Matrices — Paper 2 · A Level Mathematics - Further 9231

8 (a) Find the values of a for which the system of equations 3x + y + z = 0, ax + 6y - z = 0, ay - 2z = 0, does not have a unique solution. [3] … … … … … The matrix A is given by 3 1 1 A = 0 6 - 1 f0 0 - 2p. (b) Use the characteristic equation of A to find the inverse of A2. [4] … … … … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) ( ) ( ) 2 3 1 1 6 1 0 3 12 2 0 0 2 − =  − + −− + = − a a a a a M1 2 5 36 0 4, 9 + − =  = − a a a M1 A1 3 8(b) 3 2 3)( 6)( 2 0 ( 7 36 ) λ λ λ λ λ − − + + = = − B1 ( ) 2 2 3 1 36 7 36 7 − = −  = − I A A A I A M1 ( ) 2 1 4 1 1 1 0 1 1 36 0 0 9 − − −     =       A M1 A1 4 Question Answer Marks 8(c) Eigenvalues of A are 3, 6 and 2 −. B1 4 1 λ 3: 0 1 1 0 0 0 3 1 0 0 −       = =       −    i j k  M1 A1 1 6: 3 1 1 3 0 0 1 0 λ −     = − = −     −   i j k 9 2: 5 1 1 5 0 8 1 40 λ −     = − =     −   i j k A1 A1 Thus 1 1 9 0 3 5 0 0 40 −     =       P and 243 0 0 0 7776 0 0 0 32     =     −   D or 5 5 5 3 0 0 0 6 0 0 0 2     =       −   D M1 A1 7

This question in 9231/21 May/June 2020

8 (a) Find the values of a for which the system of equations 3x + y + z = 0, ax + 6y - z = 0, ay - 2z = 0, does not have a unique solution. [3] … … … … … The matrix A is given by 3 1 1 A = 0 6 - 1 f0 0 - 2p. (b) Use the characteristic equation of A to find the inverse of A2. [4] … … … … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) ( ) ( ) 2 3 1 1 6 1 0 3 12 2 0 0 2 − =  − + −− + = − a a a a a M1 2 5 36 0 4, 9 + − =  = − a a a M1 A1 3 8(b) 3 2 3)( 6)( 2 0 ( 7 36 ) λ λ λ λ λ − − + + = = − B1 ( ) 2 2 3 1 36 7 36 7 − = −  = − I A A A I A M1 ( ) 2 1 4 1 1 1 0 1 1 36 0 0 9 − − −     =       A M1 A1 4 Question Answer Marks 8(c) Eigenvalues of A are 3, 6 and 2 −. B1 4 1 λ 3: 0 1 1 0 0 0 3 1 0 0 −       = =       −    i j k  M1 A1 1 6: 3 1 1 3 0 0 1 0 λ −     = − = −     −   i j k 9 2: 5 1 1 5 0 8 1 40 λ −     = − =     −   i j k A1 A1 Thus 1 1 9 0 3 5 0 0 40 −     =       P and 243 0 0 0 7776 0 0 0 32     =     −   D or 5 5 5 3 0 0 0 6 0 0 0 2     =       −   D M1 A1 7

This question in 9231/22 May/June 2020

3 The matrix A is given by 5 - 1 7 A = 0 6 0 f7 7 5 p. (a) Find the eigenvalues of A. [4] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to find A -1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 5 1 7 0 6 0 0 7 7 5 λ λ λ − − − = − B1 3 2 16 36 144 0 ( 2)( 6)( 12) 0 λ λ λ λ λ λ − + − − =  + − − = M1 2, 6, 12 λ = − A1 A1 4 Question Answer Marks 3(b) 3 2 16 36 144 0 − + − − = A A A I B1 1 2 144 16 36 −= − + − A A A I M1 2 1 74 38 70 5 9 7 1 0 36 0 0 4 0 24 70 70 74 7 7 5 − − −         =  =         −     A A M1 A1 4

This question in 9231/23 May/June 2020

9 It is given that a is a positive constant. (a) Show that the system of equations ax + ( 2a + 5) y + ( a + 1) z = 1, - 4y = 2, 3y - z = 3, has a unique solution and interpret this situation geometrically. [3] … … … … … … … … … … … … … … … … … … … … … … … The matrix A is given by a 2a + 5 a + 1 A = 0 - 4 0 f0 3 - 1 p. (b) Show that the eigenvalues of A are a, - 1 and - 4 . [2] … … … … … … … … … (c) Find a matrix P such that a 0 0 A = P 0 - 1 0 P -1 . [5] f0 0 - 4p … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) 0 2 5 1 0 4 0 4 0 3 1 + + ≠ − = − a a a a or 11 8 2 1 2 9 2 = + = − = − x a y z solves equations and finds at least two correct values. The three planes intersect at a single point. B1 3 9(b) 2 5 1 0 4 0 0 0 3 1 λ λ λ − + + −− = −− a a a M1 Equates determinant to zero. ( ) )( 1 ) 0 , 4 1 4 ( , λ λ λ λ −− =  = −− − −− a a A1 AG 2 Question Answer Marks Guidance 9(c) ( )( ) 4 1 1 : 0 4 0 0 0 0 3 1 0 0 λ −− −−      = −− =       −−     a a a a a i j k M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 3( 1) 1 1: 1 2 5 1 0 0 0 3 0 3( 1) 1 a a a a a λ +         = − + + + =         − − + −     i j k  A1 3 12 1 4 : 4 2 5 1 (3 12) 1 0 3 3 3 12 1 a a a a a a λ +         = − + + + = − + −         +     i j k  A1 1 1 1 0 0 1 0 1 1     = −     −   P A1 OE 5 Question Answer Marks Guidance 9(d) 3 2 ( (5 ) ( )( 1)( 4) 4 0 4 5 ) a a a a λ λ λ λ λ λ + − + − − + + = − = B1 Finds characteristic equation. 1 2 4 (5 ) (4 5 ) a a a −= + − + − A A A I M1 A1 Multiplies through by 1 − A . 2 2 2 2 2 17 1 0 16 0 0 15 1 a a a   − −     −  =   A B1 2 2 2 1 2 5 1 1 0 0 0 4 0 (4 5 ) 0 1 0 0 3 2 17 1 4 0 16 0 (5 ) 0 15 1 1 0 0 1 a a a a a a a a a −   − −  + −    + +         = − + −         −   −     A M1 Substitutes for and A in correct equation. 1 4 5 8 4 4 1 0 0 4 0 3 4 a a a a a a − + +     = −     − −   A A1 6

This question in 9231/22 Oct/Nov 2020

8 (a) Find the value of a for which the system of equations 13 x + 18 y - 28 z = 0, - 4x - ay + 8z = 0, 2x + 6y - 5z = 0, does not have a unique solution. [2] … … … … The matrix A is given by 13 18 - 28 A = - 4 - 1 8 f 2 6 - 5p. 2 (b) Find the eigenvalue of A corresponding to the eigenvector 0 [1] f1p. … … … (c) Find a matrix P and a diagonal matrix D such that A = PDP -1 . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (d) Use the characteristic equation of A to find A -1 in terms of A. [2] … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

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

This question in 9231/23 May/June 2021

8 (a) Find the value of a for which the system of equations 3x + ay = 0, 5x - y = 0, x + 3y + 2z = 0, does not have a unique solution. [2] … … … … … … The matrix A is given by 3 0 0 A = 5 - 1 0 f1 3 2p. (b) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … (c) Use the characteristic equation of A to show that ( A + 6I) 2 = A 4 ( A + bI) 2 , where b is an integer to be determined. [4] … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

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

This question in 9231/21 May/June 2022

8 (a) Find the value of a for which the system of equations 3x + ay = 0, 5x - y = 0, x + 3y + 2z = 0, does not have a unique solution. [2] … … … … … … The matrix A is given by 3 0 0 A = 5 - 1 0 f1 3 2p. (b) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … (c) Use the characteristic equation of A to show that ( A + 6I) 2 = A 4 ( A + bI) 2 , where b is an integer to be determined. [4] … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

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

This question in 9231/22 May/June 2022

6 The matrix A is given by 2 - 3 - 7 A = 0 5 7 f0 0 - 2p. (a) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to show that A 4 = aA 2 + bI , where a and b are integers to be determined. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 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. A 3 = 5 A 2 + 4 A − 20I M1 Substitutes for A and makes A 3 the subject. Tolerate I missing. A 4 = 5A3 + 4 A 2 − 20 A = 5 5A 2 + 4 A − 20I + 4 A 2 − 20 A M1 Multiplies by A and substitutes for A 3 . ( ) A 4 = 29 A 2 − 100I A1 Has I in answer or states the values of a and b. 4

This question in 9231/21 Oct/Nov 2022

7 (a) It is given that m is an eigenvalue of the non-singular square matrix A, with corresponding eigenvector e. Show that m-1 is an eigenvalue of A -1 for which e is a corresponding eigenvector. [2] … … … … The matrix A is given by 2 0 3 A = 15 - 4 3 f 3 0 2p. (b) Given that - 1 is an eigenvalue of A, find a corresponding eigenvector. [2] … … … … … 0 1 (c) It is also given that 1 and 2 are eigenvectors of A. Find the corresponding eigenvalues. [2] f0p f1p … … … … … … … … … … (d) Hence find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . [2] … … … … … … … (e) Use the characteristic equation of A to show that A -1 = pA 2 + qI , where p and q are rational numbers to be determined. [4] … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 7(a) Ae = e leading to e = A -1e M1 Multiples by A -1 on 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.    1  M1 for their eigenvectors (all three must be non- and D = Thus P = 4 1 2    0 − 4 0  zero) correctly matched to reciprocals of their    1   −1 0 1   0 0 5  eigenvalues. 2 7(e) (+ 1)(+ 4 )(− 5) = 3 − 21− 20 = 0 B1 Characteristic equation. A 3 − 21A − 20I = 0 M1 Substitutes for A. Tolerate missing I. 20 A −=1 A 2 − 21I leading to A −=1 201 A2 − 2021 I M1 A1 Multiplies both sides of equation by A − 1, need I for A1. 4

This question in 9231/22 Oct/Nov 2022

6 The matrix A is given by 2 - 3 - 7 A = 0 5 7 f0 0 - 2p. (a) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to show that A 4 = aA 2 + bI , where a and b are integers to be determined. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 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. A 3 = 5 A 2 + 4 A − 20I M1 Substitutes for A and makes A 3 the subject. Tolerate I missing. A 4 = 5A3 + 4 A 2 − 20 A = 5 5A 2 + 4 A − 20I + 4 A 2 − 20 A M1 Multiplies by A and substitutes for A 3 . ( ) A 4 = 29 A 2 − 100I A1 Has I in answer or states the values of a and b. 4

This question in 9231/23 Oct/Nov 2022

7 The matrix A is given by - 6 2 13 A = 0 - 2 5 f 0 0 8p. (a) Find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to find A -1 . [4] … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 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   − 16 0 0  M1 A1 Or correctly matched permutations of columns.    1  M1 for their (non-zero) eigenvectors matched to their and D = Thus P = 0 2 1    0 − 2 0  eigenvalues.    1   0 0 2   0 0 8  7 7(b) A3 − 52 A − 96I = 0 M1 Substitutes A into characteristic equation. 96 A −=1 A 2 − 52I M1 Multiples through by A −1.  36 −16 36  B1 2   A = 0 4 30      0 0 64  7(b)  − 16 − 16 83  A1 −1  1 5  A =  0 − 2 16   1   0 0 8  4

This question in 9231/21 Oct/Nov 2023

6 The matrix P is given by 1 - 1 1 P = 0 2 1 f 0 0 - 1 p. (a) State the eigenvalues of P. [1] … … (b) Use the characteristic equation of P to find P -1 . [4] … … … … … … … … … … … … … … … … … … … … … The 3 # 3 matrix A has distinct non-zero eigenvalues a, 1,2 2 with corresponding eigenvectors 1 - 1 1 0 , 2 , 1 , f 0 p f 0 p f - 1 p respectively. (c) Find A -1 in terms of a. [5] … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) 1, 2, − 1 B1 1 6(b) P 3 − 2 P 2 − P + 2I = 0 B1 States that P satisfies its characteristic equation. 2P −=1 I + 2P − P 2 M1 Multiplies through by P −1 .  1 −3 −1   1 12 23  M1 A1 2   −1  1 1  P = 0 4 1  P =  0 2 2        0 0 1 0 0 −1     4 6(c)  1a 0 0  B1   D = 0 2 0    1   0 0 2   1a 0 0  M1 Applies A −1 = PDP −1. − 1   −1 A = P 0 2 0 P    1   0 0 2   1a −2 12   1 12 23  M1 A1 Multiplies two adjacent matrices.  1   1 1  =  0 4 2   0 2 2   1    0 0 − 2  0 0 −1   1a 1−22aa 3 −2 3a a  A1  3   0 2 2   1   0 0 2  5

This question in 9231/22 Oct/Nov 2023

7 The matrix A is given by - 6 2 13 A = 0 - 2 5 f 0 0 8p. (a) Find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to find A -1 . [4] … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 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   − 16 0 0  M1 A1 Or correctly matched permutations of columns.    1  M1 for their (non-zero) eigenvectors matched to their and D = Thus P = 0 2 1    0 − 2 0  eigenvalues.    1   0 0 2   0 0 8  7 7(b) A3 − 52 A − 96I = 0 M1 Substitutes A into characteristic equation. 96 A −=1 A 2 − 52I M1 Multiples through by A −1.  36 −16 36  B1 2   A = 0 4 30      0 0 64  7(b)  − 16 − 16 83  A1 −1  1 5  A =  0 − 2 16   1   0 0 8  4

This question in 9231/23 Oct/Nov 2023

8 (a) Find the set of values of a for which the system of equations 6x + ay = 3, 2x - y = 1, x + 5y + 4z = 2 has a unique solution. [2] … … … … … … … … … … … (b) Show that the system of equations in part (a) is consistent for all values of a. [3] … … … … … … … … … … … … The matrix A is given by 6 0 0 A = 2 - 1 0 f 1 5 4 p. 2 -1(c) Find a matrix P and a diagonal matrix D such that 14A + 24 I = PDP . [7] ` j … … … … … … … … … … … … … … … … … … … … … … … … (d) Use the characteristic equation of A to show that 2 4 2 14A + 24I = A A + bI , ` j ` j where b is an integer to be determined. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

16 marks

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

This question in 9231/21 May/June 2024

8 (a) Find the set of values of a for which the system of equations 6x + ay = 3, 2x - y = 1, x + 5y + 4z = 2 has a unique solution. [2] … … … … … … … … … … … (b) Show that the system of equations in part (a) is consistent for all values of a. [3] … … … … … … … … … … … … The matrix A is given by 6 0 0 A = 2 - 1 0 f 1 5 4 p. 2 -1(c) Find a matrix P and a diagonal matrix D such that 14A + 24 I = PDP . [7] ` j … … … … … … … … … … … … … … … … … … … … … … … … (d) Use the characteristic equation of A to show that 2 4 2 14A + 24I = A A + bI , ` j ` j where b is an integer to be determined. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

16 marks

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

This question in 9231/22 May/June 2024

8 The matrix A is given by - 2 0 0 A = f 0 7 9p. 4 1 7 (a) Show that the characteristic equation of A is m 3 - 12 m 2 + 12m + 80 = 0 and find the eigenvalues of A. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to show that A 4 = pA 2 + qA + rI , where p, q and r are integers to be determined. [4] … … … … … … … … … … … … … … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that ( A - 3I) 4 = PDP -1 . [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) −−2  0 0 B1 Sets determinant equal to zero. 0 7 −  9 = 0 4 1 7 −  49 − 14+  − 9 ( −−2 ) 2 2 M1A1 Expands, AG ( 7 − ) − 9 = 0  ( −−2 )( ) ( )  3 − 122 + 12+ 80 = 0 = −2, 4,10 B1 4 8(b) A 3 − 12 A 2 + 12 A + 80I = 0  A 4 − 12 A 3 + 12 A 2 + 80 A = 0 M1A1 Substitutes A and multiplies through by A. A 4 = 12 12 A 2 − 12 A − 80I − 12 A 2 − 80 A = 132 A 2 − 224 A − 960I M1A1 Substitutes A3. ( ) 4 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

This question in 9231/22 Oct/Nov 2024

8 (a) It is given that m is an eigenvalue of the non-singular square matrix A, with corresponding eigenvector e. Show that e is an eigenvector of A3 with corresponding eigenvalue m 3 . [2] … … … … … … … … The matrix A is given by -1 3 4 A = f 0 1 0 p. 0 - 2 5 (b) Show that the eigenvalues of A are -1, 1 and 5. [2] … … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that A - 2I = PDP -1 . [6] … … … … … … … … … … … … … … … (d) Use the characteristic equation of A to show that ( A - 2 I) 3 = aA 2 + bA + cI where a, b and c are constants to be determined. [3] … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 8(a) Ae = e  A 3 e = A ( Ae ) M1 Multiples by A 2 on LHS. Trying to invert an eigenvector or missing eigenvector is M0.  A 3 e = A (e ) = 2 ( Ae ) = 3 e 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 8(d) ( A − 2I )3 = A3 − 6 A 2 + 12 A − 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 − 2 I . ( A − 2I )3 = − ( A − 2I ) 2 + 9 ( A − 2I ) + 9I M1 Substitutes A − 2 I and makes ( A − 2I )3 the subject. = − A2 + 13A − 13I A1 3

This question in 9231/22 May/June 2025

8 (a) Find the values of a for which the system of equations 3 x + 3 y + 8 z = 1 , 2 ax + 3y + 4z = 2 , ay - z = 3, does not have a unique solution. [3] … … … … … … … … The matrix A is given by 3 3 8 2 A = f 0 3 4 p. 0 0 -1 (b) Given that B = A -1 , use the characteristic equation of A to show that B 2 = p I + q A , where p and q are constants to be determined. [4] … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) 3 2 3 8 M1 A1 Sets determinant equal to zero and forms 3 2 9 term quadratic equation. a 3 4 = 0  8 a − 3a − 2 = 0 0 a −1 3 A1 1 17 a = 16 ( ) 3 38(b) (− 2 ) (− 3)(+ 1) =  3 − 72 2 + 92 = 0 B1 Finds characteristic equation. 9 2 I = 72 A 2 − A 3  92 B = 72 A − A 2 M1 Using C-H Theorem and replaces  with A. B 2 = 79 I − 92 A M1 A1 Multiplies by B 2 = A −2 CAO 4 8(c) Eigenvalues of A are 32 , 3 and −.1 B1 Upper diagonal matrix or characteristic equation. i j k  − 92  1 M1 A1 Uses vector product (or equations) to find 3    corresponding eigenvectors. = 2 : 0 3 8 =  0  0 3 5    0 2 2  0  0 i j k  12  2 i j k  −20   2  A1 A1 3    5     = 3: − 2 3 8 =  6  ~ 1 = −1: 2 3 8 =  −10  ~  1         0 0 4  0  0 0 4 4  10   −1   1 2 2   23 0 0  M1 A1 Or correctly matched permutations of    1  columns. Column of zeros or repeated 0 1 1 Thus P = and D =  0 3 0        column in P, or repeated eigenvalues in D is  0 0 −1   0 0 −1  M0. 7

This question in 9231/23 May/June 2025