Matrices (Further Pure 1)

4(a) cos60 −sin60  1 − 3 B1 Rotation 60° anticlockwise about the origin.   = 2 2  sin60 cos60   3 1    2 2 d 0 B1 One-way stretch in the x-direction, scale factor d.   0 1 d 0  1 − 3  d −d 3 M1 A1 Correct order. M=    2 2 = 2 2  0 1  3 1   3 1      2 2 2 2 4 4(b) d = 1d2 M1 Uses value of detM. 2 d ≠0d =2 A1 AG 2 © UCLES 2021 Page 8 of 14 4(c) 1  1 3 B1 FT Inverse of their M. M −1 =  2  2− 3 1    2 1 1 1  1 31 1 M1 Multiplies on the left by the inverse of their M. N=M −1   =  2    1 2 1 2  2  − 2 3 1   1 2 1 2  11+ 3 1+ 3 A1 CAO =   4 1− 3 1− 3   3 4(d) 1 1x  x+ y  B1 x X   1 1    y   =   1x+ 1 y   Transforms  y   to  Y   . 2 2 2 2 1x+ 1mx=m ( x+mx ) M1 A1 Uses y=mx and Y =mX . 2 2 1+m=2m+2m2 2m2 +m−1=0 A1 y= 1x and y=−x A1 WWW 2 5 © UCLES 2021 Page 9 of 14

4(a) cos60 −sin60  1 − 3 B1 Rotation 60° anticlockwise about the origin.   = 2 2  sin60 cos60   3 1    2 2 d 0 B1 One-way stretch in the x-direction, scale factor d.   0 1 d 0  1 − 3  d −d 3 M1 A1 Correct order. M=    2 2 = 2 2  0 1  3 1   3 1      2 2 2 2 4 4(b) d = 1d2 M1 Uses value of detM. 2 d ≠0d =2 A1 AG 2 © UCLES 2021 Page 8 of 14 4(c) 1  1 3 B1 FT Inverse of their M. M −1 =  2  2− 3 1    2 1 1 1  1 31 1 M1 Multiplies on the left by the inverse of their M. N=M −1   =  2    1 2 1 2  2  − 2 3 1   1 2 1 2  11+ 3 1+ 3 A1 CAO =   4 1− 3 1− 3   3 4(d) 1 1x  x+ y  B1 x X   1 1    y   =   1x+ 1 y   Transforms  y   to  Y   . 2 2 2 2 1x+ 1mx=m ( x+mx ) M1 A1 Uses y=mx and Y =mX . 2 2 1+m=2m+2m2 2m2 +m−1=0 A1 y= 1x and y=−x A1 WWW 2 5 © UCLES 2021 Page 9 of 14

4(a) 2 k k1 0 2+k k  M1 A1 Multiplies two matrices correctly.  0 1 1    0 1 1    −1 2 0    5 −1 3     0 1   =   −1 2 0     8 −1   1 0 11 0  2 0  2 k k1 0 1 0  0 1 1   6 −1 4   Or   −1 2 0     5 −1 3     0 1   =  8 −k−2 −k+6     0 1   1 0 11 0 1 0  10 −1  A1    −k+14 −k−2 3 4(b) −1 3 5 3 5 −1 M1 A1 Sets determinant equal to zero and forms linear 2 −k +k =0 leading to −2−2k+k =0 equation. 0 1 1 1 1 0 k =−2 A1 3 © UCLES 2021 Page 10 of 16 4(c) 10 −1x 10x− y M1 x    =   Transforms  . 16 0 y  16x  y 1  Allow q  where q is x, t or a nonzero number. m 10x−mx= X and 16x=mX M1 A1 Uses y=mx and Y =mX . Expect 16x=m ( 10x−mx ) . 16=10m−m2 [m2 −10m+16=0] A1 OE y=2x and y=8x A1 5 Question Answer Marks Guidance

4(a) One-way stretch [scale factor 3 parallel to the x-axis] followed by a rotation, B2 Award B1 if given in the wrong order. [anticlockwise centred at the origin, through an angle θ.] 2 © UCLES 2021 Page 8 of 14 4(b) 3cosθ −sinθ  B1 M=   3sinθ cosθ  3cosθ −sinθ x 3xcosθ− ysinθ  B1 x X    =   Transforms   to  . 3sinθ cosθ y 3xsinθ+ ycosθ  y Y  3xsinθ+mxcosθ=3mxcosθ−m2xsinθ M1 A1 Uses y=mx and Y =mX . m2sinθ−2mcosθ+3sinθ=0 A1 4cos2θ−12sin2θ=0 M1 A1 Sets discriminant equal to 0. 1 M1 Uses an appropriate trigonometric identity. tanθ=± or 3 ( ) 4cos2θ−12 1−cos2θ =0 leading to 16cos2θ−12=0 leading to cosθ=± 3 2 θ= 1π and θ= 5π A1 Allow 30° and 150° 6 6 9 © UCLES 2021 Page 9 of 14

1(a) Enlargement, scale factor 6. B1 1 1(b) detA=6−8=−2 M1 Finds detA. Area =2×13=26cm2 A1 2 1(c)  2 −4 M1 Finds A −1. A −1 =−1   2  −2 3   2 −4  −6 12 A1 AEF B=−3  =   Could be solved by equations.  −2 3   6 −9 2 1(d) 3 4x 3x+4y B1 X    =   Finds   2 2y 2x+2y Y  3x+4y= x 2x+4y=0 M1 x X   leading to  Uses   =   to form simultaneous equations. 2x+2y= y 2x+ y=0 y Y  y=−2xleading to −6x=0 leading tox=0, y=0 M1 A1 2 4 Solves equations or states thatdet  ≠0, AG. 2 1 4 © UCLES 2021 Page 6 of 16

4(a) One-way stretch [scale factor 3 parallel to the x-axis] followed by a rotation, B2 Award B1 if given in the wrong order. [anticlockwise centred at the origin, through an angle θ.] 2 © UCLES 2021 Page 8 of 14 4(b) 3cosθ −sinθ  B1 M=   3sinθ cosθ  3cosθ −sinθ x 3xcosθ− ysinθ  B1 x X    =   Transforms   to  . 3sinθ cosθ y 3xsinθ+ ycosθ  y Y  3xsinθ+mxcosθ=3mxcosθ−m2xsinθ M1 A1 Uses y=mx and Y =mX . m2sinθ−2mcosθ+3sinθ=0 A1 4cos2θ−12sin2θ=0 M1 A1 Sets discriminant equal to 0. 1 M1 Uses an appropriate trigonometric identity. tanθ=± or 3 ( ) 4cos2θ−12 1−cos2θ =0 leading to 16cos2θ−12=0 leading to cosθ=± 3 2 θ= 1π and θ= 5π A1 Allow 30° and 150° 6 6 9 © UCLES 2021 Page 9 of 14

7(a) k 6 4 6 4 k M1 A1 Sets detA0. SOI 2 3 012k600 8 9 7 9 7 8 k 5 A1 3 7(b) 1 k 6 9k48 3k16 M1 A1 Finds one correct entry.   6012k 8 9 6012k 45k 1 2 3 6 1 A1   6012k 8 9 6012k 2(5k) 1 2 3 123k 4k A1   6012k k 6 6012k 45k 4 © UCLES 2022 Page 14 of 15 7(c) 1 2 3 M1 A1 Finds BA. 1 0 0  1 2 3 BA  4 k 6    0 1 0  4 k 6 7 8 9 a d M1 Sets C as 32 matrix. 1 2 3   a2b3c d 2e3f  2 1   b e     4 k 6  4akb6c 4d ke6f  k 4 c f  0 1 A1 Accept any a, b, c, d, e and f such that C  1 0   a2b3c d 2e3f  2 1     .  0 0   4akb6c 4d ke6f  k 4 4 7(d) 2 1x 2x y  B1 x X     Transforms   to  . k 4y kx4y y Y  kx4mxm2xmx M1 A1 Uses y mx and Y mX . m2 2mk 0 A1 44k 0 M1 Sets discriminant positive. k1 k8 A1 6 © UCLES 2022 Page 15 of 15

7(a) k 6 4 6 4 k M1 A1 Sets detA0. SOI 2 3 012k600 8 9 7 9 7 8 k 5 A1 3 7(b) 1 k 6 9k48 3k16 M1 A1 Finds one correct entry.   6012k 8 9 6012k 45k 1 2 3 6 1 A1   6012k 8 9 6012k 2(5k) 1 2 3 123k 4k A1   6012k k 6 6012k 45k 4 © UCLES 2022 Page 14 of 15 7(c) 1 2 3 M1 A1 Finds BA. 1 0 0  1 2 3 BA  4 k 6    0 1 0  4 k 6 7 8 9 a d M1 Sets C as 32 matrix. 1 2 3   a2b3c d 2e3f  2 1   b e     4 k 6  4akb6c 4d ke6f  k 4 c f  0 1 A1 Accept any a, b, c, d, e and f such that C  1 0   a2b3c d 2e3f  2 1     .  0 0   4akb6c 4d ke6f  k 4 4 7(d) 2 1x 2x y  B1 x X     Transforms   to  . k 4y kx4y y Y  kx4mxm2xmx M1 A1 Uses y mx and Y mX . m2 2mk 0 A1 44k 0 M1 Sets discriminant positive. k1 k8 A1 6 © UCLES 2022 Page 15 of 15

2(a) 10 B1 SOI 2 2 2 52 205 M1 A1 Uses 2 2 2 2 2. Alternative method for question 2(a) Put yx2 and form cubic not involving x M1 A1 y(y10)2 25y2 20y4(need not be simplified). y35y2 120y40 2 2 2 (5)5 A1 3 2(b) 1   M1 A1 Finds determinant. 1     1  1     12 2 2 2  1  1     1 12 2 2 2152(2)0 M1 A1 Substitutes and uses 2 4 © UCLES 2022 Page 8 of 17

5(a) A shear in the x-direction. B1 Accept ‘shear’. 1 5(b) 1 a B1 States base case. A  so true when n1. 0 1 1 ka B1 States inductive hypothesis. Assume that it is true for nk, so Ak  . 0 1  1 ka1 a 1 aak 1 (k1)a M1 A1 Multiplies Ak with A. Then Ak1      0 1 0 1 0 1  0 1  If it is true for [n1 and] nk then it is also true for nk1. Hence, A1 Everything correct and states conclusion. [by induction,] true for all positive integers. 5 5(c) 1 nab b bn bn M1 A1 Uses formula given in part (b). AnB    0 1  1 a 1 a  1 a a 1  bn bnx (bn)(x y) B1 x X      Transforms   by multiplying matrices to find  .  1 a 1 a y  1 a (x y)  y Y  1(1m)m(bn)(1m) M1 Uses ymx and Y mX a y x A1 y 1 x A1 a(bn) 6 © UCLES 2022 Page 13 of 17

5(a) Shear (in the x-direction) followed by a rotation (anticlockwise about the origin B2 Award B1 if given in the wrong order. through 45°). 2 5(b)  k+1 k−1  k+1 k−1 M1 A1 Finds M−1. − −     1 2 2 2 2 M−1 =  =  1 1 1   1 1  − −      2 2   2 2  2 5(c) 1 2 1(k−1) 2 B1 M=2 2  1 2 1(k+1) 2    2 2 1 k−1x x+(k−1)y B1 x X 1 2  = 1 2  Transforms   to  . 2 1 k+1y 2   x+(k+1)y   y Y  x+m(k+1)x=m ( x+m(k−1)x ) M1 A1 Uses y=mx and Y =mX . (k−1)m2 −km−1=0 A1 k2 +4k−40 M1 Sets discriminant negative. ( ) ( ) A1 2 −1− 2 k2 −1+ 2 © UCLES 2022 Page 10 of 15 5(c) Alternative method for question 5(c) 1 2 1(k−1) 2 B1 M=2 2  1 2 1(k+1) 2    2 2 B1 Sets up to find characteristic equation. 1 2− 1(k−1) 2 [M - I =] 2 2 =0 1 2 1(k+1) 2− 2 2 ( 1 2− )( 1(k+1) 2− ) −1(k−1) 21 2=0 M1 A1 Evaluates determinant. 2 2 2 2 22− 2(k+2)+2=0 A1 k2 +4k−40 M1 Sets discriminant negative. ( ) ( ) A1 2 −1− 2 k2 −1+ 2 7 © UCLES 2022 Page 11 of 15

3(a) Shear [in the y-direction] followed by a stretch, [parallel to the y-axis, scale factor k]. B2 Award B1 if given in the wrong order. 2 3(b) 1 0 −1  1 0 1 0 −1 1 0 B1   =  ,   =  k 1 −k 1 0 k  0 1  k  1 01 0 B1 Correct order. M−1=   −k 1 0 1  k 2 3(c)  1 0 B1 M=  k2 k  1 0x  x  B1 x X   =  Transforms   to   k2 ky k2x+ky y Y  k2x+ky= y M1 X x Sets  =  Y  y k2 A1 AG. y= x 1−k 4 3(d) 1= k M1 Uses DEF = detM ABC k=−1 A1 2 © UCLES 2022 Page 8 of 15

5(a) Shear (in the x-direction) followed by a rotation (anticlockwise about the origin B2 Award B1 if given in the wrong order. through 45°). 2 5(b)  k+1 k−1  k+1 k−1 M1 A1 Finds M−1. − −     1 2 2 2 2 M−1 =  =  1 1 1   1 1  − −      2 2   2 2  2 5(c) 1 2 1(k−1) 2 B1 M=2 2  1 2 1(k+1) 2    2 2 1 k−1x x+(k−1)y B1 x X 1 2  = 1 2  Transforms   to  . 2 1 k+1y 2   x+(k+1)y   y Y  x+m(k+1)x=m ( x+m(k−1)x ) M1 A1 Uses y=mx and Y =mX . (k−1)m2 −km−1=0 A1 k2 +4k−40 M1 Sets discriminant negative. ( ) ( ) A1 2 −1− 2 k2 −1+ 2 © UCLES 2022 Page 10 of 15 5(c) Alternative method for question 5(c) 1 2 1(k−1) 2 B1 M=2 2  1 2 1(k+1) 2    2 2 B1 Sets up to find characteristic equation. 1 2− 1(k−1) 2 [M - I =] 2 2 =0 1 2 1(k+1) 2− 2 2 ( 1 2− )( 1(k+1) 2− ) −1(k−1) 21 2=0 M1 A1 Evaluates determinant. 2 2 2 2 22− 2(k+2)+2=0 A1 k2 +4k−40 M1 Sets discriminant negative. ( ) ( ) A1 2 −1− 2 k2 −1+ 2 7 © UCLES 2022 Page 11 of 15

4(a)  a b2 cos sin M1 Uses correct matrix for rotation.     c2 a   sin cos b2 c2 which is impossible since b and c are real and b0. A1 AG 2 4(b)  a b2x axb2y B1 x X      Transforms   to  .  c2 a  y  c2xay   y Y  c2xamxm  axb2mx  M1 A1 Uses ymx and Y mX . c2 ammab2m2 c2 b2m2 A1 y c x and yc x A1 b b 5 4(c) 1 55 0 M1 A1* Award M1 if matrices correct but order is M   wrong. 0 10 5 5 52 DA1 Dep: previous A1   0 5 3 4(d) 12detM M1 Using their M. 300 cm2 A1FT 2 © UCLES 2023 Page 9 of 16

4(a)  a b2 cos sin M1 Uses correct matrix for rotation.     c2 a   sin cos b2 c2 which is impossible since b and c are real and b0. A1 AG 2 4(b)  a b2x axb2y B1 x X      Transforms   to  .  c2 a  y  c2xay   y Y  c2xamxm  axb2mx  M1 A1 Uses ymx and Y mX . c2 ammab2m2 c2 b2m2 A1 y c x and yc x A1 b b 5 4(c) 1 55 0 M1 A1* Award M1 if matrices correct but order is M   wrong. 0 10 5 5 52 DA1 Dep: previous A1   0 5 3 4(d) 12detM M1 Using their M. 300 cm2 A1FT 2 © UCLES 2023 Page 9 of 16

4(a) Rotation [anticlockwise] B1 about the origin through angle2. B1 Shear [in the x-direction] followed by a rotation [anticlockwise about the B1 origin through angle2]. 3 4(b) 1 1 B1 1 k 1 k cos2 sin2  cos2 sin2     ,     0 1 0 1  sin2 cos2 sin2 cos2 1 k cos2 sin2 B1 Correct order M1    0 1 sin2 cos2 2 4(c) cos21 kcos2sin2  B1 MI   sin2 ksin2cos21 cos21(ksin2cos21)ksin2θcos2θsin22 0 M1 Evaluates detMI 22cos2ksin20 A1 Brackets removed correctly and = 0 4sin22ksincos M1 Uses 1cos22sin2 andsin22sincos or all necessary double angle formulae. tan 1k A1 2 5 © UCLES 2023 Page 9 of 16 4(d)  1 3 3 B1 M 2 2 . 1 3 5    2 2  1 3 3x 1x 3 3y B1FT x X    1 2 2 3 2 5 2     y      1 2 2 3x 2 5 2 y    Transforms  y   to  Y   1 3  3 3  M1 X x  x 3yx   x 3y0  x 3y0  Sets    2 2  2 2  Y  y and 1 5  1 3  3x y y  3x y0  3x3y0   2 2  2 2  3x3y0 A1 AG. 4 Question Answer Marks Guidance

5(a) 1 3 2 3 2 1 M1 A1 Sets determinant of A equal to zero. −k +3 =0−1−k+3=0k=2 2 5 3 5 3 2 1 2 3 0 −2 −2 4  M1 Multiplying two matrices correctly, correct −2 −1 1   −2 −1 1  dimensions.   2 1 3  −1 3  =  −1 −1   1 1 3    1 1 3  3 2 5 0 0  −2 0   3 −7 M1 A1 Completing matrix multiplication, AG.   −9 3  5 5(b)  3 −7x 3x−7y  B1 x X   =  Transforms   to  . −9 3 y −9x+3y y Y  −9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX . −9+3m=3m−7m2 7m2 =9 A1 y= 3 x and y=− 3 x A1 7 7 5 © UCLES 2023 Page 10 of 15 5(c)  0 B1 D=  0   0 B1 E=  0 1 0 1 B1 F=  1 0  3 −7  0 0  M1 Setting up simultaneous equations using their D and  = −9  E. −9 3  0  1 0 3 0 7 0 A1 Condone =3,= 7 if it is clear that they refer to D=  E= 9  9 0 3 0 1 the correct matrices. 5 Question Answer Marks Guidance

3(a) Shear followed by a stretch. B2 Award B1 if given in the wrong order. 2 3(b) OPQR = detM = k B1 1 1 0 M1 A1 M−1 =   k−1 k 3 © UCLES 2023 Page 7 of 15 3(c) k 0 x  B1 Sets y= 1 x. k−1    1 1 1 x  k−1 k 0 x   kx   kx  M1 Shows that Y = 1 X.   = =  k−1 1 1 1 x   x+ 1 x   k x  k−1 k−1 k−1  x  A1 k  1 x   k−1 Alternative method for 3(c) k 0x  kx  B1 x X   =  Transforms   to   1 1y x+ y y Y  X =kx and mX =x+ y M1 Uses y = mx and Y = mX mkx=x+mx 1 A1 AG m= k−1 1 y= x k−1 3 © UCLES 2023 Page 8 of 15

5(a) 1 3 2 3 2 1 M1 A1 Sets determinant of A equal to zero. −k +3 =0−1−k+3=0k=2 2 5 3 5 3 2 1 2 3 0 −2 −2 4  M1 Multiplying two matrices correctly, correct −2 −1 1   −2 −1 1  dimensions.   2 1 3  −1 3  =  −1 −1   1 1 3    1 1 3  3 2 5 0 0  −2 0   3 −7 M1 A1 Completing matrix multiplication, AG.   −9 3  5 5(b)  3 −7x 3x−7y  B1 x X   =  Transforms   to  . −9 3 y −9x+3y y Y  −9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX . −9+3m=3m−7m2 7m2 =9 A1 y= 3 x and y=− 3 x A1 7 7 5 © UCLES 2023 Page 10 of 15 5(c)  0 B1 D=  0   0 B1 E=  0 1 0 1 B1 F=  1 0  3 −7  0 0  M1 Setting up simultaneous equations using their D and  = −9  E. −9 3  0  1 0 3 0 7 0 A1 Condone =3,= 7 if it is clear that they refer to D=  E= 9  9 0 3 0 1 the correct matrices. 5 Question Answer Marks Guidance

4(a) [One way] stretch, rotation B1 Stretch followed by rotation B1 Correct order. Stretch parallel to the x-axis, scale factor 14 B1 Rotation, 1π [anticlockwise] about the origin. B1 3 4 © Cambridge University Press & Assessment 2024 Page 8 of 16 4(b)  1 1 3 1  1 1 3 14 0 1 1 1 0  B1 Both inverses correct.  2 2   2 2  ,      1 3 1   1 3 1   0 1 140 14     2 2 2 2  1 0 1 1 3 B1 Correct order. M1  14  2 2  0 1   1 3 1   2 2 2 4(c)  7 1 3 B1 M 2   7 3 1    2  7 1 3x 7x1 3y B1 x X    7 3 2 1 2     y      7 3x 2  1 2 y    Transforms  y   to  Y   7 3x 1mxm  7x1 3mx  M1 Uses ymx and Y mX . 2 2 7 3 1m7m1 3m2  1 3m2 13m7 30 A1 2 2 2 2 y2 3x and y 7 3x A1 3 5 4(d) 2814 ABC M1 Uses DEF  detM ABC . 2cm2 A1 Allow with units missing. 2 © Cambridge University Press & Assessment 2024 Page 9 of 16

4(a) [One way] stretch, rotation B1 Stretch followed by rotation B1 Correct order. Stretch parallel to the x-axis, scale factor 14 B1 Rotation, 1π [anticlockwise] about the origin. B1 3 4 © Cambridge University Press & Assessment 2024 Page 8 of 16 4(b)  1 1 3 1  1 1 3 14 0 1 1 1 0  B1 Both inverses correct.  2 2   2 2  ,      1 3 1   1 3 1   0 1 140 14     2 2 2 2  1 0 1 1 3 B1 Correct order. M1  14  2 2  0 1   1 3 1   2 2 2 4(c)  7 1 3 B1 M 2   7 3 1    2  7 1 3x 7x1 3y B1 x X    7 3 2 1 2      y     7 3x 2  1 2 y    Transforms  y   to  Y   7 3x1mxm  7x1 3mx  M1 Uses ymx and Y mX . 2 2 7 31m7m1 3m2  1 3m2 13m7 30 A1 2 2 2 2 y2 3x and y7 3x A1 3 5 4(d) 2814 ABC M1 Uses DEF  detM ABC . 2cm2 A1 Allow with units missing. 2 © Cambridge University Press & Assessment 2024 Page 9 of 16

1(a) 5 2 6 2 M1 A1 Evaluates determinant, forms quadratic k  k(5k6)(6k2)5k2 2 expression. 3 k 1 k (Allow 1 slip in the calculation of the determinant for M1 only) No real value of k  Non-singular A1 Convincing conclusion using the discriminant or determinant. 3 1(b) 3k11 or  1  1 M1 Uses AA1 Iordet(A1)(detA)1to 2 5k2 2 find equation in k. Could also use a minor determinant e.g. k1300 k 0 A1 Only k = 0, A0 for any additional solutions. 2 © Cambridge University Press & Assessment 2024 Page 6 of 16

3(a) [One-way] stretch, shear B1 Both types. Stretch followed by shear B1 Correct order. Stretch parallel to the x-axis, scale factor 7 B1 Shear, x-axis fixed, with (0,1) mapped to (2,1). B1 4 © Cambridge University Press & Assessment 2024 Page 8 of 16 3(b) 7 2 B1 M  0 1 7 2x 7x2y B1 FT x X     Transforms   to  . 0 1y  y  y Y  mxm7x2mx M1 Uses ymx and Y mX . 2m2 6m0 A1 y0 and y3x A1 SCB1 if M0 and both straight lines correct 5 3(c) Area of PQR  |7|Area of DEF M1 Area of DEF5cm2 A1 SCB1 for an answer of 245 (35  7) 2 © Cambridge University Press & Assessment 2024 Page 9 of 16

1(a) k 0 B1 [Stretch parallel to the x-axis, scale factor k   (k0)]. (Allow without identification.) 0 1 1 k B1 [Shear, x-axis fixed, with (0,1) mapped to   0 1 (k,1).] (Allow without identification.) 1 kk 0 k k M1 A1 Correct order for M1, must have identified M=  =  which matrix gives which transformation, AG. 0 10 1 0 1 4 1(b) k kx kx+ky B1 x X   =  Transforms   to   0 1y  y  y Y  kx+ky=x M1 X x Sets  =  Y  y 1−k A1 y= x oe k 3 1(c) 11 −k B1 (An alternative is possible.) M−1=   k0 k  1 © Cambridge University Press & Assessment 2024 Page 6 of 15 1(d) |k|=3k2 M1 Uses that detM=k. Without modulus is SC B1. k 0k =1 A1 3 2 Question Answer Marks Guidance

4(a) 1 2 3 0 −2 −2 4  M1 A1 Multiplying two matrices correctly, −2 −1 1   −2 −1 1  correct dimensions.   2 1 3  −1 3  =   −1 −1   1 1 3    1 1 3  3 2 5 0 0 −2 0      1 2 3 0 −2  0 −2 −2 −1 1   −1 −3 −4  Or   2 1 3  −1 3  =   −1 3   1 1 3   12 9 21  3 2 5 0 0 0 0       3 −7 B1 Convincingly completing matrix =   multiplication, AG. −9 3  3 4(b)  3 −7x 3x−7y  B1 x X    =   Transforms   to  . −9 3 y −9x+3y y Y  −9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX . −9+3m=3m−7m2 7m2 =9 A1 y= 3 x and y=− 3 x A1 7 7 5 4(c) Stretch B1 parallel to the x-axis, scale factor 3. B1 2 © Cambridge University Press & Assessment 2024 Page 8 of 16 4(d) 11 0 B1 M−1 =   30 3 1 3 −71 0  3 −71 0 M1 Correct order N=    or N=   3  3−9 3 0 3 −9 3 0 1  1 −7 A1 =   −3 3  3 Question Answer Marks Guidance

1(a) k 0 B1 [Stretch parallel to the x-axis, scale factor k   (k0)]. (Allow without identification.) 0 1 1 k B1 [Shear, x-axis fixed, with (0,1) mapped to   0 1 (k,1).] (Allow without identification.) 1 kk 0 k k M1 A1 Correct order for M1, must have identified M=  =  which matrix gives which transformation, AG. 0 10 1 0 1 4 1(b) k kx kx+ky B1 x X   =  Transforms   to   0 1y  y  y Y  kx+ky=x M1 X x Sets  =  Y  y 1−k A1 y= x oe k 3 1(c) 11 −k B1 (An alternative is possible.) M−1=   k0 k  1 © Cambridge University Press & Assessment 2024 Page 6 of 15 1(d) |k|=3k2 M1 Uses that detM=k. Without modulus is SC B1. k 0k =1 A1 3 2 Question Answer Marks Guidance

4(a) A rotation, followed by a shear. B2 Award B1 if given in the wrong order. [anticlockwise, centred at the origin, through an angle], [x-axis fixed, with (0,1) mapped to (2,1).] 2 © Cambridge University Press & Assessment 2025 Page 11 of 18 4(b) cos+2sin 2cos−sin B1 M=   sin cos  cos+2sin 2cos−sinx xcos+2xsin−ysin+2ycos B1 x X   =  Transforms   to  .  sin cos y  xsin+ ycos  y Y  xsin+ ycos= y AND xcos+2xsin− ysin+2ycos=x M1* Uses X =x and Y = y. xsin xcos+2xsin+ (2cos−sin)=x 1−cos A1 Combines equations correctly to eliminate y or x. (cos+2sin)(1−cos)+2sincos−sin2=1−cos DM1 Simplifies. A1 cos−cos2+2sin−sin2=1−cossin+cos=1 = 1π A1 2 7 © Cambridge University Press & Assessment 2025 Page 12 of 18

4(a) A rotation, followed by a shear. B2 Award B1 if given in the wrong order. [anticlockwise, centred at the origin, through an angle], [x-axis fixed, with (0,1) mapped to (2,1).] 2 © Cambridge University Press & Assessment 2025 Page 11 of 18 4(b) cos+2sin 2cos−sin B1 M=   sin cos  cos+2sin 2cos−sinx xcos+2xsin−ysin+2ycos B1 x X   =  Transforms   to  .  sin cos y  xsin+ ycos  y Y  xsin+ ycos= y AND xcos+2xsin− ysin+2ycos=x M1* Uses X =x and Y = y. xsin xcos+2xsin+ (2cos−sin)=x 1−cos A1 Combines equations correctly to eliminate y or x. (cos+2sin)(1−cos)+2sincos−sin2=1−cos DM1 Simplifies. A1 cos−cos2+2sin−sin2=1−cossin+cos=1 = 1π A1 2 7 © Cambridge University Press & Assessment 2025 Page 12 of 18

1(a) 14 0 B1 Stretch parallel to the x-axis, scale factor k   (k 0).  0 1  1 − 3 B1 Rotation anticlockwise about the origin through  2 2  1π. angle  3 1  3   2 2  1 − 3 14 0  7 − 3 M1 Correct order. M= 2 2   = 2   3 1   0 1  7 3 1   2 2   2   14 − 3 A1 AG 2M= .   14 3 1  4 © Cambridge University Press & Assessment 2025 Page 9 of 19 1(b)  7 − 3 x  7x− 3 y  B1 x X   7 3 1 2    y   =  7 3x+ 2 1 y   Transforms  y   to  Y    2   2  7 3x+1mx=m ( 7x− 3mx ) M1 Uses y =mx and Y =mX . 2 2 3 7 3+1m=7m− m2 2 2 3m2 −13m+14 3=0 A1 Correct quadratic AEF 7 A1 Correct solutions to the quadratic m=2 3 m= 3 A1 Correct invariant lines. If M0 then SCB1 for two y=2 3x y= 7 x 3 correct lines. 5 1(c)  1 3  1 3 M1 A1 SCB1 for finding the inverse of 2M M−1 = 1  2 2 = 1 14 14  14  2  −7 3 7  − 3 1  2 © Cambridge University Press & Assessment 2025 Page 10 of 19

5(a) 3 0 B1 One-way stretch in the x-direction, scale factor 3.   0 1 0 1 B1 Reflection in the line y= x.   1 0 0 13 0 0 1 B1 Correct order and final answer. M=  =  1 00 1 3 0 3 © Cambridge University Press & Assessment 2025 Page 12 of 18 5(b) Reflection in the line y= x B1 followed by B1 a [one-way] stretch in the x-direction, scale factor1. 3 B1 Allow scale or factor. 3 5(c) 0 1 B1FT Inverse of their M. M−1 = 3  This may be seen in part (b) 1 0 1 2 0 11 2 1 2 M1 A1 Multiplies on the left by the inverse of their M. N=M−1  = 3  = 3  May also use simultaneous equations. 3 2 1 03 2 1 2 3 5(d) 1 2x  x+2y  B1 x X    =  Transforms   to  . 3 2y 3x+2y y Y  3x+2mx=m(x+2mx) M1 Uses y=mx and Y =mX . 3+2m=m(1+2m)2m2 −m−3=0 A1 Forms quadratic in m. 3 A1 m= m=−1 2 y= 3x and y=−x A1 2 5 © Cambridge University Press & Assessment 2025 Page 13 of 18

2(a) Shear, [x-axis fixed] B1 With (0,1) mapped to (3,1 ) . B1 Or other suitable example/description. 2 2 © Cambridge University Press & Assessment 2025 Page 9 of 17 2(b) Shear, [y-axis fixed], B1 with (1,0) mapped to ( 1,3) . B1 Or other suitable example/description. 2 2 2(c) 13 3 M1 Finds AB or uses product of AB= 4 2  or detAB=detAdetB determinants. Or suitable reasoning that 3 1  2  area does not change. detAB=1 A1 2 2(d) 13 3x 13x+ 3 y B1 x X  4 2  = 4 2  Transforms   to   .  3 2 1 y   3 2 x+ y   y Y  3x+mx=m (13x+3mx ) M1A1 Uses Y =mX . 2 4 2 3+m=13m+3m2m2+3m−1=0 A1 2 4 2 2 y=−2x and y= 1 x. A1 If only values of m are given, A1 A0. 2 5 © Cambridge University Press & Assessment 2025 Page 10 of 17

4(a)(i) Stretch B1 parallel to the x-axis, scale factor k. B1 Allow in the x direction, x axis 2 © Cambridge University Press & Assessment 2025 Page 11 of 19 4(a)(ii) Enlargement, B1 centre the origin, scale factor k. B1 2 4(b)  0 1  0 m M1 A1 Multiplies two matrices correctly.  k 02 −1 1  2 −1 1 Allow one error for the M mark.  −1 1   =  −k m    0 m1 1 2  1 1 2  1 1  k m  m m 2m  A1   −2k+m k+m −k+2m      2k+m −k+m k+2m  k+m −k+2m −2k+m −k+2m −2k+m k+m M1 A1 Finds determinant. Allow two errors for m −m +2m the M mark −k+m k+2m 2k+m k+2m 2k+m −k+m =m(k+m)(k+2m)−m(−k+m)(−k+2m)−m(−2k+m)(k+2m) A1 AG. There must be evidence to show that +m(2k+m)(−k+2m)+2m(−2k+m)(−k+m)−2m(2k+m)(k+m) the determinant is equal to zero =6m2k+6m2k−12m2k =0 6 © Cambridge University Press & Assessment 2025 Page 12 of 19

2(a) Shear, [x-axis fixed] B1 With (0,1) mapped to (3,1 ) . B1 Or other suitable example/description. 2 2 © Cambridge University Press & Assessment 2025 Page 9 of 17 2(b) Shear, [y-axis fixed], B1 with (1,0) mapped to ( 1,3) . B1 Or other suitable example/description. 2 2 2(c) 13 3 M1 Finds AB or uses product of AB= 4 2  or detAB=detAdetB determinants. Or suitable reasoning that 3 1  2  area does not change. detAB=1 A1 2 2(d) 13 3x 13x+ 3 y B1 x X  4 2  = 4 2  Transforms   to   .  3 2 1 y   3 2 x+ y   y Y  3x+mx=m (13x+3mx ) M1A1 Uses Y =mX . 2 4 2 3+m=13m+3m2m2+3m−1=0 A1 2 4 2 2 y=−2x and y= 1 x. A1 If only values of m are given, A1 A0. 2 5 © Cambridge University Press & Assessment 2025 Page 10 of 17

5(a)(i) cos (1π−1π ) =cos (1π ) cos (1π ) +sin (1π ) sin (1π ) B1 Uses cos(A−B)=cosAcosB+sinAsinB , 4 6 4 6 4 6 Every step must be seen because AG. ( ) = 1 2 3+1 2 = 1 6+ 2 4 4 4 1 5(a)(ii) sin (1π−1π ) =sin (1π ) cos (1π ) −cos (1π ) sin (1π ) B1 Uses sin(A−B)=sinAcosB−cosAsinB , 4 6 4 6 4 6 Every step must be seen because AG. ( ) = 1 2 3−1 2= 1 6− 2 4 4 4 1 5(b) Reflection and rotation. B1 Correct order. B1 Reflection then rotation. Reflection in the x-axis. B1 Accept reflection in y=0. A rotation, 1 π [anticlockwise] about the origin. B1 Accept 15°. 12 4 © Cambridge University Press & Assessment 2025 Page 13 of 20 5(c) 1 ( 6+ 2 ) 1 ( 2− 6 ) −1 1 ( 6+ 2 ) 1 ( 6− 2 ) B1 Both correct inverses required. 4 4  4 4  = ,  1 ( 6− 2 ) 1 ( 6+ 2 )   1 ( 2− 6 ) 1 ( 6+ 2 )   cos ( π ) sin ( π ) 4 4  4 4   12 12  −sin ( π ) cos ( π ) 1 0  −1 1 0   12 12    =  0 −1 0 −1 1 ( 6+ 2 ) 1 ( 6− 2 ) B1 Correct order of their inverses. M−1 =   1 0   4 4  1 0  cos ( π ) sin ( π ) 0 −1  1 ( 2− 6 ) 1 ( 6+ 2 )  M−1 =  12 12  4 4  0 −1   −sin ( 1 π 2 ) cos ( 1 π 2 )  1 0  6+ 2 6− 2 Or M−1 = 4    0 −1 4    2− 6 6+ 2   2 © Cambridge University Press & Assessment 2025 Page 14 of 20 5(d) cos ( π ) sin ( π )  B1 M= 12 12  sin ( π ) −cos ( π )  12 12  cos ( π) sin ( π) x cos ( π) x+sin ( π) y B1 FT their M  12 12   = 12 12  x X   sin ( 1 π 2 ) −cos ( 1 π 2 ) y   sin ( 1 π 2 ) x−cos ( 1 π 2 ) y  Transforms  y   to  Y   sin (π) x−mcos (π) x=m ( cos (π) x+sin (π) mx ) M1 Uses y=mx and Y =mX . 12 12 12 12 sin (π) m2 +2mcos (π) −sin (π) =0 A1 AG. 12 12 12 ( ) ( ) ( ) 6− 2 m2 +2 6+ 2 m− 6− 2 =0 −2cos( π ) 4cos2( π )+4sin2( π ) M1 Applies quadratic formula andcos2(π) +sin2(π) =1. m= 12 12 12 12 12 2sin( π ) OE. 12 ( ) ( ) May see m=− 2+ 3  6+ 2 . m=−cot (π) cosec (π) A1 Or a=−1, b=1 12 12 6 © Cambridge University Press & Assessment 2025 Page 15 of 20

3(a)  k 0 0 1  B1 Stretch parallel to the x-axis, scale factor k.  1 0 2 1  B1 Shear, x-axis fixed, with (0,1) mapped to ( 2 ,1 ) . M =  1 0 2 1   k 0 0 1  M1 Correct order. M =  1 0   k k + + 2 1   0 0 1 0   0 0 + + 2 1   1 1  =  k 0 2 1  . A1 AG Some intermediate step must be shown. At least one element not in final form. 4 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(b) © Cambridge University Press & Assessment 2026 Page 12 of 23  k 0 2 1   x y  =  k x + y 2 y  B1 Transforms  x y  Must see matrix applied to both coordinates. kx+2y=x M1 x x Uses M =  . Must be considering a y y fixed point. y = 12 ( 1 − k ) x A1 3 3(c) Area of P = d e t M  Area of S 1= 1|k| M1 Uses 2 d e t M = k with P and S the correct way round. k =  2 A1 CAO 2 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3(a)  k 0 0 1  B1 Stretch parallel to the x-axis, scale factor k.  1 0 2 1  B1 Shear, x-axis fixed, with (0,1) mapped to ( 2 ,1 ) . M =  1 0 2 1   k 0 0 1  M1 Correct order. M =  1 0   k k + + 2 1   0 0 1 0   0 0 + + 2 1   1 1  =  k 0 2 1  . A1 AG Some intermediate step must be shown. At least one element not in final form. 4 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(b) © Cambridge University Press & Assessment 2026 Page 12 of 23  k 0 2 1   x y  =  k x + y 2 y  B1 Transforms  x y  Must see matrix applied to both coordinates. kx+2y=x M1 x x Uses M =  . Must be considering a y y fixed point. y = 12 ( 1 − k ) x A1 3 3(c) Area of P = d e t M  Area of S 1= 1|k| M1 Uses 2 d e t M = k with P and S the correct way round. k =  2 A1 CAO 2 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a) Shear, rotation. B1 Both words. Correct order. B1 Two transformations in the correct order with at least one correct and not both the same transformation. [x-axis fixed], with © Cambridge University Press & Assessment 2026 Page 10 of 20 ( 0 , 1 ) mapped to ( 2 , 1 ) . B1 For the description of the shear. Ignore any comment relating to scale or shear factor 14 π [anticlockwise] about the origin. B1 For rotation. Allow 45° 4 2(b) M = 12 2  1 1 1 3  B1 OE and must have a scalar multiple of 1 2. 2 12 2  1 1 1 3   x y  = 12 2  x x + + 3 y y  B1FT Transforms  x y  to  X Y  . Follow through for their M 12 2 ( x + 3 m x ) = m2 2 ( x + m x ) M1 Uses y=mx and Y = m X . x + 3 m x = m ( x + m x )  1 + 3 m = m + m 2  m 2 − 2 m − 1 = 0 A1 OE. y = ( 1 + 2 ) x and y = ( 1 − 2 ) x A1 If ambiguity over invariant points instead of invariant lines, then M0 but SCB1 for the correct equations. 5 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) 1 k 3 k 3 1 M1 Finds determinant and −k − =−1−k−k(−3−3k) 1 −1 3 −1 3 1 A1 forms correct quadratic equation. =3k2 +2k−1=0 © Cambridge University Press & Assessment 2026 Page 15 of 25 k = 13 , − 1 A1 No method need be seen for solving quadratic equation. 3 4(b) 1 k −11 0  1 0  M1 A1 Multiplies two matrices correctly. 0 1 0   3 1 k   Could see   3 1 k  0 −1  =  0 −1  1 0 1   4 k+1 −2  3 1 −11 0  1 0   0 1 1 0 0 1   3 0 + 2 k − − − k 1 1  . =  k + 2 3 − − k 1 − 1  A1 AG. Some intermediate working must be shown. For first pairing of matrices must see 4 - 2 For second pairing must see 0 + 2 3 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(c) © Cambridge University Press & Assessment 2026 Page 16 of 25  1 2 − 1 1   x y  =  2 x x − + y y  B1FT Transforms  x y  to  X Y  . Allow FT if left in terms of k. 2 x + m x + c = m ( x − m x − c ) + c M1 A1 Uses y = m x + c and Y =mX +c. M1 A0 if c = 0 throughout. ( m 2 + 2 ) x + m c = 0 A1 c must be present here. m2 +2=0 no real values of m. A1 Allow if c missing above. Accept ‘no solutions’. 5 Question Answer Marks Guidance