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Matrices and linear transformations (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Use 2x2 matrices for arithmetic, determinants and inverses, and as linear transformations of the plane”

8 min read

Quick answer

A 2×22\times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} has determinant ad−bcad - bc and inverse 1ad−bc(d−b−ca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}; multiplying a position vector by a matrix transforms the plane, with rotation, reflection, dilation and shear each having a standard matrix, and the determinant giving the (signed) area scale factor.

Jump to a section
  1. What this dot point is asking
  2. Matrix arithmetic, determinant and inverse
  3. Standard transformation matrices
  4. Composition and the determinant
  5. Solving systems with the inverse
  6. Identifying a transformation from its matrix
  7. Exam-style questions

What this dot point is asking

SCSA Unit 4 treats 2×22\times 2 matrices both as objects of arithmetic and as linear transformations of the plane. You must compute products, determinants and inverses, recognise and build the standard transformation matrices, compose transformations by multiplication, and interpret the determinant geometrically.

Matrix arithmetic, determinant and inverse

For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} the determinant is det⁡A=ad−bc\det A = ad - bc. The matrix is invertible exactly when det⁡A≠0\det A \neq 0, and

Inverse of a 2x2 matrix

A−1=1ad−bc(d−b−ca),ad−bc≠0.A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \qquad ad - bc \neq 0.

Matrix multiplication is associative but not commutative: in general AB≠BAAB \neq BA. The identity is I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, and AA−1=A−1A=IA A^{-1} = A^{-1} A = I.

Standard transformation matrices

A point (xy)\begin{pmatrix} x \\ y \end{pmatrix} is mapped to A(xy)A\begin{pmatrix} x \\ y \end{pmatrix}. The standard matrices are:

  • Rotation by angle θ\theta anticlockwise about the origin: (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.
  • Reflection in the line y=xtan⁡θy = x\tan\theta (a line at angle θ\theta to the xx-axis): (cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}.
  • Dilation with factors kk in xx and ℓ\ell in yy: (k00ℓ)\begin{pmatrix} k & 0 \\ 0 & \ell \end{pmatrix}.
  • Shear parallel to the xx-axis with factor kk: (1k01)\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}.

Composition and the determinant

Applying transformation SS then TT corresponds to the product TSTS (the first-applied matrix sits on the right, next to the vector). The determinant has a direct geometric reading.

The determinant as an area factor

∣det⁡A∣|\det A| is the factor by which areas are scaled by the transformation. A negative determinant means the transformation reverses orientation (a reflection is included), and det⁡A=0\det A = 0 collapses the plane onto a line.

Composing a rotation and a reflection

Find the combined matrix when we first reflect in the xx-axis, then rotate 90∘90^\circ anticlockwise.

Step 1: Write down each transformation matrix

The standard matrix for reflection in the xx-axis negates the yy-coordinate and leaves xx unchanged. The standard matrix for anticlockwise rotation by angle θ\theta is built from cos⁡θ\cos\theta and sin⁡θ\sin\theta; at 90∘90^\circ those values are 00 and 11 respectively.

R=(100−1),T=(cos⁡90∘−sin⁡90∘sin⁡90∘cos⁡90∘)=(0−110)R = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad T = \begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

Step 2: Form the composite matrix in the correct order

When transformations are applied in sequence, the matrix of the first-applied transformation sits on the right in the product, because vectors are multiplied on the right. Since we apply RR first and then TT, the combined matrix is TRTR.

TR=(0−110)(100−1)=(0110)TR = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Step 3: Interpret the result

The matrix (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} maps (x,y)→(y,x)(x, y) \to (y, x), which is reflection in the line y=xy = x. Its determinant is (0)(0)−(1)(1)=−1(0)(0) - (1)(1) = -1, confirming that the composition reverses orientation and preserves area, consistent with the net effect being a single reflection.

Final answer: The combined transformation matrix is (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, which represents reflection in the line y=xy = x.

Solving systems with the inverse

A linear system Ax=bA\mathbf{x} = \mathbf{b} with det⁡A≠0\det A \neq 0 has the unique solution x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. If det⁡A=0\det A = 0 the system has either no solution or infinitely many.

Identifying a transformation from its matrix

The reverse skill is also examined: given a matrix, name the transformation. Compare the entries with the standard templates. A matrix of the form (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} with determinant +1+1 is a rotation by θ\theta. A symmetric matrix of the form (cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} with determinant −1-1 is a reflection. A diagonal matrix is a dilation, and a matrix with a single off-diagonal entry is a shear. Checking the determinant first tells you whether orientation is preserved (positive) or reversed (negative), which immediately narrows the possibilities.

Common traps
Multiplying in the wrong order
For "do SS then TT" the matrix is TSTS, not STST. Because matrices do not commute, the order changes the answer.
Sign slips in the inverse
The inverse swaps aa and dd and negates bb and cc. A common error is negating the wrong pair or forgetting to divide by det⁡A\det A.
Assuming an inverse always exists
If det⁡A=0\det A = 0 there is no inverse, so x=A−1b\mathbf{x} = A^{-1}\mathbf{b} is not available.
Forgetting the modulus on the determinant
The area scale factor is ∣det⁡A∣|\det A|; the sign only tells you about orientation, not whether area grows or shrinks.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2022-style7 marks
Calculator-assumed. Let A=(2134)A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}. (a) Find det⁡A\det A and A−1A^{-1}. (b) Hence solve the system 2x+y=52x + y = 5, 3x+4y=103x + 4y = 10.
Show worked answer →

Determinant, inverse and solving a system.

(a) det⁡A=(2)(4)−(1)(3)=8−3=5\det A = (2)(4) - (1)(3) = 8 - 3 = 5. So A−1=15(4−1−32)A^{-1} = \dfrac{1}{5}\begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}.

(b) Write the system as A(xy)=(510)A\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \end{pmatrix}, so (xy)=A−1(510)=15(4−1−32)(510)=15(20−10−15+20)=15(105)=(21)\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 20 - 10 \\ -15 + 20 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 10 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}.

So x=2x = 2, y=1y = 1. Markers reward the determinant, the inverse with correct swaps and signs, and the matrix solution.

WACE 2021-style5 marks
Calculator-free. A transformation has matrix T=(0−110)T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. (a) Describe the transformation geometrically. (b) State the area scale factor and whether orientation is preserved.
Show worked answer →

Identifying a transformation from its matrix.

(a) Comparing with the rotation matrix (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, we read cos⁡θ=0\cos\theta = 0 and sin⁡θ=1\sin\theta = 1, so θ=90∘\theta = 90^\circ. It is a rotation of 90∘90^\circ anticlockwise about the origin.

(b) det⁡T=(0)(0)−(−1)(1)=1\det T = (0)(0) - (-1)(1) = 1. The area scale factor is ∣det⁡T∣=1|\det T| = 1 (areas unchanged), and since the determinant is positive, orientation is preserved.

Markers reward matching to the rotation matrix, the angle 90∘90^\circ, the determinant 11, and the orientation conclusion.

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