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

What this dot point is asking

QCAA wants you to use $2\times 2$ 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 $2\times 2$ matrix multiplies a column vector to produce a new vector. For

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

Determinant and inverse

The determinant of $M$ is

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

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

Determinant as area scaling

The absolute value $|\det M|$ is the factor by which areas are scaled under the transformation: a unit square of area $1$ maps to a parallelogram of area $|\det M|$. The sign of $\det M$ records orientation: a negative determinant means the transformation includes a reflection (orientation is reversed).

Standard transformation matrices

Rotation anticlockwise by angle $\theta$ about the origin:

Reflection in the line $y = x\tan\alpha$ (here in the $x$-axis as $\alpha = 0$):

Dilation (scaling) by factors $k_x$ and $k_y$:

A shear parallel to the $x$-axis is $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ with determinant $1$, so shears preserve area.

Composing transformations

To apply transformation $A$ then transformation $B$ to a vector, compute $B(A\mathbf{v}) = (BA)\mathbf{v}$. The combined matrix is the product $BA$, 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, $\det(BA) = \det B \, \det A$, 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 $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} p \\ q \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} r \\ s \end{pmatrix}$, then $M = \begin{pmatrix} p & r \\ q & s \end{pmatrix}$. This lets you build the matrix of any described transformation directly, and it explains every standard matrix: the rotation matrix has columns $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ because those are the images of the unit vectors after turning through $\theta$.

Invariant points and lines

A transformation may fix certain points or lines. An invariant point satisfies $M\mathbf{v} = \mathbf{v}$, which rearranges to $(M - I)\mathbf{v} = \mathbf{0}$. 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 $M\mathbf{v} = \lambda\mathbf{v}$ for some scalar $\lambda$. For a reflection, the mirror line is invariant point-by-point, while the perpendicular direction is reversed; for a rotation other than $0^\circ$ or $180^\circ$, 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 $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ leaves every vector unchanged, and $M^{-1}$ undoes $M$, since $M^{-1}M = I$. Geometrically the inverse reverses the transformation: the inverse of a rotation by $\theta$ is a rotation by $-\theta$, and the inverse of a dilation by $k$ is a dilation by $1/k$. A singular matrix ($\det M = 0$) crushes the plane onto a line, losing information, which is exactly why it cannot be undone and has no inverse.