Questions on matrix algebra and transformations

Question 1

Write down the 2 × 2 matrices for a rotation of 90° anticlockwise about the origin and for a reflection in the line y = x.

Accepted Answer

Rotation: rows (0, −1) and (1, 0). Reflection in y = x: rows (0, 1) and (1, 0). B1 B1 for the two matrices. Both can be read off from where (1, 0) and (0, 1) land, since the images are the columns.

Question 2

Let A have rows (2, 1), (0, 3) and B have rows (1, −1), (2, 0). Find AB and BA, and comment.

Accepted Answer

AB has rows (4, −2), (6, 0); BA has rows (2, −2), (4, 2). They differ, because matrix multiplication is not commutative, so the order of a product must never be swapped mid-working. M1 for a correct multiplication method, A1 for both products, B1 for the comment on commutativity.

Question 3

Write down the 2 × 2 zero matrix 0 and identity matrix I, and state A + 0 and AI for a general 2 × 2 matrix A.

Accepted Answer

0 has every entry 0, rows (0, 0), (0, 0); I has rows (1, 0), (0, 1). B1 for both matrices. A + 0 = A, since adding zero to every entry changes nothing, and AI = A, since I is the identity element for multiplication just as 1 is for numbers. B1 for both properties.

Question 4

Write down the 3 × 3 matrices representing (i) reflection in the plane y = 0, and (ii) rotation of 90° anticlockwise about the z-axis, viewed from the positive z-axis. State, for each matrix, which coordinates of a point it leaves unchanged.

Accepted Answer

Reflection in y = 0 negates the second coordinate and fixes the other two: rows (1, 0, 0), (0, −1, 0), (0, 0, 1). Rotation about z leaves k fixed and rotates i and j in the xy-plane: rows (0, −1, 0), (1, 0, 0), (0, 0, 1). B1 for the reflection, B1 for the rotation. Each column is where a unit vector lands: the reflection fixes i and sends j to −j, the rotation sends i to j and j to −i, and k is fixed by both, giving the row and column of zeros with a 1 on the diagonal. So the reflection leaves x and z unchanged and reverses y, while the rotation leaves z unchanged and moves x and y between them. B1 for identifying which coordinates each matrix leaves unchanged. A coordinate is unchanged exactly when its row of the matrix is the matching row of the identity.

Question 5

Find the image of the point (3, 1) under the rotation of 90° anticlockwise about the origin.

Accepted Answer

Multiply: rows (0, −1), (1, 0) applied to (3, 1) give (−1, 3). B1 for the rotation matrix, M1 for the multiplication, A1 for (−1, 3).

Question 6

A transformation reflects in the x-axis and then rotates 90° anticlockwise. Find the single matrix representing the combined transformation and identify it.

Accepted Answer

The reflection acts first, so it sits on the right of the product. RM is rows (0, −1), (1, 0) times rows (1, 0), (0, −1), which gives rows (0, 1), (1, 0). That is the reflection in y = x. B1 for the reflection matrix, M1 for multiplying in the order RM, A1 for rows (0, 1), (1, 0), A1 for naming the reflection in y = x. Writing MR instead gives rows (0, −1), (−1, 0), the reflection in y = −x, so the order genuinely matters.

Question 7

A linear transformation sends (1, 0) to (2, 0) and (0, 1) to (1, 3). Write down its matrix, and find the image of (2, 5).

Accepted Answer

The images of the base vectors are the columns, giving rows (2, 1), (0, 3). Then (2, 5) maps to (2 × 2 + 1 × 5, 0 + 3 × 5) = (9, 15). B1 for the matrix, M1 for the multiplication, A1 for (9, 15).

Question 8

Using the matrix with rows (0, −1), (1, 0), show that applying the 90° rotation four times returns every point to where it started, and explain what this says about the matrix.

Accepted Answer

Squaring gives rows (−1, 0), (0, −1), which is −I, a half turn. Squaring again gives (−I)² = I, so R⁴ = I. M1 for squaring, A1 for −I, M1 for squaring again, A1 for the fourth power being the identity. Four quarter turns make a full turn, and the matrix records that as its fourth power being the identity.

Question 9

Let A have rows (2, 2), (1, 1) and B have rows (1, −1), (−1, 1). Show that AB is the zero matrix even though neither A nor B is the zero matrix, and explain what makes this possible.

Accepted Answer

Multiplying: top-left is 2 × 1 + 2 × (−1) = 0, and every other entry cancels the same way, so AB has rows (0, 0), (0, 0). M1 for the multiplication, A1 for the zero matrix. Neither A nor B is the zero matrix: A has rows (2, 2), (1, 1) and B has rows (1, −1), (−1, 1), both plainly nonzero. B1 for the observation. Both A and B are singular, with det A = 2 − 2 = 0 and det B = 1 − 1 = 0, and every column of B is sent to the zero vector by A because B's columns lie along the direction A collapses. B1 for identifying both factors as singular.

Question 10

Find the single matrix representing reflection in the x-axis followed by reflection in the y-axis, and identify the transformation it represents.

Accepted Answer

Reflection in the x-axis has rows (1, 0), (0, −1); reflection in the y-axis has rows (−1, 0), (0, 1). B1 for both matrices. The x-axis reflection acts first, so it sits on the right: rows (−1, 0), (0, 1) times rows (1, 0), (0, −1) gives rows (−1, 0), (0, −1). M1 for multiplying in that order, A1 for rows (−1, 0), (0, −1). This is −I, a rotation of 180° about the origin. A1 for naming it a half turn. Two reflections in perpendicular lines always compose to a rotation, here through twice the 90° angle between the mirror lines.

Question 11

A transformation sends (1, 0) to (0, 2) and (0, 1) to (2, 0). Write down its matrix, evaluate its determinant, and state whether the transformation preserves or reverses orientation.

Accepted Answer

The images of the base vectors are the columns, giving rows (0, 2), (2, 0). B1 for the matrix. det = 0 × 0 − 2 × 2 = −4. M1 for the determinant. The sign is negative, so orientation is reversed: this transformation is a reflection combined with a stretch, not a pure rotation or enlargement. A1 for reversed with the reason from the sign.

Question 12

The diagram shows the unit square OABC and its image OA′B′C′ under a transformation represented by the 2 × 2 matrix M. Use the diagram to write down M, and hence find the equation of the image of the line y = 2x under the same transformation.

Accepted Answer

A is the point (1, 0) and C is (0, 1), and the diagram puts their images at A′(2, 1) and C′(−1, 2). M1 for reading the images of the two unit vectors off the diagram. Those images are the columns of the matrix, so M has rows (2, −1) and (1, 2). A1 for the matrix. It checks against the fourth vertex: M sends (1, 1) to (2 − 1, 1 + 2) = (1, 3), which is where B′ is drawn. A general point of the line y = 2x is (t, 2t). Applying M gives (2t − 2t, t + 4t) = (0, 5t). M1 for applying M to a general point of the line, A1 for the image (0, 5t). The first coordinate is 0 for every t, while 5t runs through every real value as t does, so the image is the whole line x = 0, the y-axis. A1 for the equation of the image. The line y = 2x is therefore not invariant: its direction vector (1, 2) is sent to (0, 5), a different direction.

Question 13

R is a rotation of 90° anticlockwise about the x-axis and S is a rotation of 90° anticlockwise about the z-axis, both viewed from the positive end of the axis. Find the matrix representing S followed by R, use it to find the image of (1, 2, 3), and show that R followed by S is a different transformation.

Accepted Answer

Build each matrix from the images of the base vectors. R fixes i, sends j to k and k to −j, so R has rows (1, 0, 0), (0, 0, −1), (0, 1, 0). S fixes k, sends i to j and j to −i, so S has rows (0, −1, 0), (1, 0, 0), (0, 0, 1). S first means S on the right, so the combined matrix is RS, with rows (0, −1, 0), (0, 0, −1), (1, 0, 0). Applying it to (1, 2, 3) gives (−2, −3, 1). M1 for building a matrix from the images of the base vectors, A1 for R, A1 for S, M1 for the product in the order RS, A1 for RS, A1 for (−2, −3, 1), M1 for reversing the order, A1 for SR differing. Reversing the order gives SR, with rows (0, 0, 1), (1, 0, 0), (0, 1, 0), which sends (1, 2, 3) to (3, 1, 2). The two matrices differ, so the transformations differ.

Question 14

The curve y = x³ is reflected in the line y = x, and the image is then stretched parallel to the x-axis with scale factor 2. Find the equation of the final curve.

Accepted Answer

Reflection in y = x has rows (0, 1), (1, 0); the stretch has rows (2, 0), (0, 1). The reflection acts first, so the combined matrix is the stretch times the reflection: rows (2, 0), (0, 1) times rows (0, 1), (1, 0) gives rows (0, 2), (1, 0). M1 for each matrix, M1 for multiplying in the order stretch × reflection, A1 for the combined matrix. Its determinant is 0 − 2 = −2, so it is invertible, with inverse rows (0, 1), (½, 0): multiplying the two confirms the identity. M1 for the inverse. Writing (x, y) = M⁻¹(X, Y) gives x = Y and y = X/2. Substituting into y = x³ gives X/2 = Y³, that is X = 2Y³, or in the usual letters, x = 2y³. A1 for the substitution, A1 for the final equation. Check on (1, 1), which lies on y = x³: the combined matrix sends it to (2, 1), and 2 = 2 × 1³ confirms it.

Question 15

T1 is reflection in the plane z = 0 and T2 is rotation of 90° anticlockwise about the y-axis, viewed from the positive y-axis. Find the matrix representing T1 followed by T2, use it to find the image of (2, −1, 3), and show that T1 followed by T2 is not the same transformation as T2 followed by T1.

Accepted Answer

T1 negates the third coordinate: rows (1, 0, 0), (0, 1, 0), (0, 0, −1). M1 for T1. T2 fixes j, sends k to i and i to −k: rows (0, 0, 1), (0, 1, 0), (−1, 0, 0). M1 for T2. T1 acts first, so the combined matrix is T2T1: multiplying rows (0, 0, 1), (0, 1, 0), (−1, 0, 0) by rows (1, 0, 0), (0, 1, 0), (0, 0, −1) gives rows (0, 0, −1), (0, 1, 0), (−1, 0, 0). M1 for the product in the order T2T1, A1 for the combined matrix. Applying it to (2, −1, 3) gives (0 × 2 + 0 × (−1) − 1 × 3, −1, −2 × 1) = (−3, −1, −2). A1 for the image. Reversing the order, T1T2, gives rows (0, 0, 1), (0, 1, 0), (1, 0, 0), which differs from T2T1 in the sign of two entries. B1 for the reversed product and the conclusion that the two orders disagree.

Question 16

R is the standard 2 × 2 rotation matrix through angle θ, rows (cos θ, −sin θ), (sin θ, cos θ). Show that R^T R = I for every θ, and explain what this says about R⁻¹ and about lengths and angles under the transformation.

Accepted Answer

R^T has rows (cos θ, sin θ), (−sin θ, cos θ). Multiplying R^T R: the top-left entry is cos²θ + sin²θ = 1, using the Pythagorean identity, and the bottom-right entry is the same sum, also 1. M1 for forming R^T, A1 for the two diagonal entries reducing to 1. The off-diagonal entries are cos θ(−sin θ) + sin θ cos θ = 0 and its mirror, also 0, since the two terms are equal and opposite in sign. A1 for both off-diagonal entries vanishing. So R^T R = I for every θ. Since R^T R = I, R^T is a left inverse of R and, R being square, R⁻¹ = R^T: a rotation matrix is inverted just by transposing it, with no division needed. B1 for the conclusion R⁻¹ = R^T. Geometrically this is the algebraic signature of a transformation that preserves lengths and angles: undoing a rotation is exactly reflecting its rows into columns, which is not true of a stretch or a shear. B1 for the geometric meaning.