Matrices & Transformations
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Concept: Every linear transformation in a 2D plane (rotations, reflections, shears, enlargements) can be completely encoded into a 2×2 matrix. Combining transformations corresponds directly to matrix multiplication.
1. The Unit Square & Basis Vectors
To understand what a matrix \(T = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) does to any point in the Cartesian plane, we only need to track what happens to two fundamental unit vectors:
- The unit vector on the x-axis: \(\mathbf{i} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \implies T\mathbf{i} = \begin{bmatrix} a \\ c \end{bmatrix}\) (the 1st column of \(T\))
- The unit vector on the y-axis: \(\mathbf{j} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \implies T\mathbf{j} = \begin{bmatrix} b \\ d \end{bmatrix}\) (the 2nd column of \(T\))
Where the unit vectors land determines where every shape, Kitenge fabric motif, or architectural boundary lands!
💡 Interactive Visualizer: Matrix Transformation of the Unit Square
Adjust the matrix entries \(a, b, c, d\) to see the unit square transform in real time.
2. Order of Successive Transformations
If a transformation \(A\) is performed first on point \(\mathbf{p}\), followed by a second transformation \(B\), the compound result is given by:
\[\mathbf{p}'' = B(A\mathbf{p}) = (BA)\mathbf{p}\]Crucial Rule: The matrix of the transformation that occurs first is written on the right-hand side of the product.
Key Formulas
1. Transformation of a Point / Vector:
\[\mathbf{v}' = T\mathbf{v} \implies \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}\]2. Standard Transformation Matrices:
- Reflection in the line \(y = 0\) (x-axis): \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\)
- Reflection in the line \(x = 0\) (y-axis): \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\)
- Reflection in the line \(y = x\): \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\)
- Reflection in the line \(y = -x\): \(\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\)
- Rotation by \(90^\circ\) anticlockwise about the origin \((0,0)\): \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)
- Rotation by \(180^\circ\) about the origin: \(\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)
- Rotation by \(270^\circ\) anticlockwise (or \(90^\circ\) clockwise): \(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\)
- Enlargement with scale factor \(k\), centre \((0,0)\): \(\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\)
3. Matrix Determinant and Area Scale Factor (ASF):
\[\det(T) = \left| \begin{matrix} a & b \\ c & d \end{matrix} \right| = ad - bc\] \[\text{Area Scale Factor (ASF)} = |\det(T)| = |ad - bc|\] \[\text{Area of Image} = |\det(T)| \times \text{Area of Object}\]4. Inverse Matrix (Undoing a Transformation):
\[T^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, \quad \text{where } ad - bc \neq 0\]Worked Examples
Example 1 (Easy): Single Transformation of a Point
Problem: A point \(P(2, 3)\) is mapped onto \(P'\) by the transformation matrix \(M = \begin{bmatrix} 3 & -1 \\ 2 & 4 \end{bmatrix}\). Find the coordinates of \(P'\).
Step-by-Step Solution:
- Write the point \(P\) as a column vector: \(\mathbf{p} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\).
- Multiply the transformation matrix \(M\) by \(\mathbf{p}\): \[\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix}\]
- Compute the dot products: \[x' = (3 \times 2) + (-1 \times 3) = 6 - 3 = 3\] \[y' = (2 \times 2) + (4 \times 3) = 4 + 12 = 16\]
- Conclusion: The coordinates of the image point are \(P'(3, 16)\).
Example 2 (Medium): Successive Combined Transformations
Problem: A textile pattern at point \(A(1, 4)\) undergoes a transformation represented by \(T_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\), followed immediately by a transformation \(T_2 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\). Find the single matrix \(T\) representing the combined transformation, and determine the image of \(A\).
Step-by-Step Solution:
- Set up the combined matrix. Since \(T_1\) happens first and \(T_2\) happens second, the combined matrix is: \[T = T_2 T_1\]
- Perform matrix multiplication: \[T = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} (2)(0) + (0)(-1) & (2)(1) + (0)(0) \\ (0)(0) + (2)(-1) & (0)(1) + (2)(0) \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\]
- Apply \(T\) to point \(A(1, 4)\): \[\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} (0)(1) + (2)(4) \\ (-2)(1) + (0)(4) \end{bmatrix} = \begin{bmatrix} 8 \\ -2 \end{bmatrix}\]
- Conclusion: The single combined matrix is \(\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) and the final image is \(A'(8, -2)\).
Example 3 (Hard): Area Scale Factor & Inverse Recovery
Problem: A triangular piece of land in Machakos with an area of \(24\text{ m}^2\) is mapped onto a new plan using the transformation matrix \(M = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\).
(a) Find the area of the mapped image.
(b) Find the inverse matrix \(M^{-1}\) that maps the image back to the original land boundary.
Step-by-Step Solution:
- Find the determinant of \(M\): \[\det(M) = (5)(2) - (2)(3) = 10 - 6 = 4\]
- Calculate image area: \[\text{Area of Image} = |\det(M)| \times \text{Original Area} = 4 \times 24\text{ m}^2 = 96\text{ m}^2\]
- Determine \(M^{-1}\): \[M^{-1} = \frac{1}{\det(M)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 0.5 & -0.5 \\ -0.75 & 1.25 \end{bmatrix}\]
Common Mistakes
1. Reversing the Order of Matrix Multiplication
2. Element-by-Element Multiplication
3. Negative Determinant vs. Negative Area
Real World
Practice