Matrices (Further Pure 2)

1(a) a 3 1 M1 A1 Finds determinant. If solving equations x, y, z in 2 1 3 =−11a+26 terms in a. −1 2 −5 detA≠0 leads to unique solution. A1 States that detA≠0. (Since a is an integer.) (M1 A0 A0 if one of x, y, z found in terms of a.) (M1 A1 A0 if two of x, y, z found in terms of a.) The three planes intersect at a single point. B1 4 1(b) a=4 B1 1 Question Answer Marks Guidance

6(a) Eigenvalues of A are 5, −6and 1. B1 Upper diagonal matrix or characteristic equation. i j k 88 1 M1 Uses vector product (or equations) to find λ=5: 0 −22 8 =  0    0  corresponding Eigenvectors. 3     0 −11 0  0    0   A1 A1 A1 for each correct Eigenvector. i j k −154 2 A1 3     λ=−6: 11 −2 3 2 8 =  −77  3      0 0 7 0 0     i j k  56   2      λ=1: 4 −22 8 = 0  0 3     0 −7 0   −28     −1   1 2 2  25 0 0 M1 A1 Or correctly matched permutations of columns. Thus P=  0 3 0  and D=  0 36 0  (Accept scalar multiples of the eigenvectors     shown here.) P must have at least two non-zero  0 0 −1     0 0 1   columns. 7 6(b) (λ−5 )(λ+6 )(λ−1 )=λ3−31λ+30=0 =0 B1 Finds characteristic equation. A3 −31A +30I = 0 A3 =31A −30I M1 Finds A3 in terms of A. Allow missing I. 125 −682 248 M1A1 Substitutes for A.  3  A3 = 0 −216 0    0 0 1    4 © UCLES 2021 Page 10 of 13

1(a) a 3 1 M1 A1 Finds determinant. If solving equations x, y, z in 2 1 3 =−11a+26 terms in a. −1 2 −5 detA≠0 leads to unique solution. A1 States that detA≠0. (Since a is an integer.) (M1 A0 A0 if one of x, y, z found in terms of a.) (M1 A1 A0 if two of x, y, z found in terms of a.) The three planes intersect at a single point. B1 4 1(b) a=4 B1 1 Question Answer Marks Guidance

6(a) Eigenvalues of A are 5, −6and 1. B1 Upper diagonal matrix or characteristic equation. i j k 88 1 M1 Uses vector product (or equations) to find λ=5: 0 −22 8 =  0    0  corresponding Eigenvectors. 3     0 −11 0  0    0   A1 A1 A1 for each correct Eigenvector. i j k −154 2 A1 3     λ=−6: 11 −2 3 2 8 =  −77  3      0 0 7 0 0     i j k  56   2      λ=1: 4 −22 8 = 0  0 3     0 −7 0   −28     −1   1 2 2  25 0 0 M1 A1 Or correctly matched permutations of columns. Thus P=  0 3 0  and D=  0 36 0  (Accept scalar multiples of the eigenvectors     shown here.) P must have at least two non-zero  0 0 −1     0 0 1   columns. 7 6(b) (λ−5 )(λ+6 )(λ−1 )=λ3−31λ+30=0 =0 B1 Finds characteristic equation. A3 −31A +30I = 0 A3 =31A −30I M1 Finds A3 in terms of A. Allow missing I. 125 −682 248 M1A1 Substitutes for A.  3  A3 = 0 −216 0    0 0 1    4 © UCLES 2021 Page 10 of 13

8(a) 13 18 −28 M1 Finds determinant. −4 −a 8 =9a−24 2 6 −5 8 A1 24 a= Accept . 3 9 2 8(b) 13 18 −282  −2 B1      −4 −1 8 0 = 0 λ=−1.        2 6 −5    1   −1  1 © UCLES 2021 Page 13 of 15 8(c) 13−λ 18 −28 B1 Sets det ( A−λI )=0. −4 −1−λ 8 =0 2 6 −5−λ −λ3 +7λ2 −7λ−15=0 (λ+1 )(λ−3 )(λ−5 )=0 M1 Expands determinant and factorises. λ=−1, λ=3, λ=5 A1 i j k  −16 1 M1 Uses vector product (or equations) to find corresponding λ=3: −4 −4 8 =  −16    1  eigenvectors.     2 6 −8   −16    1   A1 A1 A1 for each correct Eigenvector. i j k  12   −1     λ=5: −4 −6 8 = −24  2     2 6 −10   −12    1   2 1 −1  −1 0 0 M1 A1 Or correctly matched permutations of columns. Accept scalar Thus P=  0 1 2  and D=  0 3 0  multiples of eigenvectors. At least two (non-zero) Eigenvectors     for M1.     1 1 1   0 0 5 8 © UCLES 2021 Page 14 of 15 8(d) −A3 +7A2 −7A−15I=0 M1 Substitutes A into characteristic equation and multiplies through by A −1. Allow missing I. A −1 = 1 ( −A2 +7A−7I ) A1 13 18 −28   15 Accept with A= −4 −1 8 and     2 6 −5    41 48 −80   A2 = −32 −23 64 substituted to give     −8 0 17   43 78 −116 A −1 = 1  4 9 −8  .   15  22 42 −59   2 © UCLES 2021 Page 15 of 15

2 (λ+1 )(λ−1 )(λ−3 )=λ3−3λ2 −λ+3[=0] M1 A1 Finds characteristic equation. A3 −3A2 −A+3I=0A4 −3A3 −A2 +3A=0 M1 A1 Substitutes A and multiplies through by A. ( ) A4 =3 3A2 +A−3I +A2 −3A=10A2 −9I M1 A1 Substitutes A3. 6 Question Answer Marks Guidance

6(a) (λ−1 ) (λ−2)(λ+3)=λ3−7λ+6[=0] M1 A1 Finds characteristic equation. P2 −7I+6P −1=0 M1 Substitutes P and multiplies through by P −1. 1 18 24 1 −3 −4 M1 A1 P2 =  0 4 −6  leading toP −1 =  0 1 1     2   0 0 9     0 0 −1 3  5 © UCLES 2021 Page 10 of 14 6(b) 4 0 0 M1 Applies A=PDP −1. A=P  0 5 0  P −1     0 0 6   4 30 36 1 −3 −4 1 6 6 4 −12 −16 M1 A1 Multiplies two adjacent matrices.       0 10 36 0 1 1 or 0 2 6 0 5 5   2    2   0 0 −18     0 0 −1 3     0 0 −3     0 0 −2   4 3 2  A1 A1 linked to first M1.   0 5 −2     0 0 6   4 6(c) 1 6  6  B1 Accept any non-zero multiples of these.       Eigenvectors: 0 , 2 , 6        0    0     −3   Eigenvalues: 64, 125, 216 B1 2 © UCLES 2021 Page 11 of 14

2 (λ+1 )(λ−1 )(λ−3 )=λ3−3λ2 −λ+3[=0] M1 A1 Finds characteristic equation. A3 −3A2 −A+3I=0A4 −3A3 −A2 +3A=0 M1 A1 Substitutes A and multiplies through by A. ( ) A4 =3 3A2 +A−3I +A2 −3A=10A2 −9I M1 A1 Substitutes A3. 6 Question Answer Marks Guidance

8(a) 3 a 0 M1 Sets determinant equal to zero. 5 1 0 010a60 1 3 2 a3 A1 5 2 8(b) Eigenvalues of A are 3, 1 and 2. B1 Lower diagonal matrix or characteristic equation. i j k  4  M1 A1 Uses vector product (or equations) to find corresponding   eigenvectors. 3: 5 4 0  5     1 3 1 19   i j k  0   0  A1     1: 4 0 0  12  1         1 3 3 12 1     i j k 0 0 A1     2: 1 0 0  0  0         1 3 0 3 1      4 0 0 9 0 0 M1 A1 Or correctly matched permutations of columns.     Thus P  5 1 0 and D 0 1 0         19 1 1 0 0 4     7 © UCLES 2022 Page 12 of 13 8(c) 312342 60 B1 Characteristic equation. A6IA34A2 M1 Substitutes for A and makes A6I the subject. A6I2   A3 4A22  A4A4I2 M1 A1 Squares and factorises. 4 © UCLES 2022 Page 13 of 13

8(a) 3 a 0 M1 Sets determinant equal to zero. 5 1 0 010a60 1 3 2 a3 A1 5 2 8(b) Eigenvalues of A are 3, 1 and 2. B1 Lower diagonal matrix or characteristic equation. i j k  4  M1 A1 Uses vector product (or equations) to find corresponding   eigenvectors. 3: 5 4 0  5     1 3 1 19   i j k  0   0  A1     1: 4 0 0  12  1         1 3 3 12 1     i j k 0 0 A1     2: 1 0 0  0  0         1 3 0 3 1      4 0 0 9 0 0 M1 A1 Or correctly matched permutations of columns.     Thus P  5 1 0 and D 0 1 0         19 1 1 0 0 4     7 © UCLES 2022 Page 12 of 13 8(c) 312342 60 B1 Characteristic equation. A6IA34A2 M1 Substitutes for A and makes A6I the subject. A6I2   A3 4A22  A4A4I2 M1 A1 Squares and factorises. 4 © UCLES 2022 Page 13 of 13

3(a) 6 9 5 M1 Sets determinant equal to zero. 5 8 5 0 1 1 2 3760 M1 Expands. 1,2,3 A1 A1 Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution. 4 © UCLES 2022 Page 7 of 15 3(b) A37A6I0 B1 States that A satisfies its characteristic equation. A2 7I6A1 0 M1 Multiplies through by A1. A1 1A2  7I A1 6 6 3 Question Answer Marks Guidance

2(a) 1 −1 2 M1 A1 Shows that determinant is zero. 1 −1 −3 =−7−3+10−2(0)=0 1 −1 7 2 2(b) x− y+ 2z= 4, M1 A1 M1 for row operations or eliminating one x− y− 3z= −5, z= 9, x− y= 2 variable. 5 5 A1 for deriving one correct equation with two x− y+ 7z= 13, unknowns. The three planes form a sheaf. B1 All three planes have the same common line. 3 © UCLES 2022 Page 6 of 13 2(c) x− y+ 2z= 4, B1 Derives contradiction. x− y− 3z= a, z= 9 ,3z+a=4−2za=−5 5 x− y+ 7z= 13, The three planes form a triangular prism. B1 2 Question Answer Marks Guidance

6(a) Eigenvalues of A are 2, 5 and −2. B1 Lower diagonal matrix or characteristic equation. i j k −12 1 M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. =2: 0 3 7 = 0 ~ 0         0 0 −4  0  0 i j k −21  1  A1     =5: −3 −3 −7 = 21 −1         0 0 7  0   0  i j k  28   1  A1     =−2: 4 −3 −7 = −28 −1         0 7 7  28   1  1 1 1  32 0 0  M1 A1 Or correctly matched permutations of columns.     Their eigenvectors must be non-zero and Thus P= 0 −1 −1 and D= 0 3125 0     correctly matched to their eigenvalues raised to     0 0 1   0 0 −32 the fifth power for M1. 7 6(b) (−2)(−5)(+2)=3−52−4+20=0 B1 Characteristic equation. A3 =5A2 +4A−20I M1 Substitutes for A and makes A3 the subject. Tolerate I missing. A4 =5A3+4A2 −20A=5 ( 5A2 +4A−20I ) +4A2 −20A M1 Multiplies by A and substitutes for A3. A4 =29A2 −100I A1 Has Iin answer or states the values of a and b. 4 © UCLES 2022 Page 10 of 13

1(a) 1 2 3 M1 A1 Evaluates determinant or solves system of equations (eliminate at least one variable for M1): k 4 6 =6k−12 7 8 9 2 4 k−8 x=− , y= , z= . k−2 k−2 3(k−2) k2 A1 3 1(b) Two parallel planes. Other plane not parallel. B1 SC k =2,two parallel planes. Parallel planes not identical. B1 Accept diagram. 2 Question Answer Marks Guidance

7(a) Ae=e leading to e=A-1e M1 Multiples by A-1on LHS. −1e=A-1e A1 2 7(b) i j k  9   1  M1 A1 Uses vector product (or equations) to find     corresponding eigenvector. =−1: 3 0 3 = 36 ~ 4         15 −3 3 −9 −1 2 7(c)  2 0 30  0  B1      15 −4 3 1 = −4 =−4            3 0 20  0   2 0 31  5  B1      15 −4 3 2 = 10 =5            3 0 21  5  2 7(d)  1 0 1 −1 0 0 M1 A1 Or correctly matched permutations of columns.     M1 for their eigenvectors (all three must be non- Thus P= 4 1 2 and D= 0 −1 0    4  zero) correctly matched to reciprocals of their  −1 0 1     0 0 1  eigenvalues. 5 2 © UCLES 2022 Page 12 of 15 7(e) (+1)(+4)(−5)=3−21−20=0 B1 Characteristic equation. A3−21A−20I=0 M1 Substitutes for A. Tolerate missing I. 20A−1 =A2 −21I leading toA−1= 1 A2 −21I M1 A1 Multiplies both sides of equation byA−1, need I for 20 20 A1. 4

2(a) 1 −1 2 M1 A1 Shows that determinant is zero. 1 −1 −3 =−7−3+10−2(0)=0 1 −1 7 2 2(b) x− y+ 2z= 4, M1 A1 M1 for row operations or eliminating one x− y− 3z= −5, z= 9, x− y= 2 variable. 5 5 A1 for deriving one correct equation with two x− y+ 7z= 13, unknowns. The three planes form a sheaf. B1 All three planes have the same common line. 3 © UCLES 2022 Page 6 of 13 2(c) x− y+ 2z= 4, B1 Derives contradiction. x− y− 3z= a, z= 9 ,3z+a=4−2za=−5 5 x− y+ 7z= 13, The three planes form a triangular prism. B1 2 Question Answer Marks Guidance

6(a) Eigenvalues of A are 2, 5 and −2. B1 Lower diagonal matrix or characteristic equation. i j k −12 1 M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. =2: 0 3 7 = 0 ~ 0         0 0 −4  0  0 i j k −21  1  A1     =5: −3 −3 −7 = 21 −1         0 0 7  0   0  i j k  28   1  A1     =−2: 4 −3 −7 = −28 −1         0 7 7  28   1  1 1 1  32 0 0  M1 A1 Or correctly matched permutations of columns.     Their eigenvectors must be non-zero and Thus P= 0 −1 −1 and D= 0 3125 0     correctly matched to their eigenvalues raised to     0 0 1   0 0 −32 the fifth power for M1. 7 6(b) (−2)(−5)(+2)=3−52−4+20=0 B1 Characteristic equation. A3 =5A2 +4A−20I M1 Substitutes for A and makes A3 the subject. Tolerate I missing. A4 =5A3+4A2 −20A=5 ( 5A2 +4A−20I ) +4A2 −20A M1 Multiplies by A and substitutes for A3. A4 =29A2 −100I A1 Has Iin answer or states the values of a and b. 4 © UCLES 2022 Page 10 of 13

1(a) 1 2 3 M1 A1 Shows that determinant is zero or row operations to obtain e.g. 4 5 6 59862497634875 x 2y 3z 1, 7 8 9 y 2z 1, 31290 2 1(b) x 2y 3z 1, M1 A1 Uses all three equations to reduce to one equation with 4x 5y 6z 1, z11x2y y2x1 two unknowns. Reducing to two equations scores M1 3 A0. 7x 8y 9z 1, The three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along a common line. 3 © UCLES 2023 Page 6 of 15

5(a) 18 5 11 M1 Sets determinant equal to zero. 8 6 4 0 32 10 20 6 4 8 4 8 6 A1 Expands determinant. Can use other rows/columns. 18 5 11 10 20 32 20 32 10 342 20480 A1 AG (2)(4)(6)0 leading to 2,4,6 B1 4 © UCLES 2023 Page 10 of 15 5(b) i j k  24   1  M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. Must attempt to solve 2: 16 5 11  24  1     equations for M1.     8 4 4  24   1  i j k 120 1 A1 A1     4: 8 10 4  0 ~ 0         32 10 16 240 2 i j k 20 1     6: 12 5 11  40 ~ 2         8 0 4 40 2  1 1 1 32 0 0  M1 A1 Or correctly matched permutations of columns.     M0 if a column of zeros appears in P. Thus P 1 0 2 and D 0 1024 0          1 2 2  0 0 7776 6 © UCLES 2023 Page 11 of 15

1(a) 1 2 3 M1 A1 Shows that determinant is zero or row operations to obtain e.g. 4 5 6 59862497634875 x 2y 3z 1, 7 8 9 y 2z 1, 31290 2 1(b) x 2y 3z 1, M1 A1 Uses all three equations to reduce to one equation with 4x 5y 6z 1, z11x2y y2x1 two unknowns. Reducing to two equations scores M1 3 A0. 7x 8y 9z 1, The three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along a common line. 3 © UCLES 2023 Page 6 of 15

5(a) 18 5 11 M1 Sets determinant equal to zero. 8 6 4 0 32 10 20 6 4 8 4 8 6 A1 Expands determinant. Can use other rows/columns. 18 5 11 10 20 32 20 32 10 342 20480 A1 AG (2)(4)(6)0 leading to 2,4,6 B1 4 © UCLES 2023 Page 10 of 15 5(b) i j k  24   1  M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. Must attempt to solve 2: 16 5 11  24  1     equations for M1.     8 4 4  24   1  i j k 120 1 A1 A1     4: 8 10 4  0 ~ 0         32 10 16 240 2 i j k 20 1     6: 12 5 11  40 ~ 2         8 0 4 40 2  1 1 1 32 0 0  M1 A1 Or correctly matched permutations of columns.     M0 if a column of zeros appears in P. Thus P 1 0 2 and D 0 1024 0          1 2 2  0 0 7776 6 © UCLES 2023 Page 11 of 15

8(a) 12a (a1)2 2a2 7a1 M1 A1 Shows that determinant is non-zero or solves ax1 2  equations. a 6a 2a2 1a 1a 1a 2 0 1a 0 a(a1)0 or y 1a 0 2a 1 2(2a) a1 z 3 1a 1a Three planes intersect at a single point. B1 Must be clear that there is just one point where all 3 planes intersect. 3 8(b) a 6a 2a2 M1 Equates determinant to zero or working to find characteristic equation. 0 1a 0 0 0 2a 1 a(1a)(1)0a,1a,1 A1 AG. Factorisation must be clear. 2 © UCLES 2023 Page 14 of 15 8(c) i j k 2a2 a1 1 M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. a: 0 12a 0  0    0      0 2a 1a  0  0 i j k (2a)(2a2) 2 A1     1: a1 6a 2a2  0  0         0 2a 0  (a1)(2a)   1  i j k 4a2 10a4 2 A1     1a: 2a1 6a 2a2  2a2 5a2  1       0 2a 2a  2a2 5a2   1    1 2 2 a4 0 0  M1 A1 Correctly matched permutations of columns. Their     eigenvectors must be non-zero for M1. P  0 0 1  and D 0 1 0   0 1 1     0 0 (1a)4   6 8(d) a(1a)(1)3 (a2 a1)(a2 a)0 B1 Finds characteristic equation multiplied out. A4 (a2 a1)A2 (a2 a)A M1 A1 Multiplies through by A. 3 © UCLES 2023 Page 15 of 15

1 14 −4 6 M1 A1 Evaluates determinant. Can expand along any row e.g. 1 k 1 k 1 1 −(36−36)+(−126+126)+k(84−84). 1 1 k =14 +4 +6 6 −9 −21 −9 −21 6 −21 6 −9 If using row operations, they must show an inconsistent =14(−9−6k)+4(−9+21k)+6(6+21)=0 system for M1. All their row operations must be correct for A1. Two parallel planes, not identical. B1 Other plane not parallel. B1 4 Question Answer Marks Guidance

7(a) =−6, =−2, =8 B1 i j k 8 1 M1 A1 Uses vector product (or equations) to find corresponding     eigenvectors. =−6: 0 4 5 = 0 0         0 0 2 0 0 i j k 10 1 A1     =−2: −4 2 13 = 20 2         0 0 5  0  0 i j k 140 2 A1     =8: −14 2 13 = 70 1         0 −10 5 140 2 1 1 2 −1 0 0 M1 A1 Or correctly matched permutations of columns. 6     M1 for their (non-zero) eigenvectors matched to their Thus P= 0 2 1 and D= 0 −1 0    2  eigenvalues.  0 0 2     0 0 1 8   7 7(b) A3−52A−96I=0 M1 Substitutes A into characteristic equation. 96A−1 =A2 −52I M1 Multiples through by A−1. 36 −16 36 B1   A2 = 0 4 30      0 0 64 © UCLES 2023 Page 11 of 14 7(b) −1 −1 3  A1 6 6 8   A−1 = 0 −1 5  2 16  0 0 1   8  4 Question Answer Marks Guidance

6(a) 1, 2, −1 B1 1 © UCLES 2023 Page 10 of 15 6(b) P3−2P2 −P+2I=0 B1 States that P satisfies its characteristic equation. 2P−1 =I+2P−P2 M1 Multiplies through by P−1. 1 −3 −1 1 1 3  M1 A1 2 2     P2 = 0 4 1 P−1 = 0 1 1    2 2      0 0 1 0 0 −1     4 6(c) 1 0 0 B1 a   D= 0 2 0     0 0 1   2 1 0 0 M1 Applies A−1 =PDP−1. a   A−1 =P 0 2 0 P−1     0 0 1   2 1 −2 1 1 1 3  M1 A1 Multiplies two adjacent matrices. a 2 2 2    = 0 4 1 0 1 1  2  2 2     0 0 −1 0 0 −1    2 1 1−2a 3−3a A1 a 2a 2a   0 2 3  2    0 0 1   2 5 © UCLES 2023 Page 11 of 15

1 14 −4 6 M1 A1 Evaluates determinant. Can expand along any row e.g. 1 k 1 k 1 1 −(36−36)+(−126+126)+k(84−84). 1 1 k =14 +4 +6 6 −9 −21 −9 −21 6 −21 6 −9 If using row operations, they must show an inconsistent =14(−9−6k)+4(−9+21k)+6(6+21)=0 system for M1. All their row operations must be correct for A1. Two parallel planes, not identical. B1 Other plane not parallel. B1 4 Question Answer Marks Guidance

7(a) =−6, =−2, =8 B1 i j k 8 1 M1 A1 Uses vector product (or equations) to find corresponding     eigenvectors. =−6: 0 4 5 = 0 0         0 0 2 0 0 i j k 10 1 A1     =−2: −4 2 13 = 20 2         0 0 5  0  0 i j k 140 2 A1     =8: −14 2 13 = 70 1         0 −10 5 140 2 1 1 2 −1 0 0 M1 A1 Or correctly matched permutations of columns. 6     M1 for their (non-zero) eigenvectors matched to their Thus P= 0 2 1 and D= 0 −1 0    2  eigenvalues.  0 0 2     0 0 1 8   7 7(b) A3−52A−96I=0 M1 Substitutes A into characteristic equation. 96A−1 =A2 −52I M1 Multiples through by A−1. 36 −16 36 B1   A2 = 0 4 30      0 0 64 © UCLES 2023 Page 11 of 14 7(b) −1 −1 3  A1 6 6 8   A−1 = 0 −1 5  2 16  0 0 1   8  4 Question Answer Marks Guidance

8(a) 6 a 0 M1 Finds determinant or solves equations. 2 1 0 8a24 1 5 4 a3 A1 2 8(b) If a3 then system has unique solution (so consistent). B1 FT 6x 3y  3, M1 Eliminates variable using all 3 equations. Or states there are two distinct equations and three If a3 then 2x y  1, 11x4z7 unknowns. Or gives accurate geometrical x 5y 4z 2, description with a3. So the system has infinitely many solutions (so consistent). A1 Alternative method for question 8(b) Setsy0. B1 1,0,3 is a solution so system is consistent for all values of a. M1 A1 Finds a solution with y0 and states 2 8 conclusion. 3 © Cambridge University Press & Assessment 2024 Page 13 of 14 8(c) Eigenvalues of A are 6, 1 and 4. B1 Lower diagonal matrix or characteristic equation. i j k 14 M1 A1 Uses vector product (or equations) to find   corresponding eigenvectors. 6: 2 7 0  4     1 5 2 17 i j k  0   0  A1     1: 2 0 0  10  1         1 5 5  10   1  i j k  0  0 A1     4: 2 5 0  0  0         1 5 0 15 1 14 0 0 11664 0 0  M1 A1 Or correctly matched permutations of columns.     Column of zeros in P gets M0. Thus P 4 1 0 and D 0 100 0         17 1 1  0 0 6400 7 8(d) 6143 92 14240 B1 Characteristic equation. 14A24IA3 9A2 M1 Substitutes for A and makes 14A24I the subject. 14A24I2   A3 9A22 A4A9I2 M1 A1 Squares and factorises. CWO. 4 © Cambridge University Press & Assessment 2024 Page 14 of 14

8(a) 6 a 0 M1 Finds determinant or solves equations. 2 1 0 8a24 1 5 4 a3 A1 2 8(b) If a3 then system has unique solution (so consistent). B1 FT 6x 3y  3, M1 Eliminates variable using all 3 equations. Or states there are two distinct equations and three If a3 then 2x y  1, 11x4z7 unknowns. Or gives accurate geometrical x 5y 4z 2, description with a3. So the system has infinitely many solutions (so consistent). A1 Alternative method for question 8(b) Setsy0. B1 1,0,3 is a solution so system is consistent for all values of a. M1 A1 Finds a solution with y0 and states 2 8 conclusion. 3 © Cambridge University Press & Assessment 2024 Page 13 of 14 8(c) Eigenvalues of A are 6, 1 and 4. B1 Lower diagonal matrix or characteristic equation. i j k 14 M1 A1 Uses vector product (or equations) to find   corresponding eigenvectors. 6: 2 7 0  4     1 5 2 17 i j k  0   0  A1     1: 2 0 0  10  1         1 5 5  10   1  i j k  0  0 A1     4: 2 5 0  0  0         1 5 0 15 1 14 0 0 11664 0 0  M1 A1 Or correctly matched permutations of columns.     Column of zeros in P gets M0. Thus P 4 1 0 and D 0 100 0         17 1 1  0 0 6400 7 8(d) 6143 92 14240 B1 Characteristic equation. 14A24IA3 9A2 M1 Substitutes for A and makes 14A24I the subject. 14A24I2   A3 9A22 A4A9I2 M1 A1 Squares and factorises. CWO. 4 © Cambridge University Press & Assessment 2024 Page 14 of 14

8(a) 12630d or 32630d M1 Substitutes point into x2y3z. 1 2 x2y3z13 A1 x2y3z9 A1 3 © Cambridge University Press & Assessment 2024 Page 13 of 15 8(b) x 1  2  x 1  1  M1 Forms vector equation of line.             y  6  12 or y  6  6              z   0   0   z   0   0              y6 A1 AG. CWO.  x1 ,z0  1x 1 y1,z0 6 2 12 2 8(c) 1 2 3 M1 Uses det(AI) (attempt an expansion). 1 2 3 1 1  2 12 (1) 21 2  3 3 1 12 0 A1 Expands determinant, expansion of brackets 4 2 12 2 shown, AG. CWO. (Must have ‘0’ here or 3 32 71 23 1 3 30 above.) 4 4 4 2 3 32  70 4 0, 7,1 B1 2 2 3 © Cambridge University Press & Assessment 2024 Page 14 of 15 8(d) i j k 1   3  M1 A1 Uses vector product (or equations) to find 4     corresponding eigenvectors. 0: 1 2 3  3 2   18      1 2 1 1 2 0   1 1 2 1   11  i j k 21 6 i j k 3 6 A1 A1  2     2     7 2 : 5 2 2 3 2 2 1~  6  1 2 : 3 2 2 3 3 2 ~  6          1 3 2 3  7 4  1 1 5 2 3  7 4   7   3 6 6 M1 A1 Or correctly matched permutations of columns.   Do not accept Thus P 18 6 6   n   0 0 0   11 1 7    D 0 7 0 . 0 0 0   2     0 0 1   and D  0 7n 0  2  2  M1 is for matching their eigenvectors correctly  0 0 1n with their eigenvalues.  2  M0 if a column of zeros appears in P. 6 © Cambridge University Press & Assessment 2024 Page 15 of 15

1 1 5 6 M1A1 Evaluates determinant. 2 2 k 2 k 2 k 2 2 = −5 +6 =8−5(8k+6)+6(4k+6) 4 8 −3 8 −3 4 −3 4 8 k  7 A1 8 Three planes intersect at a unique point. B1 4 Question Answer Marks Guidance

4(a) −11 1 8 1 −2 M1 Multiplies matrix with eigenvector.      0 −2 0 1 = −2           −16 1 131 −2 =−2 A1 2 © Cambridge University Press & Assessment 2024 Page 7 of 14 4(b) −11− 1 8 M1 Sets determinant equal to zero. 0 −2− 0 =0 −16 1 13− (−11−)(−2−)(13−)+128(−2−)=0 A1 Expands determinant, AG. (−2−)( 2 −2−15 ) =0 3 −19−30=0 =5,−3 B1 3 © Cambridge University Press & Assessment 2024 Page 8 of 14 4(c) A3−19A−30I=0 B1 States that A satisfies its characteristic equation. 30A−1 =A2 −19I M1 Multiplies through by A−1.  −7 −5 16 −26 −5 16 M1A1 M1 for substituting A2.   1   A2 = 0 4 0 A−1 = 0 −15 0     30     −32 −5 41 −32 −5 22 4 © Cambridge University Press & Assessment 2024 Page 9 of 14

8(a) −2− 0 0 B1 Sets determinant equal to zero. 0 7− 9 =0 4 1 7− (−2−) ( (7−)2 −9 ) =0(−2−)( 49−14+2 −9 ) M1A1 Expands, AG 3−122 +12+80=0 =−2,4,10 B1 4 8(b) A3−12A2 +12A+80I=0A4 −12A3+12A2 +80A=0 M1A1 Substitutes A and multiplies through by A. A4 =12 ( 12A2 −12A−80I ) −12A2 −80A=132A2 −224A−960I M1A1 Substitutes A3. 4 © Cambridge University Press & Assessment 2024 Page 12 of 13 8(c) i j k  72   2  *M1A1 Uses vector product (or equations) to find     corresponding eigenvectors. =−2: 0 9 9 = 36 1         4 1 9 −36 −1 i j k  0   0  i j k  0  0 A1A1         =4: 0 3 9 = 36 ~ 3 =10: −12 0 0 = 108 ~ 3                 4 1 3 −12 −1 0 −3 9  36  1  2 0 0 625 0 0  DM1A1 Or correctly matched permutations of     columns. Thus P= 1 3 3 and D= 0 1 0         −1 −1 1  0 0 2401 Correctly matched but eigenvalues not changed scores M1 A0. M0 if a column of zeros appears in P. 6 © Cambridge University Press & Assessment 2024 Page 13 of 13

1 1 5 6 M1A1 Evaluates determinant. 2 2 k 2 k 2 k 2 2 = −5 +6 =8−5(8k+6)+6(4k+6) 4 8 −3 8 −3 4 −3 4 8 k  7 A1 8 Three planes intersect at a unique point. B1 4 Question Answer Marks Guidance

4(a) −11 1 8 1 −2 M1 Multiplies matrix with eigenvector.      0 −2 0 1 = −2           −16 1 131 −2 =−2 A1 2 © Cambridge University Press & Assessment 2024 Page 7 of 14 4(b) −11− 1 8 M1 Sets determinant equal to zero. 0 −2− 0 =0 −16 1 13− (−11−)(−2−)(13−)+128(−2−)=0 A1 Expands determinant, AG. (−2−)( 2 −2−15 ) =0 3 −19−30=0 =5,−3 B1 3 © Cambridge University Press & Assessment 2024 Page 8 of 14 4(c) A3−19A−30I=0 B1 States that A satisfies its characteristic equation. 30A−1 =A2 −19I M1 Multiplies through by A−1.  −7 −5 16 −26 −5 16 M1A1 M1 for substituting A2.   1   A2 = 0 4 0 A−1 = 0 −15 0     30     −32 −5 41 −32 −5 22 4 © Cambridge University Press & Assessment 2024 Page 9 of 14

8(a) Ae=eA3e=A(Ae) M1 Multiples by A2on LHS. Trying to invert an eigenvector or missing eigenvector is M0.  A3e=A(e)=2(Ae)=3e A1 2 8(b) −1− 3 4 M1 Forms det(A−I)=0. Do not accept checking each 0 1− 0 =0 given value is an eigenvalue. Do not accept row operations on A. 0 −2 5− (−1−)((1−)(5−)−0 ) −3(0)+4(0)=0 (−1−)(1−)(5−)=0=−1,1 and 5. A1 AG. 2 8(c) i j k −8 1 M1 A1 Uses vector product (or equations) of rows of A−I     to find corresponding eigenvectors. =−1: 0 3 4 = 0 0         0 2 0  0  0 i j k 20 5 i j k 16 2 A1 A1         =1: −2 3 4 = 8 ~ 2 =5: −6 3 4 = 0 ~ 0                 0 −2 4  4  1 0 −4 0 24 3 1 5 2 −3 0 0 M1 A1 Or correctly matched permutations of columns.     A column of zeros is M0. Repeated eigenvalues in D is Thus P= 0 2 0 and D= 0 −1 0     M0.     0 1 3  0 0 3 6 © Cambridge University Press & Assessment 2025 Page 17 of 18 8(d) (A−2I)3 =A3−6A2 +12A−8I B1 Expands. =5A2+A−5I−6A2+12A−8I M1 Substitutes A3 =5A2+A−5I . =−A2+13A−13I A1 Alternative method for 8(d) (+3)(+1)(−3)=3+2 −9−9=0 B1 Finds characteristic equation of A−2I. (A−2I)3 =−(A−2I)2 +9(A−2I)+9I M1 SubstitutesA−2I and makes (A−2I)3 the subject. =−A2+13A−13I A1 3 © Cambridge University Press & Assessment 2025 Page 18 of 18

8(a) Ae=eA3e=A(Ae) M1 Multiples by A2on LHS. Trying to invert an eigenvector or missing eigenvector is M0.  A3e=A(e)=2(Ae)=3e A1 2 8(b) −1− 3 4 M1 Forms det(A−I)=0. Do not accept checking each 0 1− 0 =0 given value is an eigenvalue. Do not accept row operations on A. 0 −2 5− (−1−)((1−)(5−)−0 ) −3(0)+4(0)=0 (−1−)(1−)(5−)=0=−1,1 and 5. A1 AG. 2 8(c) i j k −8 1 M1 A1 Uses vector product (or equations) of rows of A−I     to find corresponding eigenvectors. =−1: 0 3 4 = 0 0         0 2 0  0  0 i j k 20 5 i j k 16 2 A1 A1         =1: −2 3 4 = 8 ~ 2 =5: −6 3 4 = 0 ~ 0                 0 −2 4  4  1 0 −4 0 24 3 1 5 2 −3 0 0 M1 A1 Or correctly matched permutations of columns.     A column of zeros is M0. Repeated eigenvalues in D is Thus P= 0 2 0 and D= 0 −1 0     M0.     0 1 3  0 0 3 6 © Cambridge University Press & Assessment 2025 Page 17 of 18 8(d) (A−2I)3 =A3−6A2 +12A−8I B1 Expands. =5A2+A−5I−6A2+12A−8I M1 Substitutes A3 =5A2+A−5I . =−A2+13A−13I A1 Alternative method for 8(d) (+3)(+1)(−3)=3+2 −9−9=0 B1 Finds characteristic equation of A−2I. (A−2I)3 =−(A−2I)2 +9(A−2I)+9I M1 SubstitutesA−2I and makes (A−2I)3 the subject. =−A2+13A−13I A1 3 © Cambridge University Press & Assessment 2025 Page 18 of 18

8(a) 3 3 8 M1 A1 Sets determinant equal to zero and forms 3 2 term quadratic equation. a 3 4 =08a2 −3a−9 =0 2 0 a −1 a= 3 ( 1 17 ) A1 16 3 8(b) ( −3) (−3)(+1)=3−72 +9 =0 B1 Finds characteristic equation. 2 2 2 9I= 7A2 −A3  9B= 7A−A2 M1 Using C-H Theorem and replaces  with A. 2 2 2 2 B2 = 7I− 2A M1 A1 Multiplies by B2 = A−2 9 9 CAO 4 © Cambridge University Press & Assessment 2025 Page 16 of 17 8(c) Eigenvalues of A are 3,3 and −1. B1 Upper diagonal matrix or characteristic 2 equation. i j k −9 1 M1 A1 Uses vector product (or equations) to find 2     corresponding eigenvectors. = 3 2 : 0 3 8 = 0   0  0 3 2 5 2   0    0   i j k 12 2 i j k −20  2  A1 A1         =3: −3 3 8 = 6 ~ 1 =−1: 5 3 8 = −10 ~ 1 2     2             0 0 4  0  0 0 4 4  10  −1 1 2 2  2 0 0  M1 A1 Or correctly matched permutations of 3     columns. Column of zeros or repeated Thus P= 0 1 1 and D= 0 1 0    3  column in P, or repeated eigenvalues in D is     0 0 −1  0 0 −1  M0. 7 © Cambridge University Press & Assessment 2025 Page 17 of 17

1(a) 1 2 3 M1 Expands determinant and forms quadratic k 5 6 =45−12k−2(9k−42)+3 ( 2k2 −35 ) expression. 7 2k 9 6k2 −30k+24=0k=1,4 M1 A1 Solves quadratic. Need both values for A1. 3 1(b) x+ 2y+3z =1, M1 A1 For M1 must use all three equations and reduce to 3y+3z=1 e.g. y = x. Accept explicit solution using a x+ 5y+6z =2,  3x+3z=1 parameter. Accept row echelon form (with clear 7x+2y+9z =3, 1 2 3 1   operations on all three rows) 0 3 3 1 .     0 0 0 0 Three planes meeting in a common line. B1 Sheaf. 3 © Cambridge University Press & Assessment 2025 Page 9 of 17

7(a) Eigenvalues of A are 1, 2 and −3. B1 Upper diagonal matrix or characteristic equation. i j k −4 1 M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. =1: 0 1 5 = 0 0         0 0 −4 0  0 i j k 35 7 A1     =2: −1 7 11 = 5 1         0 0 5  0  0 i j k −20  1  A1     =−3: 4 7 11 = −20 1         0 5 5  20  −1 1 7 1  1 0 0  M1 A1 Or correctly matched permutations of columns.     OE. Thus P= 0 1 1 and D= 0 64 0     A column of zeros or a repeated column is M0.     0 0 −1 0 0 729 7 7(b) (−1)(−2)(+3)=3 −7+6=0 B1 Characteristic equation. ‘=0’ can be implied later. A3 =7A−6I M1 Substitutes for A and makes A3 the subject. (Could substitute for A at the end). A6 =(7A−6I)(7A−6I)=49A2 −84A+36I M1 A1 Expands (7A−6I)(7A−6I). 4 © Cambridge University Press & Assessment 2025 Page 15 of 17

1 1 −1 1 M1 A1 Evaluates determinant. k 3 1 3 1 k 1 k 3 = + + =k2 −7 2 k 1 k 1 2 1 2 k k = 7 A1 3 Question Answer Marks Guidance

6(a) a, 2a, −3a B1 1 6(b)(i) a2 3a −2a−1 B1   P2 = 0 4a2 a     0 0 9a2    1 6(b)(ii) P3−7a2P+6a3I=0 B1 States that P satisfies its characteristic equation. 6a3P−1 =7a2I−P2 M1 Multiplies through by P−1. a2 3a −2a−1 6a2 −3a 2a+1 A1   1   P2 = 0 4a2 a P−1 =  0 3a2 −a    6a3    0 0 9a2   0 0 −2a2      3 © Cambridge University Press & Assessment 2025 Page 13 of 19 6(c) 1 0 0 B1   D= 0 2 0     0 0 3 1 0 0 M1 Applies A=PDP−1.   A=P 0 2 0 P−1     0 0 3 a 2 3  6a2 −3a 2a+1 M1 A1 Multiplies two adjacent matrices. For M1, P must 1     correct. = 0 4a −3  0 3a2 −a  6a3     0 0 −9a  0 0 −2a2    1 1 −4a+1 A1 OE.  2a 6a2  0 2 1  3a   0 0 3    5 © Cambridge University Press & Assessment 2025 Page 14 of 19

8(a) 1 −1 2k M1 Finds determinant equal and forms quadratic equation. Condone one error in the determinant k 1 2 =02k2 +3k+1=0 calculation. 2 −1 1 2k2+3k+1=0k =−1,−1 A1 A1 Each value is A1. 2 3 8(b) x− y− z = 1, M1 A1 Derives contradiction clearly. 1 x+z =3, −1 x+ y+2z =2, 2 2 x+2z =2, 2x− y+ z = 3, Triangular prism. B1 Accept diagram. 3 8(c) x− y−2z =1, M1 A1 Derives contradiction. −x+ y+2z =2, 1=−2 2x− y+ z = 3, Two parallel planes. Other plane not parallel. B1 Parallel planes not identical. B1 Accept diagram. 4 © Cambridge University Press & Assessment 2025 Page 16 of 17 8(d) 1− −1 −2  M1 Finds characteristic equation of A.   det −1 1− 2 =0     2 −1 1−   −3+32 −8=0 A1 Must be an equation. Could be implied by subsequent working. Must be simplified. A3 =3A2 −8A M1 Substitutes matrix and makes A3 the subject. A4 =3A3−8A2 =3 ( 3A2 −8A ) −8A2 M1 Multiplies by A and substitutes for A3. A4 =A2 −24A A1 5 © Cambridge University Press & Assessment 2025 Page 17 of 17

1 1 −1 1 M1 A1 Evaluates determinant. k 3 1 3 1 k 1 k 3 = + + =k2 −7 2 k 1 k 1 2 1 2 k k = 7 A1 3 Question Answer Marks Guidance

6(a) a, 2a, −3a B1 1 6(b)(i) a2 3a −2a−1 B1   P2 = 0 4a2 a     0 0 9a2    1 6(b)(ii) P3−7a2P+6a3I=0 B1 States that P satisfies its characteristic equation. 6a3P−1 =7a2I−P2 M1 Multiplies through by P−1. a2 3a −2a−1 6a2 −3a 2a+1 A1   1   P2 = 0 4a2 a P−1 =  0 3a2 −a    6a3    0 0 9a2   0 0 −2a2      3 © Cambridge University Press & Assessment 2025 Page 13 of 19 6(c) 1 0 0 B1   D= 0 2 0     0 0 3 1 0 0 M1 Applies A=PDP−1.   A=P 0 2 0 P−1     0 0 3 a 2 3  6a2 −3a 2a+1 M1 A1 Multiplies two adjacent matrices. For M1, P must 1     correct. = 0 4a −3  0 3a2 −a  6a3     0 0 −9a  0 0 −2a2    1 1 −4a+1 A1 OE.  2a 6a2  0 2 1  3a   0 0 3    5 © Cambridge University Press & Assessment 2025 Page 14 of 19

8(a) 5 a 0 M1 Finds determinant. 20 −5 0 =−20a−25 2 −3 1 a  −5 A1 4 Three planes intersect at a single point. B1 3 8(b) 5x− 5 y =3, B1 Shows contradiction clearly. 4 If a =−5 then 20x− 5y =2,2=12 4 2x− 3y+ z = 1, Two distinct parallel planes. B1 Other plane not parallel. B1 3 © Cambridge University Press & Assessment 2025 Page 18 of 19 8(c) Eigenvalues of A are 5, −5 and 1. B1 Lower diagonal matrix or characteristic equation. i j k  40   1  M1 A1 Uses vector product (or equations) to find     corresponding eigenvectors. =5: 20 −10 0 = 80 ~ 2         2 −3 −4 −40 −1     i j k  0  0 A1     =−5: 20 0 0 = −120  2         2 −3 6 −60 1     i j k  0  0 A1     =1: 4 0 0 = 0  0         20 −6 0 −24 1      1 0 0 64 0 0  M1 A1 Or correctly matched permutations of     columns. Thus P = 2 2 0 and D= 0 4 0         −1 1 1 0 0 16     7 © Cambridge University Press & Assessment 2025 Page 19 of 19

8(a) −1 −2 1 M1 A1 Finds determinant in terms of a and b. a 1 −1 =4a+ab−b−2 0 b 2 a = b b + + 2 4 A1 [since b−4] 3 Trace ID: PC-80D328BA4A 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/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 8(b) © Cambridge University Press & Assessment 2026 Page 18 of 18 ( 1 ) ( 1 ) ( 2 ) 3 2 2 2 0       + − − = − − + = B1 Finds characteristic equation. A 4 − 2 A 3 − A 2 + 2 A = 0 M1 Replaces  with A and multiplies through by A . A4 =2 ( 2A2 +A−2I ) +A2 −2A=5A2 −4I M1 A1 Substitutes A 3 = 2 A 2 + A − 2 I . CWO 4 8(c) Eigenvalues of A are − 1 , 1 and 2 . B1 Upper diagonal matrix or characteristic equation. 1 : 0 0 0 2 1 3 0 0 6 1 0 0  = − i − j k =  −     M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. M0 if all three vectors are zero vectors. 1 : 0 2 0 2 1 1 2 2 0 ~ 1 0 1  = − i − j k − =  −   −  i j k  3   1  A1 A1     =2: −3 −2 1 = −3 ~ −1         0 −1 −1  3   1  Thus P =  1 0 0 1 − 0 1 1 − 1 1  and D =  − 0 0 1 0 1 0 1 0 0 2 8  M1 A1 Or correctly matched permutations of columns. A column of zeros or repeated column in P is M0. 7 Trace ID: PC-80D328BA4A 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/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

8(a) −1 −2 1 M1 A1 Finds determinant in terms of a and b. a 1 −1 =4a+ab−b−2 0 b 2 a = b b + + 2 4 A1 [since b−4] 3 Trace ID: PC-45F835F92B 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/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 8(b) © Cambridge University Press & Assessment 2026 Page 18 of 18 ( 1 ) ( 1 ) ( 2 ) 3 2 2 2 0       + − − = − − + = B1 Finds characteristic equation. A 4 − 2 A 3 − A 2 + 2 A = 0 M1 Replaces  with A and multiplies through by A . A4 =2 ( 2A2 +A−2I ) +A2 −2A=5A2 −4I M1 A1 Substitutes A 3 = 2 A 2 + A − 2 I . CWO 4 8(c) Eigenvalues of A are − 1 , 1 and 2 . B1 Upper diagonal matrix or characteristic equation. 1 : 0 0 0 2 1 3 0 0 6 1 0 0  = − i − j k =  −     M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. M0 if all three vectors are zero vectors. 1 : 0 2 0 2 1 1 2 2 0 ~ 1 0 1  = − i − j k − =  −   −  i j k  3   1  A1 A1     =2: −3 −2 1 = −3 ~ −1         0 −1 −1  3   1  Thus P =  1 0 0 1 − 0 1 1 − 1 1  and D =  − 0 0 1 0 1 0 1 0 0 2 8  M1 A1 Or correctly matched permutations of columns. A column of zeros or repeated column in P is M0. 7 Trace ID: PC-45F835F92B 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/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

8(a) EITHER: © Cambridge University Press & Assessment 2026 Page 17 of 19 6 a 2 − 1 4 b 2 − 1 2 1 − 8 = − 1 2 a b − 9 6 a − 6 b − 1 2 0 = 0 (M1A1 Finds determinant in terms of a and b. a = − 2 b( + b 2 + 0 8 ) [ b  − 8 . A1) ] OR: Alternative method for 8(a) ( 6 + 1 2 a ) x + 3 6 y = 2 5 (R1+12R2) ( 2 + 8 a ) x + ( b + 3 2 ) y = 1 9 (M1 Row eliminates (R3+8R2) z from two of the equations and forms an equation from resulting coefficients being in ratio. 6 2 + + 1 2 8 a a = b 3 + 6 3 2 A1 Correct ratio equation a = − 2 b( + b 2 + 0 8 ) A1) CAO [b−8.] Available marks 3 Trace ID: PC-D4137E1CBF 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/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 8(b) © Cambridge University Press & Assessment 2026 Page 18 of 19 − 6 2 x 14x − x − + 1 2 4 3 2 y y y − + − 1 2 z = z = 8 z = 1 2 3 , , ,  − 8 ( 2 ) = 3 B1 Must show an inconsistent system. Two distinct parallel planes. B1 B1 Accept a clear sketch for parallel planes and third plane intersecting both. Other plane not parallel. 3 8(c)  6 −12 −121 −6 1 M1 MultipliesAwith given eigenvector.        −1 4 1 0 = 0 =−6 0                2 −8 −8 1 −6 1 −6 A1 2 Trace ID: PC-D4137E1CBF 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/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 8(d) © Cambridge University Press & Assessment 2026 Page 19 of 19 6 2 1 4 1 2 8 8 1 1 2 3 2 2 3 6 7 2       − − − − − − − − = − + + − M1 Characteristic equation. 2 , 6 , 6  = − A1 2 : 4 1 1 2 2 1 1 2 1 2 8 4 3 2 1  = − i − j − k =  −    −  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 6 : 0 1 1 2 2 1 1 2 1 3 2 1 6 2 ~ 1 3 1  = − i − − j − k =  − −   − −  A1 Thus P =  3 2 − 1 − 1 − 3 1 1 0 1  and D =  2 0 0 0 6 0 0 0 − 6  M1 A1 Or correctly matched permutations of columns. M0A0 if a zero eigenvector is included. 7 Trace ID: PC-D4137E1CBF 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/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) © Cambridge University Press & Assessment 2026 Page 9 of 19 3 8 8 k 1 2 7 2 4 = 3 1 2 2 4 − k 8 8 2 4 + 7 8 8 1 2 = 0 − k ( 1 6 ) + 7 ( 8 ) M1A1 Evaluates determinant. For M1 allow sign errors. k  72 A1 Three planes intersect at a unique point. B1 4 1(b) 3x+ 7 y+ 7z= 1, B1 OE. Obtains a clear numerical contradiction. 2 −25x+7=1 8x+ y+ 2z= 2, y=2−2z−8x 1 = 6 CWO. Might see 1= 2 or 1=− 1 . Accept 25 22 −8x+4=3 8 25 8x+ 2y+ 4z= 3, 3 7 7 1 2   correct row echelon form e.g. 0 1 2 2  25   0 0 0 *   with * non zero. Triangular prism. B1 2 Trace ID: PC-F3F6C8368E 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/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a) EITHER: © Cambridge University Press & Assessment 2026 Page 10 of 19 d e t ( ) d e t a 5 1 1 ( a ) ( 1 ) 5      A − I =  − − −  = − − + (M1A1 Finds d e t ( )  A − I 2 ( a 2 ) ( 1 2 ) 5 0 a 7  =  − − + =  = A1 2 8 1 2 0 6    − + =  = A1) OR: Alternative method for 2(a)  a 5 − 1 1   x y  = 2  x y   a 5 x x − + y y = = 2 2 x y  ( a − 2 ) x = 5 x (M1A1 Uses that 2 is eigenvalue to form an equation in a. a = 7 A1 7 5 1 1 x y x y 7 5 x x y y x y ( 7 ) 5 1 6        −    =    − + = =  − = −  = A1) CWO. Available marks 4 Trace ID: PC-F3F6C8368E 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/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(b) 7 −1x x 7x−y=2x x 1 *M1A1 M0 if both of their answers are zero vectors.   =2   ~  i j −1 1 5 1 y y 5x+y=2y y 5 Accept = ~  and 5 −1 −5 5 © Cambridge University Press & Assessment 2026 Page 11 of 19 i 1 − j 1 =  − − 1 1  ~  1 1  . M1 for setting up and solving equations.  7 5 − 1 1   x y  = 6  x y   7 5 x x − + y y = = 6 6 x y   x y  ~  1 1  A1 Thus P =  1 5 1 1  32 0  DM1A1 Or correctly matched permutations of columns. and D= . M0 if a column of zeros appears in P. Accept  0 7776 2 5 and 6 5 in D. 5 Trace ID: PC-F3F6C8368E 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/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/