AS & A Level Further Mathematics (9231)
Matrices Banker
10 questions · about 5% 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
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Q1Invariant points & lines · Core · Paper 1 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.
- (a) The matrix M = (3 1; 2 2) represents a transformation T of the x-y plane. Find the equations of the invariant lines of T that pass through the origin. [4]
- (b) Find the image of the point (2, 2) under T. Hence state the factor by which T multiplies the distance from the origin of each point on y = x. [2]
Show answer
- (a) y = x and y = −2x
M1 image of (x, mx) is (3x + mx, 2x + 2mx); M1 2 + 2m = m(3 + m); A1 m² + m − 2 = 0; A1 m = 1 or m = −2, so y = x and y = −2x.
- (b) (8, 8); factor 4
B1 (3 × 2 + 2, 2 × 2 + 2 × 2) = (8, 8); B1 factor 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.
- (a) The transformation S of the plane is represented by the matrix (1 4; 1 1). Find the invariant lines of S through the origin, giving your answers in the form y = mx. [4]
- (b) Show that the line x = 0 is not invariant under S. [1]
Show answer
- (a) y = ½x and y = −½x
M1 image of (x, mx) is (x + 4mx, x + mx); M1 1 + m = m(1 + 4m); A1 4m² = 1; A1 m = ±½.
- (b) Shown
B1 (0, 1) maps to (4, 1), which is not on x = 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.
- (a) The matrix M = (3 −2; 1 0) represents the transformation T. Show that every point on the line y = x is an invariant point of T. [2]
- (b) Find the equation of the other invariant line of T through the origin. [3]
- (c) Explain why the points on the line in part (b) are not invariant points. [1]
Show answer
- (a) Shown
M1 M(t, t)ᵀ = (3t − 2t, t)ᵀ; A1 = (t, t)ᵀ for all t.
- (b) y = ½x
M1 image of (x, mx) is ((3 − 2m)x, x); M1 1 = m(3 − 2m) gives 2m² − 3m + 1 = 0; A1 m = 1 (already found) or m = ½: y = ½x.
- (c) (2, 1) maps to (4, 2): the line maps to itself but each point moves (distances from O are doubled).
B1 e.g. M(2, 1)ᵀ = (4, 2)ᵀ ≠ (2, 1)ᵀ.
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.
- (a) The transformation T is a reflection in the x-axis followed by a stretch, scale factor 3, parallel to the y-axis. Find the matrix M that represents T. [2]
- (b) State the invariant points of T and the invariant lines of T through the origin. [3]
Show answer
- (a) M = (1 0; 0 −3)
M1 stretch × reflection: (1 0; 0 3)(1 0; 0 −1); A1.
- (b) Invariant points: the x-axis; invariant lines through O: y = 0 and x = 0
B1 points with −3y = y, i.e. y = 0; B1 y = 0; B1 x = 0 (as (0, y) maps to (0, −3y)).
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.
- (a) The transformation H is represented by the matrix (1 3; 0 1). Describe H fully. [2]
- (b) Find the invariant points of H. [1]
- (c) Show that every line y = c, where c is a constant, is invariant under H, and show that the x-axis is the only invariant line of H through the origin. [3]
Show answer
- (a) Shear with the x-axis invariant, (x, y) → (x + 3y, y)
B1 shear; B1 x-axis invariant and e.g. (0, 1) maps to (3, 1).
- (b) All points on the x-axis (y = 0)
B1 (x + 3y, y) = (x, y) requires y = 0.
- (c) Shown
B1 (x, c) maps to (x + 3c, c), which lies on y = c; M1 for y = mx: m = m(1 + 3m) gives 3m² = 0 so m = 0; A1 x = 0 is not invariant since (0, 1) maps to (3, 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.
- (a) The matrix M = (a 2; 3 1), where a is a constant, represents a transformation for which the line y = 2x is invariant. Find the value of a. [3]
- (b) Find the equation of the other invariant line through the origin. [3]
Show answer
- (a) a = −3/2
M1 image of (1, 2) is (a + 4, 5); M1 5 = 2(a + 4); A1 a = −3/2.
- (b) y = −¾x
M1 image of (1, m) is (a + 2m, 3 + m); M1 3 + m = m(−3/2 + 2m), i.e. 4m² − 5m − 6 = 0; A1 (m − 2)(4m + 3) = 0 so y = −¾x.
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.
- (a) The matrix M = (1 2; 2 −2) represents a transformation T. Find the invariant lines of T through the origin. [4]
- (b) Show that these two lines are perpendicular. [1]
- (c) Find the images of the points (2, 1) and (1, −2) under T, and describe the effect of T on each invariant line. [2]
Show answer
- (a) y = ½x and y = −2x
M1 image of (x, mx) is ((1 + 2m)x, (2 − 2m)x); M1 2 − 2m = m(1 + 2m); A1 2m² + 3m − 2 = 0; A1 m = ½ or −2.
- (b) Product of gradients ½ × (−2) = −1
B1
- (c) (4, 2) and (−3, 6); points on y = ½x are moved to twice their distance from O; points on y = −2x are moved to three times their distance on the opposite side of O
B1 both images; B1 descriptions (factors 2 and −3).
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.
- (a) The transformation T is represented by M = (2 −1; 1 0). Show that every point on y = x is invariant under T. [1]
- (b) Show that y = x is the only invariant line of T that passes through the origin. [3]
- (c) Show that every line y = x + c is invariant under T. [2]
Show answer
- (a) Shown
B1 M(t, t)ᵀ = (2t − t, t)ᵀ = (t, t)ᵀ.
- (b) Shown
M1 image of (x, mx) is ((2 − m)x, x); M1 1 = m(2 − m) gives (m − 1)² = 0, so m = 1 only; A1 and x = 0 is not invariant since (0, 1) maps to (−1, 0).
- (c) Shown
M1 (x, x + c) maps to (x − c, x); A1 x = (x − c) + c, so the image lies on y = x + c.
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.
- (a) The transformation T is a reflection in the line y = √3 x. Write down the matrix that represents T, giving exact values. [2]
- (b) State the invariant lines of T through the origin. [2]
- (c) Show that every line of the form y = −x/√3 + c is invariant under T, but that its points other than the one on the mirror line are not invariant. [3]
Show answer
- (a) (−½ √3/2; √3/2 ½)
M1 mirror at 60° to the x-axis, matrix (cos 120° sin 120°; sin 120° −cos 120°); A1.
- (b) y = √3 x (a line of invariant points) and y = −x/√3
B1 the mirror line; B1 the line through O perpendicular to it.
- (c) Shown
M1 apply the matrix to a general point (x, c − x/√3): image X = −x + (√3/2)c, Y = x/√3 + c/2; M1 −X/√3 + c = x/√3 − c/2 + c = Y, so the image lies on the same line; A1 X = x only when x = √3c/4, which is the point where the line meets the mirror y = √3x, so all other points on the line move.
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.
- (a) The transformation T is represented by M = (2 1; 2 3). Find the line of invariant points of T. [2]
- (b) Find all the invariant lines of T, including those that do not pass through the origin. [5]
Show answer
- (a) y = −x
M1 2x + y = x and 2x + 3y = y; A1 both give x + y = 0.
- (b) y = −x, and every line y = 2x + c
M1 image of (x, mx + c) is ((2 + m)x + c, (2 + 3m)x + 3c); M1 require (2 + 3m)x + 3c = m((2 + m)x + c) + c for all x; A1 m² − m − 2 = 0 so m = 2 or m = −1; M1 constant terms: c(2 − m) = 0; A1 m = 2 works for every c (lines y = 2x + c), m = −1 needs c = 0 (the line y = −x).
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.
- (a) The matrix M = (3 p; 1 1), where p is a constant, represents a transformation T. Show that the line y = mx is invariant under T if and only if pm² + 2m − 1 = 0. [2]
- (b) Find the value of p for which T has exactly one invariant line through the origin, and state the equation of that line. [3]
- (c) Find the set of values of p for which T has no invariant lines through the origin. [2]
Show answer
- (a) Shown
M1 image of (x, mx) is ((3 + pm)x, (1 + m)x); A1 1 + m = m(3 + pm) rearranges to pm² + 2m − 1 = 0 (AG).
- (b) p = −1; y = x
M1 repeated root: discriminant 4 + 4p = 0; A1 p = −1 (note p = 0 gives two lines, y = ½x and x = 0); A1 −m² + 2m − 1 = 0 gives m = 1, line y = x.
- (c) p < −1
M1 4 + 4p < 0; A1 p < −1 (x = 0 is invariant only when p = 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.
- (a) The matrix A = (k 3; 2 k − 1). Find the values of k for which A is singular. [2]
- (b) Find A⁻¹ when k = 1. [2]
Show answer
- (a) k = 3 or k = −2
M1 det A = k(k − 1) − 6 = 0; A1 (k − 3)(k + 2) = 0.
- (b) A⁻¹ = (0 ½; ⅓ −1/6)
M1 det = −6, adjugate (0 −3; −2 1); A1 A⁻¹ = −(1/6)(0 −3; −2 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.
- (a) The matrices A and B are such that A = (2 1; 5 3) and AB = (4 1; 11 1). Find A⁻¹. [1]
- (b) Hence find B. [2]
- (c) Find (AB)⁻¹. [2]
Show answer
- (a) A⁻¹ = (3 −1; −5 2)
B1 det A = 1.
- (b) B = (1 2; 2 −3)
M1 B = A⁻¹(AB), multiplying on the left; A1.
- (c) (AB)⁻¹ = (−1/7)(1 −1; −11 4) = (−1/7 1/7; 11/7 −4/7)
M1 det(AB) = 4 − 11 = −7; 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.
- (a) The matrix A = (3 1; 2 4). Find A² and find constants p and q such that A² = pA + qI, where I is the 2 × 2 identity matrix. [3]
- (b) Hence express A⁻¹ in the form rA + sI and find A⁻¹. [2]
Show answer
- (a) A² = (11 7; 14 18); p = 7, q = −10
B1 A²; M1 compare entries, e.g. 7 = p and 11 = 3p + q; A1 p = 7, q = −10.
- (b) A⁻¹ = −(1/10)A + (7/10)I = (1/10)(4 −1; −2 3)
M1 multiply by A⁻¹: A = 7I − 10A⁻¹; A1 A⁻¹ = (7I − A)/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.
- (a) The matrix B = (1 2 k; 0 k 1; 3 1 2). Show that det B = −(k + 1)(3k − 5). [3]
- (b) Find B⁻¹ when k = 0. [3]
Show answer
- (a) Shown
M1 expand along first row: 1(2k − 1) − 2(0 − 3) + k(0 − 3k); A1 −3k² + 2k + 5; A1 factorised (AG).
- (b) B⁻¹ = (1/5)(−1 −4 2; 3 2 −1; 0 5 0)
M1 det = 5; M1 matrix of cofactors, transposed; A1 correct inverse.
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 C = (2 1 0; 1 3 1; 0 1 2). Find C⁻¹. [4]
- (b) Hence solve the equations 2x + y = 1, x + 3y + z = 0, y + 2z = 3. [2]
Show answer
- (a) C⁻¹ = (1/8)(5 −2 1; −2 4 −2; 1 −2 5)
M1 det C = 2(5) − 1(2) = 8; M1 cofactors; A1 adjugate (5 −2 1; −2 4 −2; 1 −2 5); A1 divide by 8.
- (b) x = 1, y = −1, z = 2
M1 (x, y, z)ᵀ = C⁻¹(1, 0, 3)ᵀ; A1 (1/8)(8, −8, 16).
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.
- (a) The matrix A = (0 −1; 1 −1). Show that A³ = I. [2]
- (b) Deduce A⁻¹ without using the usual formula for an inverse. [1]
- (c) Find A¹⁰⁰. [2]
Show answer
- (a) Shown
M1 A² = (−1 1; −1 0); A1 A³ = A²A = (1 0; 0 1).
- (b) A⁻¹ = A² = (−1 1; −1 0)
B1 A × A² = I.
- (c) A¹⁰⁰ = A = (0 −1; 1 −1)
M1 100 = 3 × 33 + 1; 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.
- (a) The 2 × 2 matrix M = (a b; c d) is not the zero matrix but M² = 0, the zero matrix. By comparing entries, show that a + d = 0 and ad − bc = 0. [4]
- (b) Find the matrix M with M² = 0 for which a = 2 and b = 4. [2]
Show answer
- (a) Shown
M1 M² = (a² + bc b(a + d); c(a + d) d² + bc); M1 if a + d ≠ 0 then b = c = 0, and then a² = d² = 0, so M = 0, contradiction; A1 so a + d = 0; A1 then a² + bc = 0 with d = −a gives ad − bc = −a² − bc = 0 (AG).
- (b) M = (2 4; −1 −2)
M1 d = −2, bc = −a² so c = −1; 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.
- (a) The matrix A = (1 k 0; 0 1 k; k 0 1). Show that det A = 1 + k³, and state the real value of k for which A is singular. [3]
- (b) Find A⁻¹ in terms of k, for k ≠ −1. [4]
- (c) Write down A⁻¹ when k = 2. [1]
Show answer
- (a) det A = 1 + k³; k = −1
M1 1(1 − 0) − k(0 − k²) + 0; A1 1 + k³ (AG); B1 k = −1 (k² − k + 1 > 0 for real k).
- (b) A⁻¹ = (1/(1 + k³))(1 −k k²; k² 1 −k; −k k² 1)
M1 cofactors; A1 at least five correct; A1 transposed correctly; A1 divided by 1 + k³.
- (c) (1/9)(1 −2 4; 4 1 −2; −2 4 1)
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.
- (a) Describe fully the transformation represented by the matrix (0 −2; 2 0) as a combination of two simple transformations. [3]
- (b) Find the image of the point (3, 1). [1]
Show answer
- (a) Rotation of 90° anticlockwise about the origin, and an enlargement with scale factor 2, centre the origin (in either order)
B1 rotation 90° anticlockwise about O; B1 enlargement scale factor 2; B1 centre O stated for both.
- (b) (−2, 6)
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.
- (a) The transformation T is a reflection in the line y = −x followed by a stretch, scale factor 2, parallel to the x-axis. Find the matrix that represents T. [3]
- (b) Find the image of the point (1, 3) under T. [1]
Show answer
- (a) (0 −2; −1 0)
B1 reflection (0 −1; −1 0); B1 stretch (2 0; 0 1); B1 product in the correct order, stretch × reflection.
- (b) (−6, −1)
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.
- (a) Describe fully the transformation represented by the matrix (1 0; 4 1). [2]
- (b) Find the images of the vertices (0, 0), (1, 0), (1, 1) and (0, 1) of the unit square, and state the area of the image. [2]
Show answer
- (a) Shear with the y-axis invariant, mapping (1, 0) to (1, 4) (i.e. (x, y) → (x, 4x + y))
B1 shear; B1 y-axis invariant with a correct point and image.
- (b) (0, 0), (1, 4), (1, 5), (0, 1); area 1
B1 all four images; B1 area 1 (determinant 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.
- (a) Show that the matrix M = (1 −√3; √3 1) represents a rotation about the origin followed by an enlargement with centre the origin. State the angle of the rotation and the scale factor of the enlargement. [3]
- (b) Hence find M⁶, explaining your reasoning. [2]
Show answer
- (a) Rotation 60° anticlockwise, enlargement scale factor 2
M1 M = 2(½ −√3/2; √3/2 ½); A1 cos θ = ½, sin θ = √3/2 so θ = 60° anticlockwise; A1 scale factor 2.
- (b) M⁶ = 64I = (64 0; 0 64)
M1 six rotations of 60° give 360°, scale factor 2⁶; A1 64I.
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.
- (a) The transformation T is a rotation of 90° clockwise about the origin, followed by a reflection in the x-axis. Find the single matrix that represents T and describe T as a single transformation. [4]
- (b) Find the matrix when the same two transformations are applied in the opposite order, and describe it. [2]
Show answer
- (a) (0 1; 1 0): reflection in the line y = x
B1 rotation (0 1; −1 0); B1 reflection (1 0; 0 −1); M1 product reflection × rotation; A1 (0 1; 1 0), reflection in y = x.
- (b) (0 −1; −1 0): reflection in the line y = −x
M1 rotation × reflection; 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.
- (a) Let R = (0 1; 1 0) and S = (3 0; 0 1). Describe the transformations represented by R and by S, and show that M = (0 3; 1 0) is equal to SR. [3]
- (b) Find a matrix S′, representing a stretch, such that M = RS′. Describe S′ fully and explain the order in which the two transformations are applied. [3]
Show answer
- (a) R: reflection in y = x; S: stretch parallel to the x-axis, factor 3; SR = (0 3; 1 0)
B1 R; B1 S; B1 product SR computed.
- (b) S′ = (1 0; 0 3): stretch parallel to the y-axis, factor 3, applied first, then the reflection in y = x
M1 S′ = R⁻¹M = RM; A1 (1 0; 0 3); B1 S′ is applied first because it is on the right.
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.
- (a) The matrix M = (3/5 4/5; 4/5 −3/5). Show that M represents a reflection, and find the equation of the mirror line. [3]
- (b) The transformation represented by M is followed by a reflection in the y-axis. Find the matrix of the combined transformation and describe it as a single transformation, giving any angle correct to 1 decimal place. [4]
Show answer
- (a) Reflection in y = ½x
M1 invariant points: (−2/5)x + (4/5)y = 0 gives x = 2y; A1 det M = −1 and M² = I, so M is a reflection; A1 mirror line y = ½x.
- (b) (−3/5 −4/5; 4/5 −3/5): rotation of 126.9° anticlockwise about the origin
B1 reflection in y-axis (−1 0; 0 1); M1 (−1 0; 0 1)M; A1 (−3/5 −4/5; 4/5 −3/5); A1 cos θ = −3/5, sin θ = 4/5, θ = 126.9° anticlockwise about O.
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.
- (a) The matrix M = (3 1; −2 4) maps a triangle of area 6 cm² to a triangle T′. Find the area of T′. [2]
- (b) State, with a reason, whether M preserves or reverses the orientation of shapes. [1]
Show answer
- (a) 84 cm²
M1 det M = 12 + 2 = 14; A1 6 × 14 = 84.
- (b) Preserves, because det M > 0
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.
- (a) The matrix (k 2; 3 k) maps a square of area 5 units² onto a parallelogram of area 35 units². Find the possible values of k, giving exact answers. [4]
- (b) For these values of k, state whether the transformation reverses orientation. [1]
Show answer
- (a) k = √13 or k = −√13
M1 |k² − 6| = 7; M1 k² − 6 = 7 or k² − 6 = −7; A1 k² = 13; A1 k² = −1 has no real solution, so k = ±√13.
- (b) No: det = +7 > 0 for both values
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.
- (a) Triangle ABC has vertices A(1, 1), B(4, 2) and C(2, 5). Find the area of triangle ABC. [2]
- (b) Triangle ABC is mapped to triangle A′B′C′ by the matrix (2 −1; 1 3). Find the coordinates of A′ and the area of triangle A′B′C′. [3]
Show answer
- (a) 5.5 units²
M1 ½|(4 − 1)(5 − 1) − (2 − 1)(2 − 1)|; A1 ½ × 11 = 5.5.
- (b) A′(1, 4); area 38.5 units²
B1 A′(1, 4); M1 det = 6 + 1 = 7; A1 5.5 × 7 = 38.5.
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.
- (a) A region R of area 2 units² is transformed by the matrix M = (3 2; 1 2). The image is transformed again by M, and so on. Find the least number of applications of M after which the area of the image exceeds 10⁶ units². [3]
- (b) The matrix N = (2 4; 1 2). Show that N maps every point of the plane onto the line y = ½x, and state the area of the image of R under N. [3]
Show answer
- (a) 10
M1 det M = 4, so area after n applications is 2 × 4ⁿ; M1 4ⁿ > 500 000: 4⁹ = 262 144, 4¹⁰ = 1 048 576; A1 n = 10.
- (b) Shown; area 0
M1 (x, y) maps to (2x + 4y, x + 2y) = (2(x + 2y), x + 2y); A1 Y = ½X; B1 det N = 0 so the image has area 0.
