
(b) The three moves together act as \mathbf{r}' = M\mathbf{r} + \mathbf{t}. Write down \mathbf{t} and find the final positions of A, B, C and D, to 3 significant figures.
[5]
(c) Find the area enclosed by the final formation, and check it using the length of A'B'.
[3]
(d) The show designer wants one position of the plan that no drone's move changes: a point F with M\mathbf{f} + \mathbf{t} = \mathbf{f}. Find F, to 3 significant figures.
[3]
(e) The whole three-move sequence is repeated many times. Describe what happens to the formation relative to F, and find the smallest number of repeats for which the side length of the formation first exceeds 100 m.
[3]
Show solutionHide solution
(a) Rotation first, enlargement second: M = 1.5I \times R_{30^\circ}.
[1]
R_{30^\circ} = \begin{pmatrix} 0.8660 & -0.5000 \\ 0.5000 & 0.8660 \end{pmatrix} (degree mode).
[1]
M = 1.5R_{30^\circ} = \begin{pmatrix} 1.30 & -0.750 \\ 0.750 & 1.30 \end{pmatrix}.
[1]
(b) The translation is the last move, so \mathbf{t} = \begin{pmatrix} 20 \\ -5 \end{pmatrix}.
[1]
A' = M\mathbf{0} + \mathbf{t} = (20,\,-5), [1] B' = M\begin{pmatrix} 10 \\ 0 \end{pmatrix} + \mathbf{t} = (33.0,\,2.50),
[1]
C' = (25.5,\,15.5), [1] D' = (12.5,\,7.99).
[1]
(GDC: put the corners as columns of a 2 \times 4 matrix, multiply by M, add \mathbf{t} to each column.)
(c) \det M = 1.5^2 = 2.25 and the original area is 100\ \text{m}^2, so the area is 2.25 \times 100 = 225\ \text{m}^2.
[2]
Check: |A'B'| = 15 = 15 m, and 15^2 = 225.
[1]
(d) (M - I)\mathbf{f} = -\mathbf{t}, so \mathbf{f} = (M - I)^{-1}(-\mathbf{t}).
[1]
M - I = \begin{pmatrix} 0.2990 & -0.7500 \\ 0.7500 & 0.2990 \end{pmatrix} with \det = 0.6519 ≠ 0.
[1]
\mathbf{f} = \begin{pmatrix} -3.422 \\ 25.30 \end{pmatrix}, so F = (-3.42,\,25.3).
[1]
(e) Since M is 1.5 times a rotation, each repeat turns the formation 30^\circ about F and multiplies every distance from F by 1.5 > 1: the formation spirals away from F and grows (F repels instead of attracting).
[1]
After n repeats the side length is 10(1.5)^n > 100, so 1.5^n > 10 and n > 5.679; 10(1.5)^5 = 75.94 and 10(1.5)^6 = 113.9, so n = 6.
[2]
Final answer
(a) M=\begin{pmatrix}1.30&-0.750\\0.750&1.30\end{pmatrix}. (b) \mathbf t=\begin{pmatrix}20\\-5\end{pmatrix}; A'(20,\,-5), B'(33.0,\,2.50), C'(25.5,\,15.5), D'(12.5,\,7.99). (c) 225\ \text{m}^2. (d) F(-3.42,\,25.3). (e) spirals outwards; n=6.
Going further
Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.
G1 A turn plus a slide is a turn
G1 Worked
[5 marks]Ilse's plotting software turns every point of a drawing through 90^\circ anticlockwise about the origin O and then slides it 4 cm in the positive x-direction. The combined map is \mathbf r'=R\mathbf r+\begin{pmatrix}4\\0\end{pmatrix}, where R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
(a) Show that exactly one point is left unchanged, and find it.
[4]
(b) Hence describe the combined map as a single transformation.
[1]
Show solutionHide solution
(a) A fixed point satisfies (I-R)\mathbf f=\begin{pmatrix}4\\0\end{pmatrix}, with I-R=\begin{pmatrix}1&1\\-1&1\end{pmatrix} (M1). \det(I-R)=2\neq0, so there is exactly one solution (A1). \mathbf f=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}4\\0\end{pmatrix} (M1), so \mathbf f=\begin{pmatrix}2\\2\end{pmatrix} (A1). (b) A rotation of 90^\circ anticlockwise about (2,2) (A1).
Final answer
\det(I-R)=2; the fixed point is (2,2); a rotation of 90^\circ anticlockwise about (2,2).
G2 Investigation
[21 marks]Priya, a textile designer, builds a whirl of squares. Square S_0 has vertices (0,0), (4,0), (4,4) and (0,4), in centimetres. Each new square is the image of the previous one under M=0.8R(30^\circ), where R(30^\circ) is the matrix of an anticlockwise rotation of 30^\circ about O.
(a) Show that \det M=0.64, and hence write down an expression for the area of S_n.
[2]
(b) Find the total area of S_0,S_1,\ldots,S_9.
[2]
(c) Find the total area if the whirl continued for ever.
[2]
(d) Find the least value of n, where n is a positive integer, for which M^n is a multiple of I, and describe fully the transformation M^n.
[3]
(e) The corner (4,0) of S_0 moves to P_n=M^n\begin{pmatrix}4\\0\end{pmatrix}. Show that P_{n+1}-P_n=M^n(M-I)\begin{pmatrix}4\\0\end{pmatrix}, and find the total length of the path P_0P_1P_2\ldots if it continued for ever.
[4]
Priya now uses the map \mathbf r'=M\mathbf r+\begin{pmatrix}2\\1\end{pmatrix} instead.
(f) Find the fixed point F of this map.
[3]
(g) Show that \mathbf r'-\mathbf f=M(\mathbf r-\mathbf f), and explain why every copy spirals in towards F.
[3]
(h) Find the range of values of k, 0<k<1, for which the whirl built with kR(30^\circ) in place of M has a total area below 100 cm^2 when continued for ever.
[2]
Show solutionHide solution
(a) \det M=0.8^2\det R(30^\circ)=0.64\times1=0.64 (M1). The area of S_n is 16(0.64)^n cm^2 (A1). (b) A geometric sum with first term 16 and ratio 0.64 (M1): \dfrac{16(1-0.64^{10})}{1-0.64}=43.9 cm^2 (A1). (c) \dfrac{16}{1-0.64} (M1) =\dfrac{400}{9}=44.4 cm^2 (A1). (d) 30n must be a multiple of 360 (M1), so n=12 (A1). M^{12}=0.8^{12}I is an enlargement with scale factor 0.0687 and centre O (A1). (e) P_{n+1}-P_n=M^{n+1}\mathbf v-M^n\mathbf v=M^n(M-I)\mathbf v with \mathbf v=\begin{pmatrix}4\\0\end{pmatrix} (M1). \lvert(M-I)\mathbf v\rvert=2.0174 (A1). Each application of M multiplies a length by 0.8, so the segments are 2.0174(0.8)^n (M1) and the total length is \dfrac{2.0174}{1-0.8}=10.1 cm (A1). (f) (I-M)\mathbf f=\begin{pmatrix}2\\1\end{pmatrix} (M1), \mathbf f=(I-M)^{-1}\begin{pmatrix}2\\1\end{pmatrix} (M1), so F(0.843,\,4.35) (A1). (g) Subtract \mathbf f=M\mathbf f+\mathbf b from \mathbf r'=M\mathbf r+\mathbf b (M1), giving \mathbf r'-\mathbf f=M(\mathbf r-\mathbf f). Each step multiplies the distance from F by 0.8 (R1), and 0.8<1, so the distance tends to 0 while the copy turns 30^\circ about F each time (R1). (h) \dfrac{16}{1-k^2}<100 (M1), so k^2<0.84 and 0<k<0.917 (A1).
Final answer
(a) Shown; 16(0.64)^n. (b) 43.9\ \text{cm}^2. (c) 44.4\ \text{cm}^2. (d) n=12; enlargement, scale factor 0.0687, centre O. (e) Shown; 10.1 cm. (f) F(0.843,\,4.35). (g) Shown; the distance from F is multiplied by 0.8<1. (h) 0<k<0.917.
From the book
Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma
Chapter 11 teaches matrix transformations from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.
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