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Matrices and linear transformations of the plane (QCE Specialist Mathematics Unit 3)

Syllabus dot point

“Use 2x2 matrices to represent linear transformations of the plane, compute determinants and inverses, interpret the determinant as a scaling factor, and combine transformations through matrix multiplication”

8 min read

Quick answer

A 2×22\times 2 matrix M=(abcd)M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} transforms vectors by MvM\mathbf{v}. Its determinant ad−bcad - bc gives the area-scaling factor ∣det⁡M∣|\det M| (negative means orientation is reversed), and the inverse is 1ad−bc(d−b−ca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} when the determinant is non-zero. Rotation, reflection, dilation and shear have standard matrices. Composing transformations means multiplying matrices, second-applied on the left, and matrix multiplication is not commutative.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

What this dot point is asking

QCAA wants you to use 2×22\times 2 matrices to represent and combine linear transformations of the plane, compute determinants and inverses, and interpret the determinant geometrically as a signed area-scaling factor. Matrices are assessed in IA2 and the external assessment, and the geometric interpretation of transformations is a regular extended-response theme.

The answer

Matrix arithmetic

A 2×22\times 2 matrix multiplies a column vector to produce a new vector. For

M=(abcd),M(xy)=(ax+bycx+dy).M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.

Multiplying two matrices applies one transformation after another. Matrix multiplication is associative but not commutative, so MN≠NMMN \neq NM in general.

Determinant and inverse

The determinant of MM is

det⁡M=ad−bc.\det M = ad - bc.

When det⁡M≠0\det M \neq 0 the matrix is invertible, with

M−1=1ad−bc(d−b−ca).M^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

The inverse swaps the leading diagonal, negates the off-diagonal, and divides by the determinant. If det⁡M=0\det M = 0 the transformation collapses the plane onto a line (or point) and has no inverse.

Determinant as area scaling

The absolute value ∣det⁡M∣|\det M| is the factor by which areas are scaled under the transformation: a unit square of area 11 maps to a parallelogram of area ∣det⁡M∣|\det M|. The sign of det⁡M\det M records orientation: a negative determinant means the transformation includes a reflection (orientation is reversed).

Standard transformation matrices

Rotation anticlockwise by angle θ\theta about the origin:

R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ),det⁡R=1.R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, \qquad \det R = 1.

Reflection in the line y=xtan⁡αy = x\tan\alpha (here in the xx-axis as α=0\alpha = 0):

(100−1),det⁡=−1.\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad \det = -1.

Dilation (scaling) by factors kxk_x and kyk_y:

(kx00ky),det⁡=kxky.\begin{pmatrix} k_x & 0 \\ 0 & k_y \end{pmatrix}, \qquad \det = k_x k_y.

A shear parallel to the xx-axis is (1k01)\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} with determinant 11, so shears preserve area.

Composing transformations

To apply transformation AA then transformation BB to a vector, compute B(Av)=(BA)vB(A\mathbf{v}) = (BA)\mathbf{v}. The combined matrix is the product BABA, with the second transformation on the left. Order matters: rotating then reflecting generally differs from reflecting then rotating. The determinant of a product equals the product of determinants, det⁡(BA)=det⁡B det⁡A\det(BA) = \det B \, \det A, so the area-scaling factors multiply.

Finding the matrix from images of basis vectors

A powerful shortcut: the columns of a transformation matrix are the images of the standard basis vectors. If a linear transformation sends (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} to (pq)\begin{pmatrix} p \\ q \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix} to (rs)\begin{pmatrix} r \\ s \end{pmatrix}, then M=(prqs)M = \begin{pmatrix} p & r \\ q & s \end{pmatrix}. This lets you build the matrix of any described transformation directly, and it explains every standard matrix: the rotation matrix has columns (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) and (−sin⁡θ,cos⁡θ)(-\sin\theta, \cos\theta) because those are the images of the unit vectors after turning through θ\theta.

Invariant points and lines

A transformation may fix certain points or lines. An invariant point satisfies Mv=vM\mathbf{v} = \mathbf{v}, which rearranges to (M−I)v=0(M - I)\mathbf{v} = \mathbf{0}. The origin is always invariant under a linear transformation. An invariant line through the origin is one whose direction is unchanged in direction (though possibly stretched), satisfying Mv=λvM\mathbf{v} = \lambda\mathbf{v} for some scalar λ\lambda. For a reflection, the mirror line is invariant point-by-point, while the perpendicular direction is reversed; for a rotation other than 0∘0^\circ or 180∘180^\circ, no real line through the origin is invariant, which matches the geometric picture of everything turning.

The identity and inverse as transformations

The identity matrix I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} leaves every vector unchanged, and M−1M^{-1} undoes MM, since M−1M=IM^{-1}M = I. Geometrically the inverse reverses the transformation: the inverse of a rotation by θ\theta is a rotation by −θ-\theta, and the inverse of a dilation by kk is a dilation by 1/k1/k. A singular matrix (det⁡M=0\det M = 0) crushes the plane onto a line, losing information, which is exactly why it cannot be undone and has no inverse.

Combined rotation and reflection transformation

A point is rotated 90°90° anticlockwise about the origin, then reflected in the xx-axis. Find the single matrix for the combined transformation and apply it to (3,1)(3, 1).

Step 1: Write the matrix for the rotation

For a 90°90° anticlockwise rotation, cos⁡90°=0\cos 90° = 0 and sin⁡90°=1\sin 90° = 1, so the standard rotation matrix is:

R=(0−110)R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

Step 2: Write the matrix for the reflection

Reflection in the xx-axis maps (x,y)→(x,−y)(x, y) \to (x, -y), so the matrix is:

F=(100−1)F = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

Step 3: Compose the transformations in the correct order

When two transformations are applied in sequence, the matrix of the one applied second goes on the left. Here the rotation acts first and the reflection acts second, giving the combined matrix FRFR (not RFRF):

FR=(100−1)(0−110)=(0−1−10)FR = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}

Step 4: Apply the combined matrix to the point (3,1)(3, 1)

Multiply the matrix by the column vector:

(0−1−10)(31)=((0)(3)+(−1)(1)(−1)(3)+(0)(1))=(−1−3)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} (0)(3) + (-1)(1) \\ (-1)(3) + (0)(1) \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}

Step 5: Interpret the determinant

Compute det⁡(FR)=(0)(0)−(−1)(−1)=−1\det(FR) = (0)(0) - (-1)(-1) = -1. This equals det⁡F⋅det⁡R=(−1)(1)=−1\det F \cdot \det R = (-1)(1) = -1, confirming the combined transformation reverses orientation (negative determinant) while preserving area (absolute value equals 11).

Final answer: combined matrix =(0−1−10)= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}; image of (3,1)(3, 1) is (−1,−3)(-1, -3); orientation is reversed, area is preserved.

Common trap

The biggest trap is reversing the order of multiplication. If transformation AA acts first and BB second, the combined matrix is BABA, not ABAB, because the matrix nearest the vector acts first. Writing ABAB silently applies the transformations in the wrong order. A second error is forgetting the 1ad−bc\frac{1}{ad-bc} factor when writing an inverse, or attempting to invert a matrix with zero determinant, which is impossible.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 1 (technique). Let M=(3124)M = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}. (a) Determine det⁡M\det M. (b) Determine M−1M^{-1}. (c) State the factor by which MM scales areas.
Show worked answer →

(a) det⁡M=(3)(4)−(1)(2)=12−2=10.\det M = (3)(4) - (1)(2) = 12 - 2 = 10.

(b) M−1=110(4−1−23)=(0.4−0.1−0.20.3).M^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{pmatrix}.

(c) Areas scale by ∣det⁡M∣=10.|\det M| = 10.

Markers reward the determinant, the inverse with the diagonal swap and sign changes, and the area-scaling interpretation.

Practice6 marks
Paper 2 (complex familiar). A transformation reflects points in the xx-axis and then rotates them 90∘90^\circ anticlockwise about the origin. (a) Determine the single 2×22 \times 2 matrix TT representing the combined transformation. (b) Determine the image of the point (2,−3)(2, -3). (c) Justify, using the determinant, whether TT preserves or reverses orientation.
Show worked answer →

(a) Reflection in the xx-axis is F=(100−1)F = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}; rotation 90∘90^\circ anticlockwise is R=(0−110)R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. Rotation is applied second, so T=RF=(0−110)(100−1)=(0110).T = RF = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

(b) T(2−3)=(−32).T\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}.

(c) det⁡T=(0)(0)−(1)(1)=−1<0\det T = (0)(0) - (1)(1) = -1 < 0, so orientation is reversed (the reflection dominates), and area is preserved since ∣det⁡T∣=1|\det T| = 1.

Markers reward the correct composition order, the image, and the determinant justification.

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