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

3 (a) Show that the system of equations x - 2y - 4z = 1, x - 2 y + kz = 1, - x + 2 y + 2 z = 1, where k is a constant, does not have a unique solution. [2] … … … … … … … … (b) Given that k =- 4 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … … … … … … … … (c) Given instead that k =- 2 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … … … … … … (d) For the case where k !- 2 and k !- 4 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … … … … … … …

9 marks

Mark scheme: 3(a) 1 2 4 1 2 4 2 2(2 ) 4(0) 0 1 2 2 − − − = −− + + − = − k k k M1 A1 Shows that determinant is zero. 2 3(b) 2 4 1, 2 4 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1, 2 3  = − − = − z x y (or 2 2 1 −+ + = x y z is not parallel to 2 4 1 − − = x y z ) B1 Derives one equation with two unknowns or states that third plane is not parallel to the repeated one. Two of the planes are identical (or coincident). B1 There is a line of intersection with the other plane. B1 3 3(c) 2 4 1, 2 2 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1 1 −= B1 Derives contradiction. Two parallel planes, not identical. B1 2 3(d) 2 4 1, 2 1, 2 2 1, − − = − + = −+ + = x y z x y kz x y z ( 4 ) 0, 2 2 0 2 −− = − =  = k z z B1 Derives contradiction. The three planes form a triangular prism. B1 2

This question in 9231/21 Oct/Nov 2020

7 The matrix P is given by 1 4 2 P = 0 - 1 1 f0 0 2p. (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 eigenvalues b, - 1, 1 with corresponding eigenvectors 1 4 2 0 - 1 1 f0p, f 0p, f2p, respectively. (c) Find A in terms of b. [4] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) − B1 1 7(b) 3 2 2 2 0 − − + = P P I P B1 States that P satisfies its characteristic equation. 1 2 2 2 −= + − P I P P M1 Multiplies through by 1 − P . 1 2 1 4 3 1 0 10 1 0 1 1 0 1 2 0 0 4 1 0 0 2 −     −          = −               = P P M1 A1 4 Question Answer Marks Guidance 7(c) 1 0 0 0 1 0 0 0 1 −     = −       b A P P M1 Applies 1 − = A PDP 1 4 3 4 2 1 0 1 1 0 1 2 0 0 2 1 0 0 2     − −       = −            b M1 A1 Multiplies two adjacent matrices. 4 4 3 1 0 1 1 0 0 1 + − −     −       b b b A1 4

This question in 9231/21 Oct/Nov 2020

3 (a) Show that the system of equations x - 2y - 4z = 1, x - 2 y + kz = 1, - x + 2 y + 2 z = 1, where k is a constant, does not have a unique solution. [2] … … … … … … … … (b) Given that k =- 4 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … … … … … … … … (c) Given instead that k =- 2 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … … … … … … (d) For the case where k !- 2 and k !- 4 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … … … … … … …

9 marks

Mark scheme: 3(a) 1 2 4 1 2 4 2 2(2 ) 4(0) 0 1 2 2 − − − = −− + + − = − k k k M1 A1 Shows that determinant is zero. 2 3(b) 2 4 1, 2 4 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1, 2 3  = − − = − z x y (or 2 2 1 −+ + = x y z is not parallel to 2 4 1 − − = x y z ) B1 Derives one equation with two unknowns or states that third plane is not parallel to the repeated one. Two of the planes are identical (or coincident). B1 There is a line of intersection with the other plane. B1 3 3(c) 2 4 1, 2 2 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1 1 −= B1 Derives contradiction. Two parallel planes, not identical. B1 2 3(d) 2 4 1, 2 1, 2 2 1, − − = − + = −+ + = x y z x y kz x y z ( 4 ) 0, 2 2 0 2 −− = − =  = k z z B1 Derives contradiction. The three planes form a triangular prism. B1 2

This question in 9231/23 Oct/Nov 2020

7 The matrix P is given by 1 4 2 P = 0 - 1 1 f0 0 2p. (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 eigenvalues b, - 1, 1 with corresponding eigenvectors 1 4 2 0 - 1 1 f0p, f 0p, f2p, respectively. (c) Find A in terms of b. [4] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) − B1 1 7(b) 3 2 2 2 0 − − + = P P I P B1 States that P satisfies its characteristic equation. 1 2 2 2 −= + − P I P P M1 Multiplies through by 1 − P . 1 2 1 4 3 1 0 10 1 0 1 1 0 1 2 0 0 4 1 0 0 2 −     −          = −               = P P M1 A1 4 Question Answer Marks Guidance 7(c) 1 0 0 0 1 0 0 0 1 −     = −       b A P P M1 Applies 1 − = A PDP 1 4 3 4 2 1 0 1 1 0 1 2 0 0 2 1 0 0 2     − −       = −            b M1 A1 Multiplies two adjacent matrices. 4 4 3 1 0 1 1 0 0 1 + − −     −       b b b A1 4

This question in 9231/23 Oct/Nov 2020

1 (a) Given that a is an integer, show that the system of equations ax + 3y + z = 14, 2 x + y + 3 z = 0, - x + 2y - 5z = 17, has a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … (b) Find the value of a for which x = 1, y = 4 , z = -2 is the solution to the system of equations in part (a). [1] … … … … … … … …

5 marks

Mark scheme: 1(a) 3 1 2 1 3 11 26 1 2 5 = − + − − a a M1 A1 Finds determinant. If solving equations x, y, z in terms in a. det 0 ≠ A leads to unique solution. A1 States that det 0. ≠ A (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) 4 = a B1 1

This question in 9231/21 May/June 2021

6 The matrix A is given by 5 - 223 8 A = 0 - 6 0 f0 0 1p. (a) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to find A3. [4] … … … … … … … …

11 marks

Mark scheme: 6(a) −and 1. B1 Upper diagonal matrix or characteristic equation. 22 3 88 1 5: 0 8 0 0 0 11 0 0 0 λ       = − =       −     i j k 154 3 22 3 2 6: 11 8 77 3 0 0 7 0 0 λ   −     = − − = −           i j k 22 3 56 2 1: 4 8 0 0 0 7 0 28 1 λ         = − =         − − −      i j k M1 Uses vector product (or equations) to find corresponding Eigenvectors. A1 A1 A1 A1 for each correct Eigenvector. Thus 1 2 2 0 3 0 0 0 1     =     −   P and 25 0 0 0 36 0 0 0 1     =       D M1 A1 Or correctly matched permutations of columns. (Accept scalar multiples of the eigenvectors shown here.) P must have at least two non-zero columns. 7 6(b) ( )( )( ) 3 5 6 1 31 30 0 λ λ λ λ λ − + − = − + = =0 B1 Finds characteristic equation. 3 3 3 30 31 1 30 = − +  − = I 0 I A A A A M1 Finds 3 A in terms of A. Allow missing I. 2 3 68 3 125 248 0 216 0 0 0 1   −   = −       A M1A1 Substitutes for A. 4

This question in 9231/21 May/June 2021

1 (a) Given that a is an integer, show that the system of equations ax + 3y + z = 14, 2 x + y + 3 z = 0, - x + 2y - 5z = 17, has a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … (b) Find the value of a for which x = 1, y = 4 , z = -2 is the solution to the system of equations in part (a). [1] … … … … … … … …

5 marks

Mark scheme: 1(a) 3 1 2 1 3 11 26 1 2 5 = − + − − a a M1 A1 Finds determinant. If solving equations x, y, z in terms in a. det 0 ≠ A leads to unique solution. A1 States that det 0. ≠ A (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) 4 = a B1 1

This question in 9231/22 May/June 2021

6 The matrix A is given by 5 - 223 8 A = 0 - 6 0 f0 0 1p. (a) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use the characteristic equation of A to find A3. [4] … … … … … … … …

11 marks

Mark scheme: 6(a) −and 1. B1 Upper diagonal matrix or characteristic equation. 22 3 88 1 5: 0 8 0 0 0 11 0 0 0 λ       = − =       −     i j k 154 3 22 3 2 6: 11 8 77 3 0 0 7 0 0 λ   −     = − − = −           i j k 22 3 56 2 1: 4 8 0 0 0 7 0 28 1 λ         = − =         − − −      i j k M1 Uses vector product (or equations) to find corresponding Eigenvectors. A1 A1 A1 A1 for each correct Eigenvector. Thus 1 2 2 0 3 0 0 0 1     =     −   P and 25 0 0 0 36 0 0 0 1     =       D M1 A1 Or correctly matched permutations of columns. (Accept scalar multiples of the eigenvectors shown here.) P must have at least two non-zero columns. 7 6(b) ( )( )( ) 3 5 6 1 31 30 0 λ λ λ λ λ − + − = − + = =0 B1 Finds characteristic equation. 3 3 3 30 31 1 30 = − +  − = I 0 I A A A A M1 Finds 3 A in terms of A. Allow missing I. 2 3 68 3 125 248 0 216 0 0 0 1   −   = −       A M1A1 Substitutes for A. 4

This question in 9231/22 May/June 2021

2 The matrix A is given by - 1 2 12 A = 0 1 0 f 0 0 3p. Use the characteristic equation of A to show that A 4 = pA 2 + qI , where p and q are integers to be determined. [6] … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 ( )( )( ) 3 2 1 1 3[ 0 3 3 ] λ λ λ λ λ λ − − − + = + = − 2 3 4 2 3 3 3 0 3 3 0 − − + =  − − + = A A A I A A A A M1 A1 Substitutes A and multiplies through by . A ( ) 2 4 2 2 3 3 3 3 10 9 = + − + − = − A A A I A A A I M1 A1 Substitutes 3. A 6

This question in 9231/21 Oct/Nov 2021

6 The matrix P is given by 1 6 6 P = 0 2 6 f0 0 - 3p. (a) Use the characteristic equation of P to find P -1 . [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the matrix A such that 4 0 0 P -1 AP = 0 5 0 [4] f0 0 6 p. … … … … … … … … … … … … … … … (c) State the eigenvalues and corresponding eigenvectors of A3. [2] … … … … … … … … …

11 marks

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

This question in 9231/22 Oct/Nov 2021

2 The matrix A is given by - 1 2 12 A = 0 1 0 f 0 0 3p. Use the characteristic equation of A to show that A 4 = pA 2 + qI , where p and q are integers to be determined. [6] … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 ( )( )( ) 3 2 1 1 3[ 0 3 3 ] λ λ λ λ λ λ − − − + = + = − 2 3 4 2 3 3 3 0 3 3 0 − − + =  − − + = A A A I A A A A M1 A1 Substitutes A and multiplies through by . A ( ) 2 4 2 2 3 3 3 3 10 9 = + − + − = − A A A I A A A I M1 A1 Substitutes 3. A 6

This question in 9231/23 Oct/Nov 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

2 (a) Show that the system of equations x - y + 2z = 4, x - y - 3z = a, x - y + 7z = 13, where a is a constant, does not have a unique solution. [2] … … … … … (b) Given that a = - 5 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … (c) Given instead that a ! - 5 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … …

7 marks

Mark scheme: 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 + 2 z = 4, M1 A1 M1 for row operations or eliminating one x − y − 3 z = −5,  z = 95 , x − y = 52 variable. A1 for deriving one correct equation with two x − y + 7 z = 13, unknowns. The three planes form a sheaf. B1 All three planes have the same common line. 3 2(c) x − y + 2 z = 4, B1 Derives contradiction. x − y − 3 z = a ,  z = 95 , 3 z + a = 4 − 2 z  a = −5 x − y + 7 z = 13, The three planes form a triangular prism. B1 2

This question in 9231/21 Oct/Nov 2022

1 (a) Find the set of values of k for which the system of equations x + 2y + 3z = 1, kx + 4 y + 6 z = 0, 7x + 8y + 9z = 3, has a unique solution. [3] … … … … … … … … … … … … … … … … (b) Interpret the situation geometrically in the case where the system of equations does not have a unique solution. [2] … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1(a) 1 2 3 M1 A1 Evaluates determinant or solves system of equations (eliminate at least one variable for M1): k 4 6 = 6 k − 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

This question in 9231/22 Oct/Nov 2022

2 (a) Show that the system of equations x - y + 2z = 4, x - y - 3z = a, x - y + 7z = 13, where a is a constant, does not have a unique solution. [2] … … … … … (b) Given that a = - 5 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … (c) Given instead that a ! - 5 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] … … … … … … … …

7 marks

Mark scheme: 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 + 2 z = 4, M1 A1 M1 for row operations or eliminating one x − y − 3 z = −5,  z = 95 , x − y = 52 variable. A1 for deriving one correct equation with two x − y + 7 z = 13, unknowns. The three planes form a sheaf. B1 All three planes have the same common line. 3 2(c) x − y + 2 z = 4, B1 Derives contradiction. x − y − 3 z = a ,  z = 95 , 3 z + a = 4 − 2 z  a = −5 x − y + 7 z = 13, The three planes form a triangular prism. B1 2

This question in 9231/23 Oct/Nov 2022

1 (a) Show that the system of equations x + 2y + 3z = 1, 4x + 5y + 6z = 1, 7x + 8y + 9z = 1, does not have a unique solution. [2] … … … … … … … … (b) Show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a)       5 1 2 3 4 5 6 5 2 4 3 4 7 7 9 8 6 8 9 9 6 8 7            3 12 9 0    obtain e.g. 2 3 1, 2 1, x y z y z      2 1(b) 2 3 1, 4 5 6 1, 7 8 9 1, x y z x y z x y z            1 3 1 2 2 1 z x y y x        M1 A1 Uses all three equations to reduce to one equation with two unknowns. Reducing to two equations scores M1 A0. The three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along a common line. 3

This question in 9231/21 May/June 2023

5 The matrix A is given by 18 5 - 11 A = 8 6 - 4 f32 10 - 20p. (a) Show that the characteristic equation of A is m 3 - 4m 2 - 20m + 48 = 0 and hence find the eigenvalues of A. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/21 May/June 2023

1 (a) Show that the system of equations x + 2y + 3z = 1, 4x + 5y + 6z = 1, 7x + 8y + 9z = 1, does not have a unique solution. [2] … … … … … … … … (b) Show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a)       5 1 2 3 4 5 6 5 2 4 3 4 7 7 9 8 6 8 9 9 6 8 7            3 12 9 0    obtain e.g. 2 3 1, 2 1, x y z y z      2 1(b) 2 3 1, 4 5 6 1, 7 8 9 1, x y z x y z x y z            1 3 1 2 2 1 z x y y x        M1 A1 Uses all three equations to reduce to one equation with two unknowns. Reducing to two equations scores M1 A0. The three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along a common line. 3

This question in 9231/22 May/June 2023

5 The matrix A is given by 18 5 - 11 A = 8 6 - 4 f32 10 - 20p. (a) Show that the characteristic equation of A is m 3 - 4m 2 - 20m + 48 = 0 and hence find the eigenvalues of A. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/22 May/June 2023

8 The matrix A is given by a - 6a 2 a + 2 A = 0 1 - a 0 f0 2 - a - 1 p where a is a constant with a ! 0 and a ! 1. x 1 (a) Show that the equation A y = 2 has a unique solution and interpret this situation geometrically. f z p f 3 p [3] … … … … … … … … … (b) Show that the eigenvalues of A are a, 1- a and - 1. [2] … … … … … … … … … … … … (c) Find a matrix P and a diagonal matrix D such that A 4 = PDP -1 . [6] … … … … … … … … … … … … … … … … … (d) Use the characteristic equation of A to find A4 in terms of A and a. [3] … … … … … … … … … 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) 6 2 2 0 1 0 ( 1) 0 2 0 1 a a a a a a a         or 2 2 ( 1) 12 2 7 1 1 2 1 1 1 2 1 2(2 ) 1 3 1 1 a a a a ax a a a y a a a z a a                   equations. 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) 6 2 2 0 1 0 0 0 2 1 a a a a a            M1 Equates determinant to zero or working to find characteristic equation.   )( 1 ) 0 , (1 1 , 1 a a a a              A1 AG. Factorisation must be clear. 2 Question Answer Partial Marks Guidance 8(c) 2 2 1 : 0 1 2 0 0 0 0 2 1 0 0 1 a a a a a a                       i j k  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. (2 )(2 2) 2 1: 1 6 2 2 0 0 0 2 0 1 ( 1)(2 ) a a a a a a a a                                 i j k  A1 2 2 2 4 2 1 : 2 1 6 2 2 4 2 10 5 2 5 2 2 2 1 0 2 1 a a a a a a a a a a a a                                   i j k  A1 1 2 2 0 0 1 0 1 1            P and 4 4 0 0 0 1 0 0 0 (1 ) a a            D M1 A1 Correctly matched permutations of columns. Their eigenvectors must be non-zero for M1. 6 8(d)   3 2 2 )( 1 ) 0 (1 ( ( 1) ) a a a a a a                 B1 Finds characteristic equation multiplied out. 2 4 2 2 1) ( ( ) a a a a      A A A M1 A1 Multiplies through by . A 3

This question in 9231/23 May/June 2023

1 Show that the system of equations 14x - 4y + 6z = 5 , x + y + kz = 3, - 21x + 6 y - 9z = 14, where k is a constant, does not have a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 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 6 k ) + 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

This question in 9231/21 Oct/Nov 2023

1 Show that the system of equations 14x - 4y + 6z = 5 , x + y + kz = 3, - 21x + 6 y - 9z = 14, where k is a constant, does not have a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 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 6 k ) + 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

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

1 Find the set of values of k for which the system of equations x + 5y + 6z = 1, kx + 2 y + 2 z = 2, - 3 x + 4 y + 8 z = 3, has a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 1 5 6 M1A1 Evaluates determinant. 2 2 k 2 k 2 k 2 2 = − 5 + 6 = 8 − 5 ( 8k + 6 ) + 6 ( 4 k + 6 ) 4 8 −3 8 −3 4 −3 4 8 k  78 A1 Three planes intersect at a unique point. B1 4

This question in 9231/21 Oct/Nov 2024

4 The matrix A is given by - 11 1 8 A = f 0 - 2 0p. - 16 1 13 1 (a) Show that f1p is an eigenvector of A and state the corresponding eigenvalue. [2] 1 … … … … … … … … (b) Show that the characteristic equation of A is m 3 - 19 m - 30 = 0 and hence find the other eigenvalues of A. [3] … … … … … … … … … … … … … (c) Use the characteristic equation of A to find A -1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 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 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− 15 = 0 ( −−2 )( ) 3 − 19− 30 = 0 = 5, − 3 B1 3 4(c) A 3 − 19 A − 30I = 0 B1 States that A satisfies its characteristic equation. 30 A −=1 A 2 − 19I M1 Multiplies through by A −1 .  −7 −5 16   −26 −5 16  M1A1 M1 for substituting A 2 . 2   −1 1   A = 0 4 0  A = 0 −15 0       30    −32 −5 41   −32 −5 22  4

This question in 9231/21 Oct/Nov 2024

1 Find the set of values of k for which the system of equations x + 5y + 6z = 1, kx + 2 y + 2 z = 2, - 3 x + 4 y + 8 z = 3, has a unique solution and interpret this situation geometrically. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 1 5 6 M1A1 Evaluates determinant. 2 2 k 2 k 2 k 2 2 = − 5 + 6 = 8 − 5 ( 8k + 6 ) + 6 ( 4 k + 6 ) 4 8 −3 8 −3 4 −3 4 8 k  78 A1 Three planes intersect at a unique point. B1 4

This question in 9231/23 Oct/Nov 2024

4 The matrix A is given by - 11 1 8 A = f 0 - 2 0p. - 16 1 13 1 (a) Show that f1p is an eigenvector of A and state the corresponding eigenvalue. [2] 1 … … … … … … … … (b) Show that the characteristic equation of A is m 3 - 19 m - 30 = 0 and hence find the other eigenvalues of A. [3] … … … … … … … … … … … … … (c) Use the characteristic equation of A to find A -1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 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 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− 15 = 0 ( −−2 )( ) 3 − 19− 30 = 0 = 5, − 3 B1 3 4(c) A 3 − 19 A − 30I = 0 B1 States that A satisfies its characteristic equation. 30 A −=1 A 2 − 19I M1 Multiplies through by A −1 .  −7 −5 16   −26 −5 16  M1A1 M1 for substituting A 2 . 2   −1 1   A = 0 4 0  A = 0 −15 0       30    −32 −5 41   −32 −5 22  4

This question in 9231/23 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 9231/21 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