KCSE 2025 Maths P2 Q23 — Matrix Transformations, Combined Matrix & Area Scale Factor

KCSE 2025 Form 4 Matrices & Transformations

Published

The Question

“Triangle ABC has vertices A(2, 4), B(2, 2) and C(4, 3). It is mapped onto triangle A''B''C'' by the transformation T1 followed by the transformation T2, where T1 is the matrix with rows (2.5, 0) and (0, 1), and T2 is the matrix with rows (1, -1.5) and (0, -1). (a)(i) Find the image A'B'C' of triangle ABC under T1. (a)(ii) Describe fully the transformation represented by T1. (b)(i) Find the single transformation matrix that maps ABC onto A''B''C''. (b)(ii) Determine the area of triangle A''B''C''.”

1

Find the image under T1 (a)(i)

Write the three vertices A, B and C as columns of a single matrix, then pre-multiply by T1 so all three images come out at once. The top row of T1 multiplies each x-coordinate by 2.5, while the bottom row leaves each y-coordinate unchanged. Read the image coordinates off the resulting columns.

(2.5001)(224423)=(5510423)\begin{pmatrix} 2.5 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 2 & 4 \\ 4 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 5 & 10 \\ 4 & 2 & 3 \end{pmatrix}
A′(5,4),B′(5,2),C′(10,3)A'(5,4), \quad B'(5,2), \quad C'(10,3)
2

Describe T1 fully (a)(ii)

Look at what T1 actually did: every x-coordinate was multiplied by 2.5 while every y-coordinate stayed the same. Stretching only in the x-direction, with y fixed, is a one-way stretch parallel to the x-axis. The scale factor is 2.5, and because points already on the y-axis have x = 0 they do not move, so the y-axis is the invariant line.

A full description of a stretch needs three things: the type (one-way stretch), the scale factor (2.5), and the invariant line (the y-axis).

3

Combine the two matrices (b)(i)

T1 is applied first and T2 second, so the single matrix is T2 multiplied by T1 in that order, because the matrix applied first sits on the right. Multiply the two matrices row-by-column to get one combined transformation matrix that does both steps at once.

M=T2T1=(1−1.50−1)(2.5001)M = T_2 T_1 = \begin{pmatrix} 1 & -1.5 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2.5 & 0 \\ 0 & 1 \end{pmatrix}
M=((1)(2.5)+(−1.5)(0)(1)(0)+(−1.5)(1)(0)(2.5)+(−1)(0)(0)(0)+(−1)(1))M = \begin{pmatrix} (1)(2.5)+(-1.5)(0) & (1)(0)+(-1.5)(1) \\ (0)(2.5)+(-1)(0) & (0)(0)+(-1)(1) \end{pmatrix}
M=(2.5−1.50−1)M = \begin{pmatrix} 2.5 & -1.5 \\ 0 & -1 \end{pmatrix}
4

Area of the original triangle (b)(ii)

Before scaling, find the area of triangle ABC. Side AB is vertical, running from (2, 4) to (2, 2), so its length is 2 units; take it as the base. The perpendicular distance from C(4, 3) to that base is the horizontal gap from x = 2 to x = 4, which is 2 units. Use the half-base-times-height formula.

AreaABC=12×2×2=2 square units\text{Area}_{ABC} = \tfrac{1}{2} \times 2 \times 2 = 2 \text{ square units}
5

Scale the area by the determinant (b)(ii)

A transformation multiplies areas by the absolute value of its matrix determinant. Work out the determinant of the combined matrix M, take its size, and multiply the original area by it to get the area of the final image triangle.

det⁡M=(2.5)(−1)−(−1.5)(0)=−2.5\det M = (2.5)(-1) - (-1.5)(0) = -2.5
Area scale factor=∣−2.5∣=2.5\text{Area scale factor} = |{-2.5}| = 2.5
AreaA′B′C′=2.5×2=5 square units\text{Area}_{A'B'C'} = 2.5 \times 2 = 5 \text{ square units}

Final Result

Under T1 the image is A'(5, 4), B'(5, 2) and C'(10, 3). T1 is a one-way stretch parallel to the x-axis, scale factor 2.5, with the y-axis invariant. The single combined matrix is the matrix with rows (2.5, -1.5) and (0, -1). The area of triangle A''B''C'' is 5 square units.

Why this method works

Writing points as columns lets one matrix multiplication transform the whole triangle, and because matrices act on the right first, the order T2T1 correctly captures T1 happening before T2. A stretch is fully described only by its type, factor and invariant line, since those three facts pin down exactly where every point goes. The determinant works as an area scale factor because it measures how much a unit square is enlarged or shrunk by the transformation; its sign only tells you whether orientation was flipped, which is why you take the absolute value for area.

✓

Determinant check: T1 has determinant 2.5 and T2 has determinant (1)(-1) - (-1.5)(0) = -1, and 2.5 times -1 is -2.5, matching the determinant of the combined matrix M.