AS & A Level Further Mathematics (9231)
Matrices (eigenvalues and eigenvectors) Banker
11 questions · about 6% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Diagonalisation · Core · Paper 2 style · 6 marksMatrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrices P = (1 1 0; 0 1 1; 0 0 1) and D = diag(2, −1, 3). Find P⁻¹. [2]
- (b) Hence find the matrix A = PDP⁻¹. [2]
- (c) State the eigenvalues of A and an eigenvector corresponding to the eigenvalue 3. [2]
Show answer
- (a) P⁻¹ = (1 −1 1; 0 1 −1; 0 0 1)
M1 det P = 1 and cofactors; A1.
- (b) A = (2 −3 3; 0 −1 4; 0 0 3)
M1 correct order of multiplication; A1.
- (c) 2, −1, 3; (0, 1, 1)ᵀ
B1 eigenvalues are the entries of D; B1 third column of P.
- (a) The 3 × 3 matrix A has eigenvalues 2, −1 and 4, with corresponding eigenvectors e₁, e₂ and e₃. State the eigenvalues of the matrix A + 3I and a corresponding eigenvector for each. [2]
- (b) State the eigenvalues of A² and of A⁻¹. [2]
- (c) Given that A = PDP⁻¹ with D = diag(2, −1, 4), find det A. [1]
Show answer
- (a) 5, 2, 7 with eigenvectors e₁, e₂, e₃
M1 (A + 3I)e = (λ + 3)e; A1.
- (b) A²: 4, 1, 16; A⁻¹: ½, −1, ¼
B1 B1
- (c) −8
B1 det A = det D = 2 × (−1) × 4.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (5 −2; 1 2). Show that (2, 1)ᵀ is an eigenvector of A and state the corresponding eigenvalue. [2]
- (b) Find the other eigenvalue and a corresponding eigenvector. [2]
- (c) Write down a matrix P and a diagonal matrix D such that A = PDP⁻¹. [1]
Show answer
- (a) A(2, 1)ᵀ = (8, 4)ᵀ = 4(2, 1)ᵀ; eigenvalue 4
M1 multiplication; A1 λ = 4.
- (b) λ = 3, (1, 1)ᵀ
M1 trace 7 or characteristic equation λ² − 7λ + 12 = 0; A1.
- (c) P = (2 1; 1 1), D = diag(4, 3)
B1
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 2 0; 0 3 1; 0 0 −1). Write down the eigenvalues of A. [1]
- (b) Find a corresponding eigenvector for each eigenvalue. [4]
- (c) Hence find a matrix P and a diagonal matrix D such that A = PDP⁻¹. [2]
Show answer
- (a) 1, 3, −1
B1 diagonal entries of a triangular matrix.
- (b) λ = 1: (1, 0, 0)ᵀ; λ = 3: (1, 1, 0)ᵀ; λ = −1: (1, −1, 4)ᵀ (or any non-zero multiples)
M1 solve (A − λI)x = 0 or use vector products of rows; A1 (1, 0, 0)ᵀ; A1 (1, 1, 0)ᵀ; A1 (1, −1, 4)ᵀ.
- (c) P = (1 1 1; 0 1 −1; 0 0 4), D = diag(1, 3, −1) (columns of P in the same order as D)
B1 P with eigenvectors as columns; B1 D matching order.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 0 0; 1 3 0; 0 1 5). Find the eigenvalues of A and a corresponding eigenvector for each. [5]
- (b) Write down matrices P and D such that P⁻¹AP = D, where D is diagonal. [2]
Show answer
- (a) λ = 2: (3, −3, 1)ᵀ; λ = 3: (0, −2, 1)ᵀ; λ = 5: (0, 0, 1)ᵀ
B1 eigenvalues 2, 3, 5; M1 method for an eigenvector; A1 A1 A1 each eigenvector (any non-zero multiple).
- (b) P = (3 0 0; −3 −2 0; 1 1 1), D = diag(2, 3, 5)
B1 P; B1 D in matching order.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (4 1; 2 3). Find the eigenvalues of A and corresponding eigenvectors. [4]
- (b) Hence find Aⁿ, where n is a positive integer, as a single matrix with entries in terms of n. [4]
Show answer
- (a) λ = 2: (1, −2)ᵀ; λ = 5: (1, 1)ᵀ
M1 λ² − 7λ + 10 = 0; A1 λ = 2, 5; A1 (1, −2)ᵀ; A1 (1, 1)ᵀ.
- (b) Aⁿ = (1/3)(2ⁿ + 2 × 5ⁿ 5ⁿ − 2ⁿ; 2 × 5ⁿ − 2ⁿ⁺¹ 2ⁿ⁺¹ + 5ⁿ)
B1 P = (1 1; −2 1), D = diag(2, 5); B1 P⁻¹ = (1/3)(1 −1; 2 1); M1 Aⁿ = PDⁿP⁻¹ multiplied out; A1 all four entries.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A has eigenvalues 1, 2 and 3 with corresponding eigenvectors (1, 0, 0)ᵀ, (1, 1, 0)ᵀ and (0, 1, 1)ᵀ. Find A. [4]
- (b) Hence find A⁻¹, using A⁻¹ = PD⁻¹P⁻¹. [2]
Show answer
- (a) A = (1 1 −1; 0 2 1; 0 0 3)
B1 P = (1 1 0; 0 1 1; 0 0 1), D = diag(1, 2, 3); B1 P⁻¹ = (1 −1 1; 0 1 −1; 0 0 1); M1 A = PDP⁻¹; A1.
- (b) A⁻¹ = (1 −½ ½; 0 ½ −1/6; 0 0 ⅓)
M1 D⁻¹ = diag(1, ½, ⅓); A1.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 1 0; 0 −1 3; 0 0 1). Find matrices P and D, where D is diagonal, such that A = PDP⁻¹. [5]
- (b) Find P⁻¹ and hence find A⁵. [4]
Show answer
- (a) P = (1 1 −3; 0 −3 3; 0 0 2), D = diag(2, −1, 1)
B1 eigenvalues 2, −1, 1; M1 method for eigenvectors; A1 (1, 0, 0)ᵀ; A1 (1, −3, 0)ᵀ; A1 (−3, 3, 2)ᵀ, with P and D in matching order.
- (b) P⁻¹ = (1 1/3 1; 0 −1/3 1/2; 0 0 1/2); A⁵ = (32 11 30; 0 −1 3; 0 0 1)
M1 P⁻¹ by cofactors (det P = −6); A1 P⁻¹; M1 A⁵ = PD⁵P⁻¹ with D⁵ = diag(32, −1, 1); A1 A⁵.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 k 2; 0 2 1; 0 0 4), where k is a constant, has (1, 1, 0)ᵀ as an eigenvector. Find k and the corresponding eigenvalue. [2]
- (b) Find matrices P and D such that A = PDP⁻¹, where D is a diagonal matrix. [4]
- (c) Write down a diagonal matrix similar to A² − 2A. [1]
Show answer
- (a) k = 1, eigenvalue 2
M1 A(1, 1, 0)ᵀ = (1 + k, 2, 0)ᵀ = λ(1, 1, 0)ᵀ; A1 λ = 2, k = 1.
- (b) P = (1 1 5; 0 1 3; 0 0 6), D = diag(1, 2, 4)
B1 eigenvalue 1 with (1, 0, 0)ᵀ; M1 (A − 4I)x = 0: −3x + y + 2z = 0, −2y + z = 0; A1 (5, 3, 6)ᵀ; A1 P and D consistent.
- (c) diag(−1, 0, 8)
B1 λ² − 2λ for λ = 1, 2, 4.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) Two cities, Lahore and Karachi, are modelled as a closed system. Each year 20% of the residents of Lahore move to Karachi and 30% of the residents of Karachi move to Lahore; everyone else stays. If the populations (in millions) in a given year are the column vector (L, K)ᵀ, the populations a year later are T(L, K)ᵀ. Write down the matrix T. [1]
- (b) Find the eigenvalues of T and corresponding eigenvectors. [3]
- (c) Hence find Tⁿ and the limit of Tⁿ as n → ∞. [3]
- (d) Initially Lahore has 4 million and Karachi has 6 million residents. Find the long-term populations predicted by the model. [1]
Show answer
- (a) T = (0.8 0.3; 0.2 0.7)
B1
- (b) λ = 1: (3, 2)ᵀ; λ = 0.5: (1, −1)ᵀ
M1 λ² − 1.5λ + 0.5 = 0; A1 λ = 1, 0.5; A1 eigenvectors.
- (c) Tⁿ = (1/5)(3 + 2(0.5)ⁿ 3 − 3(0.5)ⁿ; 2 − 2(0.5)ⁿ 2 + 3(0.5)ⁿ); limit (0.6 0.6; 0.4 0.4)
M1 Tⁿ = PDⁿP⁻¹ with P = (3 1; 2 −1), P⁻¹ = (1/5)(1 1; 2 −3); A1 Tⁿ; A1 (0.5)ⁿ → 0.
- (d) Lahore 6 million, Karachi 4 million
B1 (0.6 × 10, 0.4 × 10).
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 3; 1 4). Find the characteristic equation of A. [2]
- (b) Use the Cayley-Hamilton theorem to express A⁻¹ in the form pA + qI, and hence find A⁻¹. [3]
Show answer
- (a) λ² − 6λ + 5 = 0
M1 (2 − λ)(4 − λ) − 3 = 0; A1.
- (b) A⁻¹ = −(1/5)A + (6/5)I = (1/5)(4 −3; −1 2)
M1 A² − 6A + 5I = 0; M1 multiply by A⁻¹: A − 6I + 5A⁻¹ = 0; A1.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 2; 3 −1). Find the characteristic equation of A. [2]
- (b) Use the Cayley-Hamilton theorem to find A⁸ and A⁻¹. [3]
Show answer
- (a) λ² − 7 = 0
M1 (1 − λ)(−1 − λ) − 6 = 0; A1.
- (b) A⁸ = 2401I; A⁻¹ = (1/7)A = (1/7 2/7; 3/7 −1/7)
M1 A² = 7I; A1 A⁸ = (7I)⁴ = 2401I; A1 A⁻¹ = A/7.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (k 1; 2 3), where k is a constant, satisfies A² = 5A + pI for some constant p. Find the characteristic equation of A in terms of k. [2]
- (b) Use the Cayley-Hamilton theorem to find k and p. [2]
- (c) Hence express A⁻¹ in terms of A and I. [1]
Show answer
- (a) λ² − (k + 3)λ + 3k − 2 = 0
M1 (k − λ)(3 − λ) − 2 = 0; A1.
- (b) k = 2, p = −4
M1 compare A² = (k + 3)A − (3k − 2)I; A1.
- (c) A⁻¹ = (5I − A)/4
B1 A² − 5A = −4I gives A(A − 5I) = −4I, so A⁻¹ = (5I − A)/4.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 1 0; 0 2 1; 0 0 3). Find the characteristic equation of A in the form λ³ + aλ² + bλ + c = 0. [2]
- (b) Use the Cayley-Hamilton theorem to show that A⁻¹ = (1/6)(A² − 6A + 11I), and hence find A⁻¹. [4]
Show answer
- (a) λ³ − 6λ² + 11λ − 6 = 0
M1 (1 − λ)(2 − λ)(3 − λ) = 0; A1 expanded.
- (b) A⁻¹ = (1 −½ 1/6; 0 ½ −1/6; 0 0 ⅓)
M1 A³ − 6A² + 11A − 6I = 0; A1 multiply by A⁻¹ and rearrange (AG); M1 A² = (1 3 1; 0 4 5; 0 0 9); A1 A⁻¹.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (3 −1; 2 1). Show that A² = 4A − 5I. [2]
- (b) Hence find integers p and q such that A⁴ = pA + qI, and find A⁴. [4]
Show answer
- (a) Shown
M1 characteristic equation λ² − 4λ + 5 = 0; A1 by Cayley-Hamilton (or direct calculation A² = (7 −4; 8 −1)).
- (b) p = 24, q = −55; A⁴ = (17 −24; 48 −31)
M1 A³ = 4A² − 5A = 11A − 20I; M1 A⁴ = 11A² − 20A; A1 24A − 55I; A1 A⁴.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 2 0; 0 −2 1; 0 0 3). Show that A satisfies A³ − 2A² − 5A + 6I = 0, and state the eigenvalues of A. [3]
- (b) Hence express A⁻¹ in terms of A² , A and I, and find A⁻¹. [3]
Show answer
- (a) Shown; eigenvalues 1, −2, 3
M1 characteristic polynomial (1 − λ)(−2 − λ)(3 − λ); A1 expands to −(λ³ − 2λ² − 5λ + 6), Cayley-Hamilton (AG); B1 eigenvalues.
- (b) A⁻¹ = −(1/6)(A² − 2A − 5I) = (1 1 −⅓; 0 −½ 1/6; 0 0 ⅓)
M1 A(A² − 2A − 5I) = −6I; M1 A² = (1 −2 2; 0 4 1; 0 0 9); A1 A⁻¹.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 1 0; 1 2 1; 0 1 2). Show that the characteristic equation of A is λ³ − 6λ² + 10λ − 4 = 0, and find the eigenvalues of A in exact form. [4]
- (b) Use the Cayley-Hamilton theorem to find A⁻¹. [3]
Show answer
- (a) Shown; λ = 2, 2 + √2, 2 − √2
M1 expand det(A − λI) = (2 − λ)[(2 − λ)² − 1] − (2 − λ); A1 (2 − λ)[(2 − λ)² − 2] expanded (AG); M1 factor λ − 2; A1 λ = 2, 2 ± √2.
- (b) A⁻¹ = ¼(A² − 6A + 10I) = (¾ −½ ¼; −½ 1 −½; ¼ −½ ¾)
M1 A⁻¹ = (1/4)(A² − 6A + 10I); M1 A² = (5 4 1; 4 6 4; 1 4 5); A1.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 1 1; 0 2 1; 0 0 −1). Find the characteristic equation of A and hence show that A³ = 2A² + A − 2I. [3]
- (b) Find constants p, q and r such that A⁵ = pA² + qA + rI. [4]
- (c) Hence find A⁵. [1]
Show answer
- (a) λ³ − 2λ² − λ + 2 = 0; shown
M1 (1 − λ)(2 − λ)(−1 − λ) = 0; A1 expanded; A1 Cayley-Hamilton gives A³ − 2A² − A + 2I = 0 (AG).
- (b) p = 10, q = 1, r = −10
M1 A⁴ = 2A³ + A² − 2A; A1 A⁴ = 5A² − 4I; M1 A⁵ = 5A³ − 4A; A1 10A² + A − 10I.
- (c) A⁵ = (1 31 11; 0 32 11; 0 0 −1)
B1 using A² = (1 3 1; 0 4 1; 0 0 1).
- (a) Show that the matrix of coefficients of the systemx + y + z = 62x − y + 3z = 9x + 2y − z = 2is non-singular. [2]
- (b) Solve the system and interpret the solution geometrically. [3]
Show answer
- (a) det = 5 ≠ 0
M1 1(1 − 6) − 1(−2 − 3) + 1(4 + 1); A1 = 5 ≠ 0.
- (b) x = 1, y = 2, z = 3; the three planes meet at the single point (1, 2, 3)
M1 elimination or inverse matrix; A1 solution; B1 interpretation.
- (a) Show that the systemx + y + z = 1x − y + 2z = 32x + 3z = 5has a singular matrix of coefficients. [2]
- (b) Show that the system is inconsistent and describe the arrangement of the three planes. [2]
Show answer
- (a) det = 0
M1 1(−3 − 0) − 1(3 − 4) + 1(0 + 2); A1 = −3 + 1 + 2 = 0.
- (b) Adding the first two equations gives 2x + 3z = 4, which contradicts 2x + 3z = 5; triangular prism
M1 combination; A1 inconsistent with triangular prism (no two planes parallel).
- (a) Consider the systemkx + y + z = 1x + 2y + 3z = 22x + 3y + 4z = 3Find the value of k for which the system does not have a unique solution. [2]
- (b) For this value of k, show that the system is consistent and describe the configuration of the three planes. [2]
- (c) Find a vector equation of the common line. [3]
Show answer
- (a) k = 1
M1 det = k(8 − 9) − 1(4 − 6) + 1(3 − 4) = 1 − k; A1 k = 1.
- (b) Third equation = first + second, so consistent; the planes form a sheaf (meet in a common line)
M1 row relation; A1 sheaf.
- (c) r = (0, 1, 0)ᵀ + t(1, −2, 1)ᵀ
M1 eliminate x: y + 2z = 1; M1 let z = t; A1 x = t, y = 1 − 2t, z = t.
- (a) Consider the systemx − 2y + 3z = 42x − 4y + 6z = kx + y + z = 2Explain why the matrix of coefficients is singular for every value of k. [1]
- (b) Describe the geometry of the three planes when k = 8 and when k ≠ 8. [2]
- (c) For k = 8, find the general solution. [3]
Show answer
- (a) Row 2 is twice row 1, so the determinant is 0
B1
- (b) k = 8: the first two planes coincide and the third cuts them in a line (infinitely many solutions); k ≠ 8: the first two planes are parallel and distinct, so there is no solution (the third plane cuts each in a separate line)
B1 B1
- (c) r = (1, 0, 1)ᵀ + s(−5, 2, 3)ᵀ
M1 eliminate x: 3y − 2z = −2; M1 parameter; A1 correct line (any equivalent form).
- (a) Show that the systemx + y + z = 3x + 2y + 4z = kx + 3y + 9z = k²has a unique solution for every value of k. [2]
- (b) Find the solution in terms of k. [4]
- (c) Find the value of k for which z is least. [1]
Show answer
- (a) det = 2 ≠ 0
M1 1(18 − 12) − 1(9 − 4) + 1(3 − 2); A1 = 2, independent of k.
- (b) x = k² − 3k + 9, y = −(3k² − 8k + 15)/2, z = (k² − 2k + 3)/2
M1 eliminate x twice: y + 3z = k − 3 and y + 5z = k² − k; M1 2z = k² − 2k + 3; A1 z and y; A1 x.
- (c) k = 1
B1 z = ((k − 1)² + 2)/2 is least at k = 1.
- (a) The system of equationsx + 2y + z = 42x − y + kz = 33x + y + 2z = mhas no unique solution. Find the value of k. [2]
- (b) With this value of k, find the value of m for which the system is consistent, and describe the geometry when the system is inconsistent. [3]
- (c) For k = 1 and m = 7, find the solution in the form r = a + tb. [3]
Show answer
- (a) k = 1
M1 det = 1(−2 − k) − 2(4 − 3k) + 1(2 + 3) = 5k − 5; A1 k = 1.
- (b) m = 7; if m ≠ 7 the planes form a triangular prism (no common point, no two planes parallel)
M1 equation 3 left side = equation 1 + equation 2; A1 m = 4 + 3 = 7; B1 triangular prism, normals not parallel.
- (c) r = (−1, 0, 5)ᵀ + t(3, 1, −5)ᵀ
M1 subtract: x − 3y = −1; M1 parameter y = t; A1 x = 3t − 1, z = 5 − 5t.
- (a) Consider the systemx + ky + z = 1x + y + kz = 1kx + y + z = 1Show that the determinant of the matrix of coefficients is (1 − k)²(k + 2). [3]
- (b) Find the unique solution when k ≠ 1 and k ≠ −2. [2]
- (c) Describe the geometric arrangement of the planes when k = 1 and when k = −2. [3]
Show answer
- (a) Shown
M1 expand: 1(1 − k) − k(1 − k²) + 1(1 − k); M1 (1 − k)(2 − k − k²); A1 (1 − k)(1 − k)(k + 2) (AG).
- (b) x = y = z = 1/(k + 2)
M1 by symmetry or subtracting equations x = y = z; A1 (k + 2)x = 1.
- (c) k = 1: all three equations are x + y + z = 1, the same plane (infinitely many solutions); k = −2: adding the equations gives 0 = 3, inconsistent, and no two planes are parallel, so they form a triangular prism
B1 k = 1 coincident planes; M1 sum of equations for k = −2; A1 prism.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) Find the eigenvalues of the matrix A = (5 4; 1 2). [2]
- (b) Find an eigenvector corresponding to each eigenvalue. [2]
Show answer
- (a) 1 and 6
M1 λ² − 7λ + 6 = 0; A1.
- (b) λ = 1: (1, −1)ᵀ; λ = 6: (4, 1)ᵀ
B1 B1 (any non-zero multiples).
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (3 0 0; 2 1 0; 4 −1 −2). Write down the eigenvalues of A. [1]
- (b) Find an eigenvector corresponding to the eigenvalue 3. [3]
Show answer
- (a) 3, 1, −2
B1
- (b) (5, 5, 3)ᵀ
M1 (A − 3I)x = 0; M1 2x − 2y = 0 and 4x − y − 5z = 0; A1 (5, 5, 3)ᵀ or a multiple.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (1 1 0; 1 0 1; 0 1 1). Show that the characteristic equation of A is λ³ − 2λ² − λ + 2 = 0. [3]
- (b) Hence find the eigenvalues of A and a corresponding eigenvector for each. [4]
Show answer
- (a) Shown
M1 det(A − λI) = (1 − λ)[−λ(1 − λ) − 1] − 1(1 − λ); A1 simplification; A1 −(λ³ − 2λ² − λ + 2) = 0 (AG).
- (b) λ = 2: (1, 1, 1)ᵀ; λ = 1: (1, 0, −1)ᵀ; λ = −1: (1, −2, 1)ᵀ
B1 eigenvalues 2, 1, −1; A1 A1 A1 eigenvectors.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 1; 4 −1). Find the eigenvalues of A and corresponding eigenvectors. [4]
- (b) Show that if e is an eigenvector of A with eigenvalue λ, then e is an eigenvector of A² + 2A with eigenvalue λ² + 2λ. [2]
- (c) Hence state the eigenvalues of A² + 2A and deduce the value of det(A² + 2A). [2]
Show answer
- (a) λ = 3: (1, 1)ᵀ; λ = −2: (1, −4)ᵀ
M1 λ² − λ − 6 = 0; A1 3, −2; A1 A1 eigenvectors.
- (b) Shown
M1 A²e = A(λe) = λ²e; A1 (A² + 2A)e = (λ² + 2λ)e (AG).
- (c) 15 and 0; det = 0
B1 9 + 6 and 4 − 4; B1 product 0.
Matrices are written row by row, with rows separated by semicolons; for example (1 2; 3 4) has first row 1, 2 and second row 3, 4. Column vectors are written as (x, y, z)ᵀ.
- (a) The matrix A = (2 a 0; 1 1 1; 0 1 2), where a is a constant. Show that the characteristic equation of A is (λ − 2)(λ² − 3λ + 1 − a) = 0. [3]
- (b) Given that 4 is an eigenvalue of A, find a and the other eigenvalues. [3]
- (c) Find an eigenvector corresponding to the eigenvalue 4. [2]
Show answer
- (a) Shown
M1 expand det(A − λI) along the first row: (2 − λ)[(1 − λ)(2 − λ) − 1] − a(2 − λ); M1 take out factor (2 − λ); A1 (2 − λ)(λ² − 3λ + 1 − a) = 0 (AG).
- (b) a = 5; eigenvalues 2 and −1
M1 16 − 12 + 1 − a = 0; A1 a = 5; A1 λ² − 3λ − 4 = 0 gives 4, −1, plus 2.
- (c) (5, 2, 1)ᵀ
M1 (A − 4I)x = 0; A1.
