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How do matrices encode and combine linear transformations of the plane?

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

WACE Specialist Unit 4 matrices: 2x2 matrix arithmetic, determinants and inverses, the matrices for rotation, reflection, dilation and shear, composition of transformations by matrix multiplication, and the geometric meaning of the determinant.

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  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

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 detA=adbc\det A = ad - bc. The matrix is invertible exactly when detA0\det A \neq 0, and

Matrix multiplication is associative but not commutative: in general ABBAAB \neq BA. The identity is I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, and AA1=A1A=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: (cosθsinθsinθcosθ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.
  • Reflection in the line y=xtanθy = x\tan\theta (a line at angle θ\theta to the xx-axis): (cos2θsin2θsin2θcos2θ)\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}.
  • Dilation with factors kk in xx and \ell in yy: (k00)\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.

Solving systems with the inverse

A linear system Ax=bA\mathbf{x} = \mathbf{b} with detA0\det A \neq 0 has the unique solution x=A1b\mathbf{x} = A^{-1}\mathbf{b}. If detA=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 (cosθsinθsinθcosθ)\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 (cos2θsin2θsin2θcos2θ)\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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksCalculator-assumed. Let A=(2134)A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}. (a) Find detA\det A and A1A^{-1}. (b) Hence solve the system 2x+y=52x + y = 5, 3x+4y=103x + 4y = 10.
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Determinant, inverse and solving a system.

(a) detA=(2)(4)(1)(3)=83=5\det A = (2)(4) - (1)(3) = 8 - 3 = 5. So A1=15(4132)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)=A1(510)=15(4132)(510)=15(201015+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 20215 marksCalculator-free. A transformation has matrix T=(0110)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.
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Identifying a transformation from its matrix.

(a) Comparing with the rotation matrix (cosθsinθsinθcosθ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, we read cosθ=0\cos\theta = 0 and sinθ=1\sin\theta = 1, so θ=90\theta = 90^\circ. It is a rotation of 9090^\circ anticlockwise about the origin.

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

Markers reward matching to the rotation matrix, the angle 9090^\circ, the determinant 11, and the orientation conclusion.

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